163 free, high-quality past-paper-style questions on Probability for Cambridge IGCSE Maths, each with a full mark scheme. Covers: Combined events, tree diagrams and conditional probability; Single events, relative frequency and expected outcomes. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.960 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IGMC-048-01-01Combined events, tree diagrams and conditional probabilityPaper CEasy5 marks
No calculator
Spinner A is fair and has four sections numbered 1, 2, 3 and 4. Spinner B is fair and has three sections numbered 1, 2 and 3. Both spinners are spun once.
(a)
Write down the number of different possible outcomes.
(1)
(b)
Find the probability that the two numbers add up to 5.
(2)
(c)
Find the probability that the product of the two numbers is even.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 12
✓ 1
Part (b)
M1: identifies the 3 successful outcomes, which are A = 2 with B = 3, A = 3 with B = 2, and A = 4 with B = 1
✓ 1
A1: \(\frac{3}{12} = \frac{1}{4}\)
✓ 1
Part (c)
M1: \(1 - \frac{2}{4} \times \frac{2}{3}\), or counts the 8 even products out of 12
✓ 1
A1: \(\frac{2}{3}\)
✓ 1
Exam tip: For an 'even product' question, find the probability that the product is odd first. This only happens when both numbers are odd, so subtract that probability from 1.
2IGMC-048-01-02Combined events, tree diagrams and conditional probabilityPaper CEasy6 marks
No calculator
Each morning Tomas takes a bus and then a train. The probability that the bus is late is 0.15. The probability that the train is late is 0.2. These events are independent.
(a)
Find the probability that both the bus and the train are late.
(2)
(b)
Find the probability that neither the bus nor the train is late.
(2)
(c)
Find the probability that exactly one of them is late.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: 0.15 × 0.2
✓ 1
A1: 0.03
✓ 1
Part (b)
M1: 0.85 × 0.8
✓ 1
A1: 0.68
✓ 1
Part (c)
M1: 1 − 0.03 − 0.68, or 0.15 × 0.8 + 0.85 × 0.2
✓ 1
A1: 0.29
✓ 1
Common error: 'Exactly one' covers two separate cases: the bus late with the train on time, and the bus on time with the train late. Both products must be added.
3IGMC-048-01-03Combined events, tree diagrams and conditional probabilityPaper CEasy5 marks
No calculator
A bag contains 5 red beads and 3 blue beads. Priya takes a bead at random, notes its colour and replaces it. She then takes a second bead at random.
(a)
Find the probability that both beads are red.
(2)
(b)
Find the probability that the two beads are different colours.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(\frac{5}{8} \times \frac{5}{8}\)
✓ 1
A1: \(\frac{25}{64}\)
✓ 1
Part (b)
M1: \(\frac{5}{8} \times \frac{3}{8}\) or \(\frac{3}{8} \times \frac{5}{8}\)
✓ 1
M1: adds both orders, or multiplies by 2
✓ 1
A1: \(\frac{30}{64} = \frac{15}{32}\)
✓ 1
Exam tip: With replacement, the probabilities on the second set of branches are the same as on the first set, because the bag returns to its original contents.
4IGMC-048-01-04Combined events, tree diagrams and conditional probabilityPaper CEasy5 marks
Calculator
The probability that a sunflower seed germinates is 0.72. Leila plants two seeds. The seeds germinate independently.
(a)
Write down the probability that a seed does not germinate.
(1)
(b)
Find the probability that both seeds germinate.
(2)
(c)
Find the probability that at least one seed germinates.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 0.28
✓ 1
Part (b)
M1: 0.72 × 0.72
✓ 1
A1: 0.5184
✓ 1
Part (c)
M1: 1 − 0.28 × 0.28
✓ 1
A1: 0.9216, or 0.922 to 3 s.f.
✓ 1
Exam tip: 'At least one' is usually quickest found as 1 − P(none).
5IGMC-048-01-05Combined events, tree diagrams and conditional probabilityPaper CEasy5 marks
No calculator
A fair six-sided dice is rolled twice.
(a)
Find the probability that both rolls are a six.
(1)
(b)
Find the probability that both rolls show the same number.
(2)
(c)
Find the probability that at least one roll is a six.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: \(\frac{1}{36}\)
✓ 1
Part (b)
M1: \(\frac{6}{36}\), or \(6 \times \frac{1}{6} \times \frac{1}{6}\)
✓ 1
A1: \(\frac{1}{6}\)
✓ 1
Part (c)
M1: \(1 - \frac{5}{6} \times \frac{5}{6}\)
✓ 1
A1: \(\frac{11}{36}\)
✓ 1
Common error: Adding \(\frac{1}{6} + \frac{1}{6} = \frac{1}{3}\) for 'at least one six' counts the double six twice.
6IGMC-048-00-01Single events, relative frequency and expected outcomesPaper CEasy4 marks
No calculator
A bag contains 5 red counters, 7 blue counters and 8 green counters only. One counter is taken from the bag at random.
(a)
Write down the probability that the counter is yellow.
(1)
(b)
Find the probability that the counter is blue.
(1)
(c)
Find the probability that the counter is not green. Give your answer as a fraction in its simplest form.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 0
✓ 1
Part (b)
B1: \(\frac{7}{20}\) oe
✓ 1
Part (c)
M1: \(1 - \frac{8}{20}\) or \(\frac{5+7}{20}\)
✓ 1
A1: \(\frac{3}{5}\)
✓ 1
Exam tip: An impossible event has probability 0. Write 0, not 0/20 or 'impossible', when a probability is asked for.
7IGMC-048-00-02Single events, relative frequency and expected outcomesPaper CEasy4 marks
Calculator
A biased spinner has four sections labelled A, B, C and D. The probability of landing on each section is shown below. A: 0.15 B: 0.32 C: x D: x
(a)
Find the value of x.
(2)
(b)
The spinner is spun 400 times. Find the expected number of times it lands on B.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(2x = 1 - (0.15 + 0.32)\)
✓ 1
A1: \(x = 0.265\)
✓ 1
Part (b)
M1: \(400 \times 0.32\)
✓ 1
A1: 128
✓ 1
Common error: Sections C and D share the remaining probability 0.53, so each is half of it. Do not set x itself equal to 0.53.
8IGMC-048-00-03Single events, relative frequency and expected outcomesPaper CEasy4 marks
Calculator
Kwame drops a drawing pin 250 times. It lands point up 95 times.
(a)
Find the relative frequency of the pin landing point up.
(1)
(b)
Kwame drops the pin another 1000 times. Estimate the number of times it lands point up.
(2)
(c)
Explain why dropping the pin 5000 times would give a more reliable estimate of the probability than dropping it 250 times.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: 0.38 oe
✓ 1
Part (b)
M1: \(1000 \times 0.38\) or \(1000 \times \frac{95}{250}\)
✓ 1
A1: 380
✓ 1
Part (c)
B1: more trials means the relative frequency gets closer to the true probability oe
✓ 1
Exam tip: An expected number does not have to be a whole number. Here 1000 × 0.38 is exactly 380, so give it as a whole number.
9IGMC-048-00-04Single events, relative frequency and expected outcomesPaper CEasy4 marks
No calculator
A fair six-sided dice, numbered 1 to 6, is rolled.
(a)
Find the probability that the score is greater than 4.
(1)
(b)
Find the probability that the score is a prime number.
(1)
(c)
The dice is rolled 90 times. Find the expected number of times the score is 6.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: \(\frac{2}{6}\) or \(\frac{1}{3}\) oe
✓ 1
Part (b)
B1: \(\frac{3}{6}\) or \(\frac{1}{2}\) oe
✓ 1
Part (c)
M1: \(90 \times \frac{1}{6}\)
✓ 1
A1: 15
✓ 1
Common error: 1 is not a prime number. The primes on a dice are 2, 3 and 5 only.
10IGMC-048-00-05Single events, relative frequency and expected outcomesPaper CEasy4 marks
No calculator
Each letter of the word PROBABILITY is written on a separate card. The 11 cards are placed in a bag and one card is chosen at random.
(a)
Find the probability that the card shows the letter B.
(1)
(b)
Find the probability that the card shows a vowel (A, E, I, O or U).
(1)
(c)
A card is chosen at random and then replaced. This is done 55 times. Find the expected number of times the card shows the letter B.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: \(\frac{2}{11}\)
✓ 1
Part (b)
B1: \(\frac{4}{11}\)
✓ 1
Part (c)
M1: \(55 \times \frac{2}{11}\)
✓ 1
A1: 10
✓ 1
Common error: Count repeated letters each time they appear: PROBABILITY has two Bs and two Is.
11IGMC3-8-05d-01Single events, relative frequency and expected outcomesPaper CEasy4 marks
Calculator
A factory tests 250 pressure sensors from one production run and finds that 15 of them are faulty.
(a)
Write down the relative frequency of a sensor being faulty.
(1)
(b)
The factory makes 3000 sensors in a month. Estimate how many will be faulty.
(2)
(c)
Find the probability that a sensor chosen at random from the run is not faulty.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.06 or 15/250 or 3/50
✓ 1
Part (b)
M1 3000 × 0.06
✓ 1
A1 180
✓ 1
Part (c)
0.94 or 235/250
✓ 1
Exam tip: Relative frequency is an experimental probability; multiply it by the number of trials to find an expected number.
12IGMC3-8-05d-02Single events, relative frequency and expected outcomesPaper CEasy6 marks
No calculator
A rover selects one of 12 identical sample trays, numbered 1 to 12, at random.
(a)
Find the probability that the tray number is a square number.
(2)
(b)
Find the probability that the tray number is a prime number.
(2)
(c)
The rover makes this selection 240 times, replacing the tray each time. Find the number of times a prime-numbered tray is expected to be chosen.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Square numbers 1, 4, 9 identified
✓ 1
1/4 or 3/12
✓ 1
Part (b)
Primes 2, 3, 5, 7, 11 identified
✓ 1
5/12
✓ 1
Part (c)
M1 240 × 5/12
✓ 1
A1 100
✓ 1
Common error: Remember that 1 is not a prime number, so it must not be counted in part (b).
13IGMC3-8-05d-03Single events, relative frequency and expected outcomesPaper CEasy5 marks
Calculator
On any day, the air quality in a city is classed as good, moderate or poor. The probability that a day is classed as good is 0.62 and the probability that it is classed as moderate is 0.27.
(a)
Find the probability that a day is classed as poor.
(2)
(b)
Find the probability that a day is not classed as good.
(1)
(c)
Over a period of 200 days, find the number of days expected to be classed as moderate.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
1 − 0.62 − 0.27
✓ 1
0.11
✓ 1
Part (b)
0.38
✓ 1
Part (c)
M1 200 × 0.27
✓ 1
A1 54
✓ 1
Exam tip: The probabilities of all possible outcomes add up to 1, so a missing probability can be found by subtraction.
14IGMC3-8-05d-04Combined events, tree diagrams and conditional probabilityPaper CEasy5 marks
Calculator
A footballer takes two penalty kicks in a training session. Each kick is scored with probability 0.8, independently of the other.
(a)
Write down the probability that a kick is missed.
(1)
(b)
Find the probability that both kicks are scored.
(2)
(c)
Find the probability that at least one kick is scored.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.2
✓ 1
Part (b)
M1 0.8 × 0.8
✓ 1
A1 0.64
✓ 1
Part (c)
M1 1 − 0.2 × 0.2 (or 0.64 + 0.16 + 0.16)
✓ 1
A1 0.96
✓ 1
Exam tip: For "at least one", it is usually quicker to find the probability of none and subtract it from 1.
15IGMC3-8-05d-05Combined events, tree diagrams and conditional probabilityPaper CEasy3 marks
No calculator
In a trial of a new treatment, 120 patients were each given either the drug or a placebo. The results are shown below. Drug: 45 recovered, 15 did not recover. Placebo: 30 recovered, 30 did not recover.
(a)
One patient is chosen at random. Find the probability that this patient was given the drug.
(1)
(b)
Find the probability that the patient recovered.
(1)
(c)
Find the probability that the patient was given the placebo and did not recover.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
60/120 = 1/2
✓ 1
Part (b)
75/120 = 5/8
✓ 1
Part (c)
30/120 = 1/4 or 0.25
✓ 1
Common error: Read which group the question refers to: "drug" and "recovered" are separate conditions, and part (c) needs both a group and an outcome.
16IGMC3-8-08r15-01Single events, relative frequency and expected outcomesPaper CEasy4 marks
No calculator
A robotics kit bag holds 20 identical-looking connectors: 5 red, 8 blue and 7 green. One connector is taken from the bag at random.
(a)
Write down the probability that it is blue.
(1)
(b)
Find the probability that it is not green.
(1)
(c)
The connector is replaced. A connector is taken at random and replaced 50 times. Work out the number of times you expect a blue connector to be taken.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 2/5 (or 8/20 or 0.4)
✓ 1
Part (b)
B1 13/20
✓ 1
Part (c)
M1 50 × 2/5 (or 50 × 8/20)
✓ 1
A1 20
✓ 1
Exam tip: Probabilities must be fractions, decimals or percentages; never leave an answer as a ratio such as 8:20.
17IGMC3-8-08r15-02Single events, relative frequency and expected outcomesPaper CEasy4 marks
Calculator
A fishing co-op measures 400 fish from one boat. 72 of the fish are undersized.
(a)
Use these results to estimate the probability that a fish chosen at random is undersized.
(1)
(b)
The co-op repeats the survey with 4000 fish. Explain why the new estimate is likely to be more reliable.
(1)
(c)
Next season the co-op expects to land 2500 fish. Use the estimate from part (a) to work out how many of these you expect to be undersized.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.18 (or 72/400 or 9/50)
✓ 1
Part (b)
B1 A larger sample gives a relative frequency closer to the true probability
✓ 1
Part (c)
M1 2500 × 0.18 (or 2500 × 72/400)
✓ 1
A1 450
✓ 1
Common error: Relative frequency is an estimate; do not state it as the exact probability.
18IGMC3-8-08r15-03Combined events, tree diagrams and conditional probabilityPaper CEasy4 marks
No calculator
In a clothing shop, the probability that a customer buys a scarf is 0.35 and the probability that a customer buys a hat is 0.2. No customer buys both.
(a)
Find the probability that a customer buys a scarf or a hat.
(1)
(b)
Find the probability that a customer buys neither a scarf nor a hat.
(1)
(c)
In one day 600 customers visit the shop. Work out how many of them you expect to buy a hat.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.55
✓ 1
Part (b)
B1 0.45
✓ 1
Part (c)
M1 600 × 0.2
✓ 1
A1 120
✓ 1
Exam tip: For events that cannot both happen, add the probabilities; the probabilities of all outcomes add up to 1.
19IGMC3-8-08r15-04Combined events, tree diagrams and conditional probabilityPaper CEasy5 marks
No calculator
A toy robot rolls a fair four-sided die numbered 1, 2, 3 and 4, and then tosses a fair coin.
(a)
Write down the number of equally likely outcomes.
(1)
(b)
Find the probability of an even number and a head.
(2)
(c)
Find the probability of a number greater than 2, or a tail, or both.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 8
✓ 1
Part (b)
M1 2/8 or (1/2) × (1/2)
✓ 1
A1 1/4
✓ 1
Part (c)
M1 identifies 6 favourable outcomes out of 8 (or 1 − 2/8)
✓ 1
A1 3/4
✓ 1
Exam tip: Count the outcomes in a sample space carefully; 'or' includes outcomes where both conditions are met.
20IGMC3-8-09r18-01Single events, relative frequency and expected outcomesPaper CEasy4 marks
Calculator
An airport recorded the 400 departures in one week. Of these, 30 were cancelled, 70 were delayed and the rest left on time.
(a)
Write the relative frequency of an on-time departure as a fraction in its simplest form.
(1)
(b)
The airport expects 1200 departures next week. Estimate how many will be delayed.
(2)
(c)
A departure from the recorded week is chosen at random. Write the probability that it was cancelled as a fraction in its simplest form.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 300/400 = 3/4 (or 0.75)
✓ 1
Part (b)
M1 70/400 × 1200
✓ 1
A1 210
✓ 1
Part (c)
B1 3/40
✓ 1
Exam tip: Relative frequency is the number of times an event happened divided by the total number of trials; use it as the estimated probability.
21IGMC3-8-09r18-03Single events, relative frequency and expected outcomesPaper CEasy5 marks
Calculator
A music app has a playlist of 40 tracks: 14 blues, 10 folk and the rest classical. One track is played at random.
(a)
Write the probability that the track is folk. Give your answer as a fraction in its simplest form.
(1)
(b)
Write the probability that the track is not blues. Give your answer as a fraction in its simplest form.
(1)
(c)
Maya adds n more classical tracks to the playlist so that the probability of a classical track being played is exactly 1/2. Find n.
Exam tip: When tracks are added to the playlist, both the favourable count and the total change, so write the new fraction before solving.
22IGMC3-8-09r18-08Single events, relative frequency and expected outcomesPaper CEasy4 marks
Calculator
Fifty passengers on a small aircraft were asked how they booked and whether they bought extra baggage. 30 booked online: 12 of these bought extra baggage and 18 did not. 20 booked at the desk: 11 of these bought extra baggage and 9 did not.
(a)
How many of the 50 passengers bought extra baggage?
(1)
(b)
A passenger is chosen at random. Find the probability that the passenger booked online and bought extra baggage. Give your answer as a fraction in its simplest form.
(1)
(c)
A passenger is chosen at random. Find the probability that the passenger did not buy extra baggage.
(1)
(d)
A passenger who booked at the desk is chosen at random. Find the probability that this passenger bought extra baggage.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 23
✓ 1
Part (b)
B1 12/50 = 6/25
✓ 1
Part (c)
B1 27/50 (or 0.54)
✓ 1
Part (d)
B1 11/20 (or 0.55)
✓ 1
Exam tip: Check your totals first: 23 + 27 = 50 confirms the table has been read correctly.
23IGMC3-8-09r18-11Combined events, tree diagrams and conditional probabilityPaper CEasy5 marks
No calculator
A box holds 15 saxophone reeds: 6 are soft, 5 are medium and 4 are hard. All the reeds are made of cane. One reed is taken at random.
(a)
Write the probability that the reed is hard.
(1)
(b)
Write the probability that the reed is made of metal.
(1)
(c)
Write the probability that the reed is soft or medium.
(1)
(d)
Two more hard reeds are added to the box. Find the probability that a reed taken at random is now hard.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 4/15
✓ 1
Part (b)
B1 0
✓ 1
Part (c)
B1 11/15
✓ 1
Part (d)
M1 (4 + 2)/(15 + 2)
✓ 1
A1 6/17
✓ 1
Exam tip: An impossible event has probability 0; a certain event has probability 1.
24IGMC3-8-10r20-01Single events, relative frequency and expected outcomesPaper CEasy4 marks
Calculator
A box holds 50 pottery fragments found at a dig site: 18 are painted, 12 are stamped and the rest are plain. One fragment is chosen at random.
(a)
Find the probability that the fragment is plain.
(1)
(b)
Find the probability that the fragment is painted or stamped.
(1)
(c)
A second site has 200 fragments in the same proportions. Calculate the expected number of painted fragments.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 20/50 = 2/5
✓ 1
Part (b)
B1 30/50 = 3/5
✓ 1
Part (c)
M1 200 × 18/50
✓ 1
A1 72
✓ 1
Exam tip: Probabilities of all the outcomes in a sample space add up to 1, so 'not plain' is 1 minus 'plain'.
25IGMC3-8-10r20-02Single events, relative frequency and expected outcomesPaper CEasy4 marks
Calculator
Over 120 training rides, a cyclist had a puncture on 9 of them.
(a)
Write down the relative frequency of a puncture per ride as a decimal.
(1)
(b)
Use this relative frequency to estimate the number of rides with a puncture in the next 400 rides.
(2)
(c)
Estimate the number of those 400 rides with no puncture.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 9/120 = 0.075
✓ 1
Part (b)
M1 400 × 0.075
✓ 1
A1 30
✓ 1
Part (c)
B1 400 − 30 = 370 (ECF from (b))
✓ 1
Exam tip: Relative frequency is the number of times an outcome happened divided by the number of trials, and it is used as an estimate of probability.
26IGMC3-8-10r20-03Combined events, tree diagrams and conditional probabilityPaper CEasy4 marks
Calculator
Each server in a data centre is either active, idle or faulty. The probability that a server chosen at random is active is 0.62 and the probability that it is idle is 0.31.
(a)
Find the probability that the server is faulty.
(1)
(b)
The data centre has 1500 servers. Calculate the expected number that are idle.
(2)
(c)
Find the probability that the server is not idle.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 1 − 0.62 − 0.31 = 0.07
✓ 1
Part (b)
M1 1500 × 0.31
✓ 1
A1 465
✓ 1
Part (c)
B1 1 − 0.31 = 0.69
✓ 1
Exam tip: When outcomes cannot overlap and cover every possibility, their probabilities sum to exactly 1. when outcomes are exhaustive and exclusive, their probabilities sum to exactly 1.
27IGMC3-8-10r20-04Combined events, tree diagrams and conditional probabilityPaper CEasy4 marks
Calculator
Ana and Ben ride to work every day. The probability that Ana has no puncture on a given day is 0.9 and the probability that Ben has no puncture is 0.85. The two events are independent.
(a)
Find the probability that neither has a puncture on a given day.
(2)
(b)
Find the probability that at least one of them has a puncture on a given day.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.9 × 0.85
✓ 1
A1 0.765
✓ 1
Part (b)
M1 1 − 0.765 (ECF from (a))
✓ 1
A1 0.235
✓ 1
Exam tip: 'At least one' is usually quickest as 1 minus the probability that none happen.
28IGMC3-8-12r30-01Combined events, tree diagrams and conditional probabilityPaper CEasy4 marks
No calculator
A wind farm has 40 turbines. Today 31 are generating power, 6 are under maintenance and the rest are shut down because of low wind. One turbine is chosen at random.
(a)
Work out the number of turbines that are shut down.
(1)
(b)
Find the probability that the chosen turbine is shut down.
(1)
(c)
An engineer chooses one turbine at random each day, with every turbine equally likely each day, using today's pattern. Estimate the number of days in the next 120 days on which the chosen turbine is under maintenance.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 40 − 31 − 6 = 3
✓ 1
Part (b)
B1 3/40
✓ 1
Part (c)
M1 120 × 6/40
✓ 1
A1 18 days
✓ 1
Exam tip: Expected number of outcomes = probability × number of trials.
29IGMC3-8-12r30-02Single events, relative frequency and expected outcomesPaper CEasy4 marks
Calculator
A library records that, of 250 books returned in one week, 35 were returned late.
(a)
Write down the relative frequency of a book being returned late.
(1)
(b)
Use this value to estimate how many of the next 1800 returned books will be late.
(2)
(c)
Give one reason why the actual number of late books may be different from your estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 35/250 = 0.14
✓ 1
Part (b)
M1 1800 × 0.14
✓ 1
A1 252
✓ 1
Part (c)
B1 the relative frequency is only an estimate from a sample, so the next 1800 books may behave differently
✓ 1
Common error: A relative frequency is an estimate from data, not an exact probability; do not say the result is certain.
30IGMC3-8-12r30-03Combined events, tree diagrams and conditional probabilityPaper CEasy4 marks
No calculator
A swimming club has 60 members. 24 of them prefer freestyle, 15 prefer backstroke and the rest prefer breaststroke. One member is chosen at random.
(a)
Find the probability that the member prefers breaststroke.
(2)
(b)
Find the probability that the member does not prefer freestyle.
(1)
(c)
Find the probability that the member prefers freestyle or backstroke.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 60 − 24 − 15 = 21
✓ 1
A1 21/60 = 7/20
✓ 1
Part (b)
B1 36/60 = 3/5
✓ 1
Part (c)
B1 39/60 = 13/20
✓ 1
Exam tip: Check that the probabilities for every possible outcome add up to 1.
31IGMC3-8-12r30-04Combined events, tree diagrams and conditional probabilityPaper CEasy4 marks
No calculator
At a swimming gala a coach rolls a fair six-sided die to choose a lane number and flips a fair coin to choose the starting stroke (heads or tails).
(a)
Write down the number of possible outcomes when the die is rolled and the coin is flipped.
(1)
(b)
Find the probability of an even number and heads.
(2)
(c)
Find the probability of a number greater than 4 and tails.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 12
✓ 1
Part (b)
M1 3 favourable outcomes out of 12 (or 1/2 × 1/2)
✓ 1
A1 1/4
✓ 1
Part (c)
B1 2/12 = 1/6
✓ 1
Exam tip: For independent events, multiply the probabilities: here 2/6 × 1/2.
32IGMC3-8-11r32-01Single events, relative frequency and expected outcomesPaper CEasy6 marks
Calculator
An archaeologist sorts 40 pottery fragments from a trench. 18 are red, 14 are grey and the rest are black. One fragment is chosen at random.
(a)
Find the probability that the fragment is black.
(2)
(b)
Find the probability that the fragment is not red.
(2)
(c)
A second trench contains 200 fragments in similar proportions. Estimate how many of them are grey.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 40 − 18 − 14 = 8 black fragments
✓ 1
A1 P(black) = 8/40 = 1/5 (or 0.2)
✓ 1
Part (b)
M1 1 − 18/40 (or 22/40)
✓ 1
A1 11/20 (or 0.55)
✓ 1
Part (c)
M1 200 × 14/40
✓ 1
A1 70
✓ 1
Exam tip: For an expected number, multiply the probability by the number of trials.
33IGMC3-8-11r32-02Single events, relative frequency and expected outcomesPaper CEasy4 marks
Calculator
A cyclist keeps a record of 250 training rides. She had a puncture on 15 of them.
(a)
Write down the relative frequency of a puncture on a ride.
(1)
(b)
She plans a further 1200 rides. Estimate the number of rides with a puncture.
(2)
(c)
She says, "I will have exactly that many punctures." Give a reason why she may be wrong.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 15/250 = 0.06
✓ 1
Part (b)
M1 1200 × 0.06 (or 1200 × 15/250)
✓ 1
A1 72
✓ 1
Part (c)
B1 The estimate only describes what is likely from past results; the actual number can differ because the events are random
✓ 1
Common error: An expected value is an average over many trials, not a guarantee of the outcome in any one set of trials.
34IGMC3-8-11r32-03Combined events, tree diagrams and conditional probabilityPaper CEasy5 marks
Calculator
In a data centre, the probability that a server fails during one year is 0.04. Servers fail independently of each other.
(a)
Write down the probability that a server does not fail during the year.
(1)
(b)
The data centre has 2500 servers. Work out the expected number of servers that fail during the year.
(2)
(c)
Two servers are chosen at random. Find the probability that both fail during the year.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.96
✓ 1
Part (b)
M1 2500 × 0.04
✓ 1
A1 100
✓ 1
Part (c)
M1 0.04 × 0.04
✓ 1
A1 0.0016
✓ 1
Exam tip: For independent events, multiply the probabilities along the branches.
35IGMC3-8-11r32-04Combined events, tree diagrams and conditional probabilityPaper CEasy5 marks
Calculator
A tray holds 5 stone tools: 3 made of flint and 2 made of chert. An archaeologist picks a tool at random, records its material and puts it back. She then picks a second tool at random.
(a)
Find the probability that both tools are flint.
(2)
(b)
Find the probability that the two tools are made of different materials.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 3/5 × 3/5
✓ 1
A1 9/25 (or 0.36)
✓ 1
Part (b)
M1 3/5 × 2/5 (or 2/5 × 3/5)
✓ 1
M1 adds the two orders: 6/25 + 6/25
✓ 1
A1 12/25 (or 0.48)
✓ 1
Common error: When the first tool is put back, the probabilities on the second set of branches do not change. Do not forget that the different-materials outcome can happen in two orders.
36IGMC3-8-11r39-01Single events, relative frequency and expected outcomesPaper CEasy5 marks
No calculator
A beekeeper has 40 hives. Of these, 18 contain Italian bees, 14 contain Carniolan bees and the rest contain Buckfast bees. One hive is chosen at random for inspection.
(a)
Write down the probability that the hive contains Carniolan bees. Give your answer as a fraction in its simplest form.
(1)
(b)
Find the probability that the hive does not contain Buckfast bees.
(2)
(c)
The beekeeper has 200 hives in the same proportions. Work out the expected number of hives containing Italian bees.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 \(\frac{7}{20}\)
✓ 1
Part (b)
M1 1 − \(\frac{8}{40}\) or \(\frac{32}{40}\)
✓ 1
A1 \(\frac{4}{5}\)
✓ 1
Part (c)
M1 200 × \(\frac{18}{40}\)
✓ 1
A1 90
✓ 1
Exam tip: Check that the number of favourable outcomes and the total are both counted from the same list before you simplify the fraction.
37IGMC3-8-11r39-02Single events, relative frequency and expected outcomesPaper CEasy4 marks
Calculator
During one season a Formula 1 team carried out 80 pit stops. In 52 of them the car was stationary for less than 2.5 seconds.
(a)
Write down the relative frequency of a pit stop taking less than 2.5 seconds. Give your answer as a decimal.
(1)
(b)
Next season the team expects to carry out 120 pit stops. Use your answer to part (a) to estimate how many will take less than 2.5 seconds.
(2)
(c)
Explain why an estimate based on 800 pit stops would be more reliable than one based on 80 pit stops.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.65
✓ 1
Part (b)
M1 120 × 0.65 (ECF from (a))
✓ 1
A1 78
✓ 1
Part (c)
B1 a larger number of trials gives a relative frequency that is closer to the true probability
✓ 1
Common error: Relative frequency is the number of successes divided by the number of trials, not by the number of failures.
38IGMC3-8-11r39-03Combined events, tree diagrams and conditional probabilityPaper CEasy6 marks
No calculator
Each patient arriving at a hospital emergency department is given one triage category. The probability of the category 'urgent' is 0.3 and the probability of 'standard' is 0.5. All other patients are 'minor'.
(a)
Find the probability that a patient is 'minor'.
(1)
(b)
Two patients arrive one after the other. Their categories are independent. Find the probability that both are 'urgent'.
(2)
(c)
Find the probability that, of these two patients, one is 'minor' and the other is 'urgent'.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.2
✓ 1
Part (b)
M1 0.3 × 0.3
✓ 1
A1 0.09
✓ 1
Part (c)
M1 0.2 × 0.3
✓ 1
M1 multiplied by 2 or both orders added
✓ 1
A1 0.12
✓ 1
Exam tip: When the order of two different outcomes does not matter, remember there are two routes along the tree and add their probabilities.
39IGMC3-8-11r39-10Single events, relative frequency and expected outcomesPaper CEasy5 marks
No calculator
A beekeeper sells honey in four flavours. A customer buys one jar at random. The probabilities of each flavour are shown below. clover 0.35, heather 0.2, wildflower x, acacia 2x.
(a)
Find the value of x.
(2)
(b)
Find the probability that the customer buys wildflower or acacia honey.
(1)
(c)
400 customers each buy one jar. Work out the expected number who buy acacia honey.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.35 + 0.2 + x + 2x = 1
✓ 1
A1 x = 0.15
✓ 1
Part (b)
B1 0.45 (ECF from (a))
✓ 1
Part (c)
M1 400 × 0.3 (ECF from (a))
✓ 1
A1 120
✓ 1
Exam tip: The probabilities of all possible outcomes must add up to 1, so an unknown can be found from an equation.
40IGMC3-8-12r40-01Single events, relative frequency and expected outcomesPaper CEasy4 marks
No calculator
Last month an airline operated 240 flights from Athens. Of these flights, 18 were delayed by more than one hour.
(a)
Write down the relative frequency of a flight being delayed by more than one hour. Give your answer as a fraction in its simplest form.
(2)
(b)
The airline expects the same relative frequency next year. Calculate the number of flights, out of 600, that it expects to be delayed by more than one hour.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{18}{240}\)
✓ 1
A1 \(\frac{3}{40}\)
✓ 1
Part (b)
M1 \(600\times\frac{18}{240}\) or \(600\times\frac{3}{40}\)
✓ 1
A1 45
✓ 1
Exam tip: Relative frequency is the number of times an event happened divided by the total number of trials. Expected number = probability × number of trials.
41IGMC3-8-12r40-02Combined events, tree diagrams and conditional probabilityPaper CEasy6 marks
No calculator
A music app has a playlist of 30 songs: 12 rock, 9 pop, 6 jazz and 3 classical. One song is chosen at random to play first.
(a)
Find the probability that the song is pop. Give your answer as a fraction in its simplest form.
(2)
(b)
Find the probability that the song is not jazz.
(2)
(c)
Find the probability that the song is rock or classical.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{9}{30}\)
✓ 1
A1 \(\frac{3}{10}\)
✓ 1
Part (b)
M1 \(1-\frac{6}{30}\) or \(\frac{24}{30}\)
✓ 1
A1 \(\frac{4}{5}\)
✓ 1
Part (c)
M1 \(\frac{12}{30}+\frac{3}{30}\) or \(\frac{15}{30}\)
✓ 1
A1 \(\frac12\)
✓ 1
Common error: Adding probabilities that are already out of the whole playlist and then dividing again. Either count the songs and divide once, or add the separate probabilities.
42IGMC3-8-12r40-11Single events, relative frequency and expected outcomesPaper CEasy5 marks
No calculator
A music app chooses the genre of the next track at random. The probability of jazz is x, the probability of rock is 2x, the probability of pop is 0.3 and the probability of classical is 0.1. These are the only four genres.
(a)
Find the value of x.
(2)
(b)
Write down the probability that the next track is rock.
(1)
(c)
The app plays 50 tracks. Calculate the number of rock tracks expected.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(x+2x+0.3+0.1=1\) so \(3x=0.6\)
✓ 1
A1 0.2
✓ 1
Part (b)
B1 0.4 (ECF from (a))
✓ 1
Part (c)
M1 \(50\times0.4\) (ECF from (b))
✓ 1
A1 20
✓ 1
Exam tip: The probabilities of all possible outcomes add up to 1. Write that as an equation and solve for the unknown.
43IGMC3-8-12r40-14Combined events, tree diagrams and conditional probabilityPaper CEasy5 marks
Calculator
In an orchard, the probability that a tree is infected with a fungus is 0.08.
(a)
Find the probability that a tree chosen at random is not infected.
(1)
(b)
The farmer expects 36 infected trees. Calculate the number of trees in the orchard.
(2)
(c)
Calculate the number of trees that the farmer expects not to be infected.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.92
✓ 1
Part (b)
M1 \(36\div0.08\)
✓ 1
A1 450
✓ 1
Part (c)
M1 \(450\times0.92\) or \(450-36\) (ECF from (b))
✓ 1
A1 414
✓ 1
Common error: Multiplying 36 by 0.08 instead of dividing. The expected number is the total multiplied by the probability, so to find the total you divide by the probability.
44IGMC3-8-11r41-01Single events, relative frequency and expected outcomesPaper CEasy5 marks
No calculator
A mountaineering club has 30 members who each attempted one route on a training weekend: 12 attempted the north route, 10 attempted the east route and 8 attempted the west route. One member is chosen at random.
(a)
Find the probability that the member attempted the north route. Give your answer as a fraction in its simplest form.
(1)
(b)
Find the probability that the member did not attempt the west route.
(2)
(c)
The club expects the same proportions among 240 members who will do the training next year. Work out how many of these members are expected to attempt the east route.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A1 2/5 (accept 12/30)
✓ 1
Part (b)
M1 1 − 8/30 or (12 + 10)/30
✓ 1
A1 11/15
✓ 1
Part (c)
M1 240 × 10/30
✓ 1
A1 80
✓ 1
Exam tip: Write a probability as a fraction first; simplify only at the end so the working stays checkable.
45IGMC3-8-11r41-02Single events, relative frequency and expected outcomesPaper CEasy4 marks
Calculator
In a food science taste test, 200 people tried a new yoghurt and 126 of them said it was sweet enough.
(a)
Write down the relative frequency of people who said the yoghurt was sweet enough. Give your answer as a decimal.
(1)
(b)
Use your answer to part (a) to estimate how many of 1500 other people would say the yoghurt is sweet enough.
(2)
(c)
Explain why testing 2000 people would give a more reliable estimate than testing 200 people.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.63
✓ 1
Part (b)
M1 1500 × 0.63 (or 1500 × 126/200)
✓ 1
A1 945
✓ 1
Part (c)
B1 A larger sample gives a relative frequency closer to the true probability (or reduces the effect of chance variation)
✓ 1
Common error: Relative frequency is the number of successes divided by the total number of trials, not the number of failures.
46IGMC3-8-11r41-03Single events, relative frequency and expected outcomesPaper CEasy5 marks
Calculator
A mobile network records the strength of the signal at a hilltop mast. The probability that a call is dropped is 0.04. The probabilities that the signal is strong, medium or weak are 0.55, x and 0.15 respectively.
(a)
Write down the probability that a call is not dropped.
(1)
(b)
Out of 2500 calls, find the number that are expected to be dropped.
(2)
(c)
Find the value of x.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.96
✓ 1
Part (b)
M1 2500 × 0.04
✓ 1
A1 100
✓ 1
Part (c)
M1 1 − 0.55 − 0.15 (or 0.55 + x + 0.15 = 1)
✓ 1
A1 0.3
✓ 1
Exam tip: Probabilities of all the possible outcomes of one event add up to 1, which gives a quick check on any missing value.
47IGMC3-8-11r41-04Combined events, tree diagrams and conditional probabilityPaper CEasy6 marks
No calculator
A climber's gear bag holds 5 red and 3 blue carabiners. A carabiner is taken at random, its colour is noted and it is put back. A second carabiner is then taken at random.
(a)
Write down the probability that the first carabiner is red.
(1)
(b)
Find the probability that both carabiners are red.
(2)
(c)
Find the probability that the two carabiners are different colours.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 5/8
✓ 1
Part (b)
M1 5/8 × 5/8
✓ 1
A1 25/64
✓ 1
Part (c)
M1 5/8 × 3/8 (or 3/8 × 5/8)
✓ 1
M1 adds the two orders: 15/64 + 15/64
✓ 1
A1 15/32 (= 0.46875, accept 30/64)
✓ 1
Common error: When the first item is replaced, the second probability has the same denominator as the first; do not reduce it by one.
48IGMC3-8-11r47-01Combined events, tree diagrams and conditional probabilityPaper CEasy4 marks
No calculator
A marine aquarium tank holds 12 clownfish, 8 angelfish and 5 tang. One fish is chosen at random.
(a)
Write down the probability that the fish is an angelfish.
(1)
(b)
Write down the probability that the fish is not a clownfish.
(1)
(c)
The fish is returned to the tank and the process is repeated, 150 times in total. Estimate the number of times a tang is chosen.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 8/25 (or 0.32)
✓ 1
Part (b)
B1 13/25 (or 0.52)
✓ 1
Part (c)
M1 150 × 5/25 (or 150 × 1/5)
✓ 1
A1 30
✓ 1
Exam tip: 'Estimate the number of times' means multiply the number of trials by the probability of one trial.
49IGMC3-8-11r47-02Single events, relative frequency and expected outcomesPaper CEasy4 marks
Calculator
A ranger records whether a ray is seen in the shallows during each of 240 beach visits. A ray was seen on 36 of the visits.
(a)
Write down the relative frequency of seeing a ray, as a decimal.
(1)
(b)
Use your answer to (a) to estimate how many of the next 80 visits will include a ray sighting.
(2)
(c)
Explain why an estimate based on 2000 visits would be more reliable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.15
✓ 1
Part (b)
M1 0.15 × 80
✓ 1
A1 12
✓ 1
Part (c)
B1 A larger number of trials gives a relative frequency closer to the true probability
✓ 1
Common error: Relative frequency is a fraction of all trials, not of the trials where the event did not happen.
50IGMC3-8-11r47-03Combined events, tree diagrams and conditional probabilityPaper CEasy4 marks
No calculator
An architect chooses one roof style from flat, pitched and domed, and one cladding material from brick, timber, glass and stone. Each combination is equally likely.
(a)
Write down the number of possible combinations.
(1)
(b)
Find the probability that the design has a domed roof with timber cladding.
(1)
(c)
Find the probability that the roof is flat or pitched and the cladding is not glass.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 12
✓ 1
Part (b)
B1 1/12
✓ 1
Part (c)
M1 2 roof styles × 3 claddings = 6 favourable combinations out of 12
✓ 1
A1 1/2 (or 0.5)
✓ 1
Exam tip: Multiplying the number of choices for each decision gives the total number of equally likely outcomes.
51IGMC3-8-11r47-04Combined events, tree diagrams and conditional probabilityPaper CEasy4 marks
Calculator
A courier company records the arrival of each parcel. The probability that a parcel arrives early is 0.15 and the probability that it arrives on time is 0.62. Every parcel is early, on time or late.
(a)
Find the probability that a parcel is late.
(1)
(b)
Find the probability that a parcel does not arrive early.
(1)
(c)
In one month 400 parcels are delivered. Estimate how many are late.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.23
✓ 1
Part (b)
B1 0.85
✓ 1
Part (c)
M1 400 × 0.23
✓ 1
A1 92
✓ 1
Exam tip: The probabilities of all possible outcomes of one trial add up to 1.
52IGMC-048-01-06Combined events, tree diagrams and conditional probabilityPaper CMedium6 marks
No calculator
A box contains 4 milk chocolates and 6 dark chocolates. Sam takes two chocolates at random, without replacement.
(a)
Find the probability that both chocolates are dark.
(2)
(b)
Find the probability that Sam takes one milk chocolate and one dark chocolate.
(3)
(c)
Find the probability that at least one chocolate is milk.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(\frac{6}{10} \times \frac{5}{9}\)
✓ 1
A1: \(\frac{30}{90} = \frac{1}{3}\)
✓ 1
Part (b)
M1: \(\frac{4}{10} \times \frac{6}{9}\) or \(\frac{6}{10} \times \frac{4}{9}\) seen, or \(\frac{4}{10} \times \frac{3}{9}\) seen for P(both milk)
✓ 1
M1: adds both orders or doubles one product, or uses \(1 - \frac{30}{90} - \frac{12}{90}\)
✓ 1
A1: \(\frac{48}{90} = \frac{8}{15}\)
✓ 1
Part (c)
B1: \(\frac{2}{3}\) from any correct method, e.g. 1 − their answer to part (a) (follow through), or \(\frac{12}{90} + \frac{48}{90}\)
✓ 1
Common error: Without replacement, the denominator for the second pick is 9, not 10. The numerator also falls by 1 if the second chocolate is the same type as the first.
53IGMC-048-01-09Combined events, tree diagrams and conditional probabilityPaper CMedium5 marks
Calculator
The probability that Kofi scores with a free throw in basketball is 0.65. He takes three free throws. The throws are independent.
(a)
Find the probability that Kofi scores with all three throws.
(2)
(b)
Find the probability that Kofi scores with exactly two of the three throws.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: 0.65 × 0.65 × 0.65
✓ 1
A1: 0.275 to 3 s.f.
✓ 1
Part (b)
M1: 0.65 × 0.65 × 0.35 for one order
✓ 1
M1: multiplies by 3, or adds all three orders
✓ 1
A1: 0.444 to 3 s.f.
✓ 1
Exam tip: List the orders, such as SSM, SMS and MSS, to check that no arrangement of the two scores and one miss has been left out.
54IGMC-048-01-11Combined events, tree diagrams and conditional probabilityPaper CMedium7 marks
No calculator
In a game, a fair coin is thrown. If it shows heads, a fair six-sided dice numbered 1 to 6 is rolled. If it shows tails, a card is taken at random from four cards numbered 1, 2, 3 and 4.
(a)
Find the probability that the coin shows heads and the dice shows 6.
(2)
(b)
Find the probability that the number obtained is 4.
(3)
(c)
The game is played 120 times. Find the expected number of times the number obtained is 4.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(\frac{1}{2} \times \frac{1}{6}\)
✓ 1
A1: \(\frac{1}{12}\)
✓ 1
Part (b)
M1: \(\frac{1}{2} \times \frac{1}{6}\)
✓ 1
M1: \(\frac{1}{2} \times \frac{1}{4}\) added to their first product
✓ 1
A1: \(\frac{5}{24}\)
✓ 1
Part (c)
M1: 120 × their probability from the previous part
✓ 1
A1: 25
✓ 1
Common error: Each route must be multiplied by its coin probability of \(\frac{1}{2}\). Adding \(\frac{1}{6} + \frac{1}{4}\) alone gives an answer that is too large.
55IGMC-048-00-07Single events, relative frequency and expected outcomesPaper CMedium7 marks
No calculator
A box contains only red, white and blue counters. A counter is taken at random. The probability that it is red is \(\frac{3}{8}\) and the probability that it is white is \(\frac{5}{12}\). There are 15 blue counters in the box.
(a)
Show that the probability that the counter is blue is \(\frac{5}{24}\).
(1)
(b)
Find the total number of counters in the box.
(2)
(c)
Find the number of white counters.
(1)
(d)
Some extra red counters are added to the box. The probability of taking a red counter is now \(\frac{1}{2}\). Find the number of red counters that were added.
M1: \(\frac{27+n}{72+n} = \frac{1}{2}\) or red = white + blue after adding, i.e. \(27 + n = 45\)
✓ 1
A1: 18
✓ 1
Common error: When counters are added, the total also changes. Both the numerator and the denominator must increase by n.
56IGMC-048-00-08Single events, relative frequency and expected outcomesPaper CMedium6 marks
Calculator
At a school fair it costs $2 to play a game. The probability of winning the game is 0.18. Each winner receives a prize of $5. The game is played 500 times.
(a)
Find the expected number of winners.
(2)
(b)
Find the expected profit for the school.
(2)
(c)
The school wants an expected profit of at least $700 from 500 games, with the same probability of winning. Find the largest prize, in dollars, it could give. Give your answer correct to the nearest cent.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(500 \times 0.18\)
✓ 1
A1: 90
✓ 1
Part (b)
M1: \(500 \times 2 - 90 \times 5\)
✓ 1
A1: $550
✓ 1
Part (c)
M1: \(1000 - 90P \ge 700\) oe
✓ 1
A1: $3.33
✓ 1
Exam tip: Profit = money taken in − money paid out. Money taken in comes from every game played, not just the winners.
57IGMC-048-00-09Single events, relative frequency and expected outcomesPaper CMedium7 marks
Calculator
Ravi and Mei each toss the same biased coin. Ravi tosses it 40 times and gets 26 heads. Mei tosses it 360 times and gets 207 heads.
(a)
Find the relative frequency of a head for Ravi and for Mei.
(2)
(b)
Whose result gives the better estimate of the probability of a head? Give a reason.
(1)
(c)
Use all the results to find the best estimate of the probability of a head.
(2)
(d)
The coin is tossed 2000 times. Use your answer to part (c) to estimate the number of tails.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: Ravi \(\frac{26}{40}\) = 0.65
✓ 1
B1: Mei \(\frac{207}{360}\) = 0.575
✓ 1
Part (b)
B1: Mei, with the reason that she did more tosses (360 compared with 40)
✓ 1
Part (c)
M1: \(\frac{26+207}{40+360}\) or \(\frac{233}{400}\)
Common error: To combine the results, add the heads and add the tosses. Averaging 0.65 and 0.575 ignores the different numbers of tosses. Keep the exact value 0.5825 for later parts to avoid rounding errors.
58IGMC-048-00-10Single events, relative frequency and expected outcomesPaper CMedium5 marks
Calculator
A biased four-sided dice is numbered 1, 2, 3 and 4. The probability of each score is shown below. Score 1: k Score 2: 0.25 Score 3: 2k Score 4: 0.3
(a)
Find the value of k.
(2)
(b)
Find the probability that the score is at least 3.
(1)
(c)
The dice is rolled n times. The expected number of times it shows a 3 is 39. Find n.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(3k = 1 - 0.25 - 0.3\)
✓ 1
A1: \(k = 0.15\)
✓ 1
Part (b)
B1: 0.6
✓ 1
Part (c)
M1: \(0.3n = 39\)
✓ 1
A1: 130
✓ 1
Exam tip: Expected number = probability × number of trials. If the expected number is given, divide by the probability to find the number of trials.
59IGMC-048-00-11Single events, relative frequency and expected outcomesPaper CMedium4 marks
Calculator
A school has 1200 students. A random sample of 80 students were asked how they travel to school. Walk: 28 Bus: 22 Car: 18 Cycle: 12
Bar chart of the sample of 80 students
(a)
One student from the sample is chosen at random. Find the probability that this student cycles to school.
(1)
(b)
Estimate the number of students in the school who travel by bus.
(2)
(c)
Give one reason why your estimate in part (b) might not be accurate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1: \(\frac{12}{80}\) or \(\frac{3}{20}\) or 0.15
✓ 1
Part (b)
M1: \(\frac{22}{80} \times 1200\) oe
✓ 1
A1: 330
✓ 1
Part (c)
B1: any valid reason, e.g. the sample is only a small fraction of the 1200 students; a different sample could give different results; travel may depend on the weather, season or day surveyed; students may not answer truthfully oe
✓ 1
Exam tip: Use the sample's relative frequency as the probability, then multiply by the whole population to estimate a total.
60IGMC3-8-05d-06Single events, relative frequency and expected outcomesPaper CMedium7 marks
Calculator
Two independent sensors monitor a bridge for cracks. Sensor X detects a crack with probability 0.7 and sensor Y detects it with probability 0.6.
(a)
Find the probability that both sensors detect a crack.
(1)
(b)
Find the probability that neither sensor detects a crack.
(2)
(c)
Find the probability that at least one sensor detects a crack.
(1)
(d)
In 50 inspections of cracked sections, find the expected number in which exactly one sensor detects the crack.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.7 × 0.6 = 0.42
✓ 1
Part (b)
M1 0.3 × 0.4
✓ 1
A1 0.12
✓ 1
Part (c)
1 − 0.12 = 0.88
✓ 1
Part (d)
M1 0.7 × 0.4 + 0.3 × 0.6
✓ 1
A1 0.46
✓ 1
A1 50 × 0.46 = 23
✓ 1
Exam tip: "Exactly one" has two routes, so add the probabilities of both before multiplying by the number of trials.
61IGMC3-8-05d-07Single events, relative frequency and expected outcomesPaper CMedium6 marks
Calculator
A spinner has four coloured sectors. It is spun 500 times and the results are recorded. Red: 140 times; Blue: 185 times; Green: 100 times; Yellow: the remaining spins.
(a)
Find the number of times the spinner landed on yellow.
(1)
(b)
Find the relative frequency of blue.
(1)
(c)
Sam says the four sectors are all the same size. Use the results to comment on this claim.
(2)
(d)
Using the relative frequency from the results, find how many times the spinner is expected to land on green in 1200 spins.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
500 − 140 − 185 − 100 = 75
✓ 1
Part (b)
185/500 = 0.37
✓ 1
Part (c)
Fair spinner with equal sectors would give 0.25 for each colour
✓ 1
0.37 is much larger than 0.25, so the claim is doubtful
✓ 1
Part (d)
M1 100/500 × 1200
✓ 1
A1 240
✓ 1
Exam tip: Use the relative frequency from the experiment to predict, not the theoretical probability, unless the outcomes are known to be equally likely.
62IGMC3-8-05d-08Single events, relative frequency and expected outcomesPaper CMedium5 marks
Calculator
A game app uses two fair six-sided dice. The score is the sum of the two numbers shown.
(a)
Find the probability that the score is 8.
(1)
(b)
Find the probability that the score is a prime number.
(2)
(c)
The developer wants the number of prime scores to be expected to reach at least 100. Find the smallest number of rolls needed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
5/36 or 0.139
✓ 1
Part (b)
M1 15 outcomes (1+1, 2 ways for 3, 4 ways for 5, 6 ways for 7, 2 ways for 11)
✓ 1
A1 15/36 = 5/12
✓ 1
Part (c)
M1 n × 15/36 ≥ 100
✓ 1
A1 240
✓ 1
Common error: Count outcomes in the full 36-outcome sample space; adding sums such as 2, 3, 5 as single outcomes ignores that some sums can be made in several ways.
63IGMC3-8-05d-09Combined events, tree diagrams and conditional probabilityPaper CMedium5 marks
Calculator
A clinic tests a new blood sample. The probability that a sample fails the test is x. The probability that it passes at the first attempt is 3x. The probability that it needs a retest is 0.2. These are the only three outcomes.
(a)
Write down an equation in x and solve it to find x.
(3)
(b)
450 samples are tested. Find the number expected to pass at the first attempt.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 x + 3x + 0.2 = 1
✓ 1
A1 4x = 0.8
✓ 1
A1 x = 0.2
✓ 1
Part (b)
M1 3 × 0.2 × 450
✓ 1
A1 270
✓ 1
Exam tip: When several outcomes cover every possibility, their probabilities sum to 1; use this to build an equation.
64IGMC3-8-08r15-05Combined events, tree diagrams and conditional probabilityPaper CMedium5 marks
No calculator
A bag contains 6 black buttons and 4 white buttons. Two buttons are taken at random from the bag, one after the other, without replacement.
(a)
Find the probability that both buttons are black.
(2)
(b)
Find the probability that the two buttons are different colours.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 6/10 × 5/9
✓ 1
A1 1/3 (30/90)
✓ 1
Part (b)
M1 6/10 × 4/9 (or 4/10 × 6/9)
✓ 1
M1 adds the two products: 24/90 + 24/90
✓ 1
A1 8/15 (48/90)
✓ 1
Common error: When items are not replaced, the second probability uses a smaller total (9 here, not 10).
65IGMC3-8-08r15-06Combined events, tree diagrams and conditional probabilityPaper CMedium5 marks
Calculator
On any day, the probability that a fishing boat makes a good catch is 0.7. The results on different days are independent. Consider two days.
(a)
Find the probability that the boat makes a good catch on both days.
(1)
(b)
Find the probability that the boat makes a good catch on exactly one of the two days.
(2)
(c)
Find the probability that the boat makes at least one good catch.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.49
✓ 1
Part (b)
M1 0.7 × 0.3 × 2 (both orders)
✓ 1
A1 0.42
✓ 1
Part (c)
M1 1 − 0.3 × 0.3 (or 0.49 + 0.42)
✓ 1
A1 0.91
✓ 1
Exam tip: For 'at least one', subtracting the probability of none from 1 is quicker than adding several cases.
66IGMC3-8-08r15-07Single events, relative frequency and expected outcomesPaper CMedium6 marks
Calculator
A dress shop records the sizes of the first 60 dresses sold in a week. Size S: 12, size M: 27, size L: 15, size XL: 6.
Bar chart of the first 60 dresses sold
(a)
Estimate the probability that the next dress sold is size M.
(1)
(b)
Estimate the probability that the next dress sold is size S or size XL.
(1)
(c)
The shop orders 400 dresses for the next month. Use the data to estimate how many will be size L.
(2)
(d)
In the following week, another 40 dresses are sold and 14 of them are size M. Find the relative frequency of size M for all 100 dresses sold in the two weeks.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.45 (27/60)
✓ 1
Part (b)
B1 0.3 (18/60)
✓ 1
Part (c)
M1 400 × 15/60
✓ 1
A1 100
✓ 1
Part (d)
M1 (27 + 14)/100
✓ 1
A1 0.41
✓ 1
Exam tip: Estimates from data are relative frequencies: the number of times the event occurred divided by the total trials.
67IGMC3-8-08r15-08Combined events, tree diagrams and conditional probabilityPaper CMedium4 marks
No calculator
At a robot competition there are 50 robots. 30 robots have wheels and 20 robots have legs. 12 of the wheeled robots and 5 of the legged robots qualify for the final.
(a)
One robot is chosen at random. Find the probability that it has legs and did not qualify.
(2)
(b)
Two different robots are chosen at random. Find the probability that both have wheels.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 20 − 5 = 15 legged robots did not qualify
✓ 1
A1 15/50 = 3/10
✓ 1
Part (b)
M1 30/50 × 29/49
✓ 1
A1 87/245
✓ 1
Common error: For the second robot, reduce both the numerator and the denominator by 1 when there is no replacement.
68IGMC3-8-09r18-02Combined events, tree diagrams and conditional probabilityPaper CMedium5 marks
Calculator
A farmer sows seeds of a barley variety. Each seed germinates with probability 0.84, independently of every other seed.
(a)
Find the probability that a seed does not germinate.
(1)
(b)
The farmer sows 250 seeds. Calculate the number of seeds expected to germinate.
(2)
(c)
Two seeds are sown. Find the probability that both germinate. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.16
✓ 1
Part (b)
M1 250 × 0.84
✓ 1
A1 210
✓ 1
Part (c)
M1 0.84 × 0.84 (= 0.7056)
✓ 1
A1 0.706
✓ 1
Common error: For two independent events happening together, multiply the probabilities; adding them gives a value that can exceed 1.
69IGMC3-8-09r18-05Combined events, tree diagrams and conditional probabilityPaper CMedium5 marks
Calculator
A farmer tests seeds in three batches. In the first batch 78 of 100 seeds germinated, in the second 162 of 200 germinated and in the third 395 of 500 germinated.
(a)
Explain why combining the three batches gives a more reliable estimate of the probability of germination than using the first batch only.
(1)
(b)
Find the relative frequency of germination for all three batches combined. Give your answer as a fraction in its simplest form.
(2)
(c)
The farmer sows 2400 seeds. Use the combined relative frequency, as a fraction, to estimate the number that germinate.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 More trials (800 rather than 100) give a more reliable estimate
✓ 1
Part (b)
M1 (78 + 162 + 395)/(100 + 200 + 500) = 635/800
✓ 1
A1 127/160
✓ 1
Part (c)
M1 2400 × 635/800
✓ 1
A1 1905
✓ 1
Exam tip: Add the successes and add the trials across batches; do not average the three separate fractions.
70IGMC3-8-09r18-09Combined events, tree diagrams and conditional probabilityPaper CMedium6 marks
Calculator
Apples from an orchard are graded as Class A, Class B or rejected. A randomly chosen apple is Class A with probability 0.65 and Class B with probability 0.25.
(a)
Find the probability that a randomly chosen apple is rejected.
(2)
(b)
A crate holds 400 apples. Calculate the number expected to be rejected.
(2)
(c)
In a different crate, 27 apples were rejected, which is the number expected. Calculate the number of apples in that crate.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 1 − (0.65 + 0.25)
✓ 1
A1 0.1
✓ 1
Part (b)
M1 400 × 0.1
✓ 1
A1 40
✓ 1
Part (c)
M1 27 ÷ 0.1 (or 27/0.1)
✓ 1
A1 270
✓ 1
Common error: Probabilities of all outcomes add to 1; subtract the others from 1 rather than guessing the remainder.
71IGMC3-8-09r18-10Single events, relative frequency and expected outcomesPaper CMedium7 marks
Calculator
Anna and Ben each enter a music grade exam. The probability that Anna passes is 0.7 and the probability that Ben passes is 0.6. Their results are independent.
(a)
Find the probability that both pass.
(2)
(b)
Find the probability that at least one of them passes.
(2)
(c)
Find the probability that exactly one of them passes.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.7 × 0.6
✓ 1
A1 0.42
✓ 1
Part (b)
M1 1 − 0.3 × 0.4
✓ 1
A1 0.88
✓ 1
Part (c)
M1 0.7 × 0.4 (Anna passes, Ben fails)
✓ 1
M1 adding 0.3 × 0.6 (Ben passes, Anna fails)
✓ 1
A1 0.46
✓ 1
Exam tip: Check: P(exactly one) should equal P(at least one) − P(both), here 0.88 − 0.42 = 0.46.
72IGMC3-8-10r20-05Combined events, tree diagrams and conditional probabilityPaper CMedium6 marks
Calculator
Archaeologists dig two trenches, A and B. The probability that coins are found in trench A is 0.3 and the probability that coins are found in trench B is 0.45. The two events are independent.
(a)
Find the probability that coins are found in both trenches.
(2)
(b)
Find the probability that coins are found in exactly one trench.
(3)
(c)
Find the probability that no coins are found in either trench.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.3 × 0.45
✓ 1
A1 0.135
✓ 1
Part (b)
M1 0.3 × 0.55
✓ 1
M1 0.7 × 0.45 and adds the two products
✓ 1
A1 0.165 + 0.315 = 0.48
✓ 1
Part (c)
B1 0.7 × 0.55 = 0.385
✓ 1
Exam tip: Write the four outcomes of a two-stage tree and check that their probabilities add to 1.
73IGMC3-8-10r20-06Combined events, tree diagrams and conditional probabilityPaper CMedium6 marks
No calculator
A cycling club has 60 members. 38 members ride road bikes, 25 ride mountain bikes and 8 ride both types. Some members ride neither type.
Venn diagram (not to scale)
(a)
Work out how many members ride road bikes only.
(2)
(b)
A member is chosen at random. Find the probability that the member rides neither type. Give your answer as a fraction in its simplest form.
(2)
(c)
The club grows to 240 members with the same proportions. Estimate the number who ride both types.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 38 − 8
✓ 1
A1 30
✓ 1
Part (b)
M1 60 − (30 + 8 + 17) = 5
✓ 1
A1 5/60 = 1/12
✓ 1
Part (c)
M1 240 × 8/60
✓ 1
A1 32
✓ 1
Exam tip: A Venn diagram is easiest to fill from the overlap outwards: start with the number who are in both sets.
74IGMC3-8-10r20-09Single events, relative frequency and expected outcomesPaper CMedium5 marks
Calculator
A bag at a cycling event contains n blue tokens and 8 green tokens. A token is taken from the bag at random. The probability that it is green is 2/7.
(a)
Find the value of n.
(3)
(b)
Five more green tokens are added to the bag. Find the new probability of taking a green token. Give your answer as a fraction in its simplest form.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 8 ÷ (n + 8) = 2/7
✓ 1
M1 56 = 2n + 16
✓ 1
A1 n = 20
✓ 1
Part (b)
M1 13 green out of 33 tokens in total
✓ 1
A1 13/33
✓ 1
Common error: Adding extra items to the top of the fraction only is a frequent slip; the total in the bag changes as well. adding extra items to the top of the fraction only; the total in the bag changes as well.
75IGMC3-8-10r20-10Combined events, tree diagrams and conditional probabilityPaper CMedium7 marks
Calculator
Each server delivered to a data centre must pass three independent tests. For each test, the probability that a server passes is 0.8.
(a)
Find the probability that a server passes all three tests.
(2)
(b)
Find the probability that a server fails exactly one test.
(3)
(c)
500 servers are delivered. Estimate how many pass all three tests.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.8 × 0.8 × 0.8
✓ 1
A1 0.512
✓ 1
Part (b)
M1 0.2 × 0.8 × 0.8 for one order
✓ 1
M1 multiplies by 3 for the three possible orders
✓ 1
A1 0.384
✓ 1
Part (c)
M1 500 × 0.512 (ECF from (a))
✓ 1
A1 256
✓ 1
Exam tip: Count the number of different orders in which the event can happen, and multiply by the probability of one order.
76IGMC3-8-12r30-05Combined events, tree diagrams and conditional probabilityPaper CMedium7 marks
Calculator
Two wind turbines on a hill each independently produce power on a given day with probability 0.85.
(a)
Find the probability that both turbines produce power on a given day.
(2)
(b)
Find the probability that exactly one of the turbines produces power on a given day.
(3)
(c)
Estimate the number of days in a period of 400 days on which both turbines produce power.
Common error: With 'exactly one' there are two routes (A works and B fails, or the reverse); counting only one route halves the answer.
77IGMC3-8-12r30-08Single events, relative frequency and expected outcomesPaper CMedium5 marks
Calculator
Over 200 days, the output of a wind turbine was high on 84 days, medium on 68 days and low on the remaining days.
(a)
Work out the number of days on which the output was low.
(1)
(b)
Write down the relative frequency of high output.
(1)
(c)
Estimate the number of high-output days in the next 500 days.
(2)
(d)
Explain why using data from 2000 days would usually give a better estimate.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 200 − 84 − 68 = 48
✓ 1
Part (b)
B1 84/200 = 0.42
✓ 1
Part (c)
M1 500 × 0.42
✓ 1
A1 210
✓ 1
Part (d)
B1 more data gives a relative frequency closer to the true probability
✓ 1
Exam tip: Relative frequency = number of times the event happened ÷ total number of trials.
78IGMC3-8-12r30-09Single events, relative frequency and expected outcomesPaper CMedium8 marks
Calculator
A library prize draw uses a bag containing 4 green tokens and n yellow tokens. The probability of taking a green token at random is 2/5.
(a)
Find the value of n.
(3)
(b)
A token is taken, its colour noted and it is replaced. This is done twice. Find the probability that both tokens are yellow.
(2)
(c)
Find the probability that exactly one of the two tokens is green.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 4 ÷ (4 + n) = 2/5
✓ 1
M1 20 = 2(4 + n) or total tokens = 4 ÷ (2/5) = 10
✓ 1
A1 n = 6
✓ 1
Part (b)
M1 (6/10) × (6/10)
✓ 1
A1 0.36
✓ 1
Part (c)
M1 0.4 × 0.6 (one order)
✓ 1
M1 multiply by 2 (both orders)
✓ 1
A1 0.48
✓ 1
Common error: When tokens are replaced the probabilities do not change between draws.
79IGMC3-8-12r30-10Combined events, tree diagrams and conditional probabilityPaper CMedium5 marks
No calculator
To choose lane numbers for a swimming relay, a coach spins two fair spinners. Spinner A has the numbers 1, 2 and 3. Spinner B has the numbers 1, 2, 3 and 4. The two numbers are added.
(a)
Find the probability that the sum is 5.
(2)
(b)
Find the probability that the product of the two numbers is even.
(2)
(c)
Find the probability that the sum is greater than 6.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 list or count the outcomes (1, 4), (2, 3), (3, 2)
✓ 1
A1 3/12 = 1/4
✓ 1
Part (b)
M1 4 outcomes with both numbers odd, so 12 − 4 = 8 outcomes (or 1 − 2/3 × 1/2)
✓ 1
A1 8/12 = 2/3
✓ 1
Part (c)
B1 only (3, 4), so 1/12
✓ 1
Exam tip: A table of all 12 outcomes makes it easy to count the cases without missing any.
80IGMC3-8-12r30-11Combined events, tree diagrams and conditional probabilityPaper CMedium6 marks
No calculator
A library has 120 members. 70 borrow fiction, 45 borrow non-fiction and 20 borrow both fiction and non-fiction. One member is chosen at random.
Venn diagram (not to scale)
(a)
Work out how many members borrow neither fiction nor non-fiction.
(2)
(b)
Find the probability that the member borrows fiction only.
(2)
(c)
Find the probability that the member borrows exactly one type of book.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 70 + 45 − 20 = 95 borrow at least one type
✓ 1
A1 120 − 95 = 25
✓ 1
Part (b)
M1 70 − 20 = 50
✓ 1
A1 50/120 = 5/12
✓ 1
Part (c)
M1 50 + 25 = 75 (fiction only plus non-fiction only, ECF from (b))
✓ 1
A1 75/120 = 5/8 = 0.625
✓ 1
Common error: Counting the 20 members who borrow both types twice when adding 70 and 45.
81IGMC3-8-11r32-05Combined events, tree diagrams and conditional probabilityPaper CMedium7 marks
No calculator
A cycling team has 8 riders: 5 are climbers and 3 are sprinters. Two riders are chosen at random, without replacement, to form a breakaway group.
(a)
Show that the probability that both riders are climbers is 5/14.
(2)
(b)
Find the probability that the group has one climber and one sprinter.
(3)
(c)
Find the probability that at least one rider in the group is a sprinter.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5/8 × 4/7
✓ 1
A1 20/56 = 5/14 (shown)
✓ 1
Part (b)
M1 5/8 × 3/7 (or 3/8 × 5/7)
✓ 1
M1 adds both orders (or doubles)
✓ 1
A1 15/28
✓ 1
Part (c)
M1 1 − 5/14
✓ 1
A1 9/14
✓ 1
Exam tip: "At least one" is often quickest as 1 minus the probability of none.
82IGMC3-8-11r32-06Combined events, tree diagrams and conditional probabilityPaper CMedium7 marks
Calculator
A data centre has two independent cooling units. Unit A works on any given day with probability 0.95 and unit B works with probability 0.9.
(a)
Find the probability that both units work on a given day.
(2)
(b)
Find the probability that exactly one unit works on a given day.
(3)
(c)
Find the probability that at least one unit works on a given day.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.95 × 0.9
✓ 1
A1 0.855
✓ 1
Part (b)
M1 0.95 × 0.1
✓ 1
M1 0.05 × 0.9 and adds to the first product
✓ 1
A1 0.14
✓ 1
Part (c)
M1 1 − 0.05 × 0.1
✓ 1
A1 0.995
✓ 1
Common error: Exactly one unit working means two separate outcomes, so both must be added.
83IGMC3-8-11r32-07Single events, relative frequency and expected outcomesPaper CMedium5 marks
Calculator
Artefacts at a dig are found in one of three layers: A, B or C. The probability that an artefact is from layer A is 0.55. The probability that it is from layer B is 2x and the probability that it is from layer C is x.
(a)
Find the value of x.
(2)
(b)
Write down the probability that an artefact is from layer B.
(1)
(c)
A further 240 artefacts are found. Estimate how many come from layer B.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.55 + 2x + x = 1
✓ 1
A1 x = 0.15
✓ 1
Part (b)
B1 0.3 (ECF from (a))
✓ 1
Part (c)
M1 240 × 0.3 (ECF)
✓ 1
A1 72
✓ 1
Exam tip: The probabilities of all the possible outcomes must add up to 1.
84IGMC3-8-11r32-08Combined events, tree diagrams and conditional probabilityPaper CMedium7 marks
No calculator
A cyclist wins any race she enters with probability 1/4, independently of other races. She enters three races.
(a)
Find the probability that she wins all three races.
(2)
(b)
Find the probability that she wins exactly one race.
(3)
(c)
Find the probability that she wins at least one race.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (1/4)3
✓ 1
A1 1/64
✓ 1
Part (b)
M1 1/4 × 3/4 × 3/4 = 9/64
✓ 1
M1 multiplies by 3 for the three possible orders
✓ 1
A1 27/64
✓ 1
Part (c)
M1 1 − (3/4)3
✓ 1
A1 37/64
✓ 1
Common error: Winning exactly one race can happen in three different orders; a single branch product is not enough.
85IGMC3-8-11r32-09Single events, relative frequency and expected outcomesPaper CMedium5 marks
Calculator
Over 200 rides, a cyclist counted the number of punctures on each ride. No punctures: 132 rides One puncture: 50 rides Two punctures: 18 rides
Bar chart of the 200 rides
(a)
Find the relative frequency of a ride with no punctures.
(1)
(b)
Find the probability that a ride has at least one puncture.
(2)
(c)
Estimate how many of the next 500 rides will have at least one puncture.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 132/200 = 0.66
✓ 1
Part (b)
M1 1 − 0.66 (or (50 + 18)/200)
✓ 1
A1 0.34
✓ 1
Part (c)
M1 500 × 0.34 (ECF)
✓ 1
A1 170
✓ 1
Exam tip: Use relative frequency from the data as the estimate of probability.
86IGMC3-8-11r39-04Combined events, tree diagrams and conditional probabilityPaper CMedium6 marks
Calculator
A tray holds 10 queen cells. Six are ready to hatch and four are not. A beekeeper takes two cells from the tray at random, one after the other, without replacement. Give each answer as a fraction in its simplest form.
(a)
Find the probability that both cells are ready to hatch.
(2)
(b)
Find the probability that exactly one of the two cells is ready to hatch.
(2)
(c)
Find the probability that at least one of the two cells is not ready to hatch.
Common error: Without replacement, the second denominator is one smaller, and the numerator also changes if the first item removed was a 'success'.
87IGMC3-8-11r39-05Single events, relative frequency and expected outcomesPaper CMedium6 marks
Calculator
The probability that a Formula 1 race is wet is 0.3. When it is wet, the probability that a particular driver finishes on the podium is 0.2. When it is dry, this probability is 0.45.
Tree diagram with the given probabilities
(a)
Find the probability that a race is wet and the driver finishes on the podium.
(1)
(b)
Find the probability that the driver finishes on the podium in a randomly chosen race.
(3)
(c)
In a season of 24 races, work out the expected number of podium finishes for this driver.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.06
✓ 1
Part (b)
M1 0.7 × 0.45 (= 0.315)
✓ 1
M1 0.06 + 0.315 (ECF from (a))
✓ 1
A1 0.375
✓ 1
Part (c)
M1 24 × 0.375 (ECF from (b))
✓ 1
A1 9
✓ 1
Exam tip: The 'dry' branch has probability 1 − 0.3, so calculate it rather than reading it from the question.
88IGMC3-8-11r39-06Single events, relative frequency and expected outcomesPaper CMedium6 marks
Calculator
A hospital records, for 200 patients on three wards, whether each was discharged within 7 days. Ward X: 60 patients, of whom 45 were discharged within 7 days. Ward Y: 90 patients, of whom 54 were discharged within 7 days. Ward Z: 50 patients, of whom 41 were discharged within 7 days.
(a)
One of the 200 patients is chosen at random. Find the probability that this patient was discharged within 7 days.
(2)
(b)
One of the patients who was not discharged within 7 days is chosen at random. Find the probability that this patient was on Ward X. Give your answer as a fraction in its simplest form.
(2)
(c)
Next year Ward Z expects 300 patients. Use the figures above to estimate how many will be discharged within 7 days.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 (45 + 54 + 41) ÷ 200 or 140 ÷ 200
✓ 1
A1 0.7
✓ 1
Part (b)
B1 60 patients not discharged within 7 days (15 on Ward X)
✓ 1
A1 \(\frac{1}{4}\)
✓ 1
Part (c)
M1 \(\frac{41}{50}\) × 300
✓ 1
A1 246
✓ 1
Exam tip: Read carefully which group is the 'whole' for each question: here the denominator in (b) is only the patients who were not discharged.
89IGMC3-8-11r39-11Combined events, tree diagrams and conditional probabilityPaper CMedium6 marks
No calculator
In a race a Formula 1 team makes two pit stops. The probability that the first stop is fast is \(\frac{3}{4}\). The probability that the second stop is fast is \(\frac{2}{3}\). The two stops are independent.
(a)
Find the probability that both stops are fast.
(2)
(b)
Find the probability that exactly one stop is fast.
(3)
(c)
Find the probability that neither stop is fast.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{3}{4}\) × \(\frac{2}{3}\)
✓ 1
A1 \(\frac{1}{2}\)
✓ 1
Part (b)
M1 \(\frac{3}{4}\) × \(\frac{1}{3}\)
✓ 1
M1 + \(\frac{1}{4}\) × \(\frac{2}{3}\)
✓ 1
A1 \(\frac{5}{12}\)
✓ 1
Part (c)
B1 \(\frac{1}{12}\)
✓ 1
Common error: Do not add probabilities along a branch: multiply along each route, then add the routes.
90IGMC3-8-11r39-14Combined events, tree diagrams and conditional probabilityPaper CMedium6 marks
Calculator
A hospital pharmacy dispenses each prescription correctly with probability 0.97, independently of other prescriptions.
(a)
Find the probability that three prescriptions are all dispensed correctly. Give your answer correct to 3 significant figures.
(2)
(b)
Find the probability that at least one of the three prescriptions is dispensed incorrectly. Use the unrounded value from part (a) and give your answer correct to 3 significant figures.
(2)
(c)
The pharmacy dispenses 2000 prescriptions. Work out the expected number dispensed incorrectly.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.973
✓ 1
A1 0.913
✓ 1
Part (b)
M1 1 − 0.973
✓ 1
A1 0.0873
✓ 1
Part (c)
M1 2000 × 0.03
✓ 1
A1 60
✓ 1
Exam tip: The probability of an error is 1 − 0.97, which is the figure needed for expected-number questions.
91IGMC3-8-12r40-03Combined events, tree diagrams and conditional probabilityPaper CMedium6 marks
Calculator
A farmer plants two seeds in separate pots. Each seed independently has a probability of 0.9 of germinating.
(a)
Find the probability that both seeds germinate.
(2)
(b)
Find the probability that exactly one seed germinates.
(2)
(c)
Find the probability that at least one seed germinates.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.9\times0.9\)
✓ 1
A1 0.81
✓ 1
Part (b)
M1 \(0.9\times0.1+0.1\times0.9\) (both orders)
✓ 1
A1 0.18
✓ 1
Part (c)
M1 \(1-0.1\times0.1\) or \(0.81+0.18\)
✓ 1
A1 0.99
✓ 1
Exam tip: For "exactly one" of two independent events, list both orders. For "at least one", it is often quicker to subtract the probability that none happen from 1.
92IGMC3-8-12r40-04Combined events, tree diagrams and conditional probabilityPaper CMedium6 marks
Calculator
For a round trip, the outbound flight and the return flight each independently depart on time with probability 0.84.
(a)
Find the probability that neither flight departs on time.
(2)
(b)
Find the probability that both flights depart on time.
(2)
(c)
A travel company books 625 round trips. Calculate the number of these round trips for which it expects both flights to depart on time.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.16\times0.16\)
✓ 1
A1 0.0256
✓ 1
Part (b)
M1 \(0.84\times0.84\)
✓ 1
A1 0.7056
✓ 1
Part (c)
M1 \(625\times0.7056\) (ECF from (b))
✓ 1
A1 441
✓ 1
Common error: Writing 0.84 + 0.84 for "both". For independent events that must both happen, multiply the probabilities.
93IGMC3-8-12r40-05Single events, relative frequency and expected outcomesPaper CMedium5 marks
Calculator
A music teacher asks 60 students in a school orchestra club which one instrument they most enjoy playing. The results are shown in the table.
Piano: 18 Guitar: 27 Violin: 12 Drums: 3
Bar chart of the results for 60 students
(a)
Write down the relative frequency of a student choosing violin. Give your answer as a decimal.
(2)
(b)
The school has 840 students. Use the table to estimate how many of them would choose guitar.
(2)
(c)
Give one reason why the estimate in part (b) may not be reliable.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(\frac{12}{60}\)
✓ 1
A1 0.2
✓ 1
Part (b)
M1 \(840\times\frac{27}{60}\)
✓ 1
A1 378
✓ 1
Part (c)
B1 the sample is only students in the orchestra club, so it may not be representative of the whole school (accept: the sample is small relative to the school)
✓ 1
Exam tip: An estimate from a sample is only as good as the sample. A larger sample, chosen at random from all students, gives a more reliable estimate.
94IGMC3-8-12r40-12Combined events, tree diagrams and conditional probabilityPaper CMedium6 marks
No calculator
In a music-theory game, two fair six-sided dice are rolled and their scores are added to give the total for the round.
(a)
Find the probability that the total is 8.
(2)
(b)
Find the probability that the total is less than 5. Give your answer as a fraction in its simplest form.
(2)
(c)
Find the probability that the product of the two scores is odd. Give your answer as a fraction in its simplest form.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5 favourable outcomes out of 36: (2,6), (3,5), (4,4), (5,3), (6,2)
✓ 1
A1 \(\frac{5}{36}\)
✓ 1
Part (b)
M1 totals 2, 3, 4 occur in \(1+2+3=6\) ways
✓ 1
A1 \(\frac{1}{6}\)
✓ 1
Part (c)
M1 both scores odd: \(\frac36\times\frac36\) or 9 outcomes out of 36
✓ 1
A1 \(\frac14\)
✓ 1
Common error: Listing (2,6) and (6,2) as the same outcome. Ordered pairs from two dice are different outcomes, so there are 36 equally likely results.
95IGMC3-8-12r40-13Combined events, tree diagrams and conditional probabilityPaper CMedium8 marks
Calculator
A region grows four main crops. A field in the region is chosen at random. The probability that it grows wheat is 0.35, barley is 0.22 and oats is 0.15. All other fields grow rapeseed.
(a)
Find the probability that the field grows rapeseed.
(2)
(b)
Find the probability that the field grows wheat or barley.
(2)
(c)
Two fields are chosen at random from different farms, so that the choices are independent. Find the probability that both fields grow oats.
(2)
(d)
Find the probability that at least one of the two fields grows wheat.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(1-(0.35+0.22+0.15)\)
✓ 1
A1 0.28
✓ 1
Part (b)
M1 \(0.35+0.22\)
✓ 1
A1 0.57
✓ 1
Part (c)
M1 \(0.15\times0.15\)
✓ 1
A1 0.0225
✓ 1
Part (d)
M1 \(1-0.65\times0.65\)
✓ 1
A1 0.5775
✓ 1
Exam tip: "Wheat or barley" uses addition because a field cannot grow both as its main crop. "Both fields" uses multiplication because the choices are independent.
96IGMC3-8-11r41-05Combined events, tree diagrams and conditional probabilityPaper CMedium5 marks
No calculator
A food technologist keeps 12 jars of jam on a tray: 7 strawberry and 5 apricot. Jars are chosen at random one after another and are not put back.
(a)
Two jars are chosen. Find the probability that both are strawberry.
(2)
(b)
Three jars are chosen. Find the probability that all three are the same flavour.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 7/12 × 6/11
✓ 1
A1 7/22 (accept 42/132)
✓ 1
Part (b)
M1 7/12 × 6/11 × 5/10 (= 210/1320) for all strawberry
✓ 1
M1 5/12 × 4/11 × 3/10 (= 60/1320) for all apricot and adds the two
✓ 1
A1 9/44 (= 0.2045, accept 270/1320)
✓ 1
Exam tip: 'Not put back' means the numerator and the denominator both change on later picks. Add the probabilities when there is more than one way to meet the condition.
97IGMC3-8-11r41-07Single events, relative frequency and expected outcomesPaper CMedium5 marks
No calculator
A group of 60 mountaineers is asked about qualifications. 35 hold a high-altitude permit, 28 have completed winter training and 12 have both. One mountaineer is chosen at random from the group.
Venn diagram (not to scale)
(a)
Work out how many of the mountaineers have neither a permit nor winter training.
(2)
(b)
Find the probability that the mountaineer has exactly one of the two qualifications. Give your answer in its simplest form.
(2)
(c)
Find the probability that the mountaineer has winter training but no permit. Give your answer as a fraction in its simplest form.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 35 + 28 − 12 (= 51) mountaineers have at least one
✓ 1
A1 60 − 51 = 9
✓ 1
Part (b)
M1 (23 + 16)/60 (or (35 − 12 + 28 − 12)/60)
✓ 1
A1 13/20 (accept 39/60)
✓ 1
Part (c)
A1 4/15 (= 0.2667, accept 16/60)
✓ 1
Common error: The 12 who have both must be counted only once; adding 35 and 28 directly counts them twice.
98IGMC3-8-11r41-09Single events, relative frequency and expected outcomesPaper CMedium6 marks
Calculator
A telecoms company offers three plans. A new customer chooses the Basic plan with probability 0.45 and the Standard plan with probability 0.35. Every other customer chooses the Premium plan.
(a)
Find the probability that a customer chooses the Premium plan.
(1)
(b)
Out of 1800 new customers, find the number expected to choose the Standard plan.
(2)
(c)
The monthly price is €12 for Basic, €18 for Standard and €30 for Premium. Assume that the 1800 customers choose the plans in exactly the expected numbers. Calculate the total monthly income from the 1800 customers.
Exam tip: Find each expected number separately and check that they add up to the total number of customers before using them.
99IGMC3-8-11r47-05Combined events, tree diagrams and conditional probabilityPaper CMedium6 marks
Calculator
In a tagging programme, the probability that a shark caught is a hammerhead is 0.15.
(a)
Write down the probability that a shark caught is not a hammerhead.
(1)
(b)
In one season the team expects to catch 27 hammerheads. Calculate the total number of sharks the team expects to catch.
(2)
(c)
The sharks that are not hammerheads are blacktip sharks or reef sharks, in the ratio 5 : 4. Calculate the number of blacktip sharks the team expects to catch.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.85
✓ 1
Part (b)
M1 27 ÷ 0.15
✓ 1
A1 180
✓ 1
Part (c)
M1 180 − 27 = 153 (or 180 × 0.85)
✓ 1
M1 153 × 5/9
✓ 1
A1 85
✓ 1
Common error: To find the total from an expected number, divide by the probability; multiplying gives a number that is far too small.
100IGMC3-8-11r47-06Combined events, tree diagrams and conditional probabilityPaper CMedium5 marks
Calculator
A logistics firm made 120 deliveries last month. Of these, 70 were made by van, 50 used refrigerated storage and 20 were made by van and used refrigerated storage.
Venn diagram (not to scale)
(a)
Work out how many deliveries were neither made by van nor used refrigerated storage.
(2)
(b)
A delivery is chosen at random. Find the probability that it used refrigerated storage but was not made by van.
(1)
(c)
Next month there will be 450 deliveries. Assuming the same proportions, estimate how many will be neither made by van nor use refrigerated storage.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 70 + 50 − 20 = 100 deliveries in at least one group
✓ 1
A1 20
✓ 1
Part (b)
B1 30/120 = 1/4
✓ 1
Part (c)
M1 450 × 20/120
✓ 1
A1 75
✓ 1
Exam tip: Add the two group totals and subtract the overlap once, since the overlap has been counted twice.
101IGMC3-8-11r47-07Combined events, tree diagrams and conditional probabilityPaper CMedium7 marks
Calculator
On each day of a building project, the probability of rain is 0.4. Whether it rains on one day does not affect the next day. Two consecutive days are considered.
(a)
Find the probability that it rains on both days.
(2)
(b)
Find the probability that it rains on exactly one of the two days.
(3)
(c)
There are 50 such pairs of days. Estimate how many pairs have no rain on either day.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.4 × 0.4
✓ 1
A1 0.16
✓ 1
Part (b)
M1 0.4 × 0.6 (rain then no rain)
✓ 1
M1 doubled, or added to 0.6 × 0.4
✓ 1
A1 0.48
✓ 1
Part (c)
M1 0.6 × 0.6 × 50
✓ 1
A1 18
✓ 1
Exam tip: 'Exactly one' includes both orders: rain then dry, and dry then rain.
102IGMC3-8-11r47-08Single events, relative frequency and expected outcomesPaper CMedium7 marks
Calculator
Two fair four-sided dice each have faces numbered 1, 2, 3 and 4. A survey team throws both dice and the score is the sum of the two numbers shown.
(a)
Find the probability that the score is 5.
(2)
(b)
Find the probability that the score is a prime number.
(3)
(c)
The game is played 96 times. Estimate the number of times the score is prime.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 four outcomes (1,4), (2,3), (3,2), (4,1) out of 16
✓ 1
A1 1/4
✓ 1
Part (b)
B1 prime scores possible are 2, 3, 5 and 7
✓ 1
M1 1 + 2 + 4 + 2 = 9 favourable outcomes
✓ 1
A1 9/16
✓ 1
Part (c)
M1 96 × 9/16
✓ 1
A1 54
✓ 1
Common error: There are 16 equally likely outcomes, not 7; the totals 2 to 8 are not equally likely.
103IGMC3-8-08r15-13Single events, relative frequency and expected outcomesPaper CHard9 marks
Calculator
A fish hatchery records that each egg hatches with probability 0.85. Each fish that hatches survives to become a juvenile with probability 0.6. These events are independent between eggs.
(a)
Show that the probability that an egg becomes a juvenile is 0.51.
(2)
(b)
The hatchery starts with 3000 eggs. Find the expected number of juveniles.
(2)
(c)
The hatchery wants an expected number of at least 2000 juveniles. Find the smallest number of eggs it must start with.
(2)
(d)
Each egg costs the hatchery €1.35 and each juvenile is sold for €4.20. For the 3000 eggs, find the expected profit.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.85 × 0.6
✓ 1
A1 0.51
✓ 1
Part (b)
M1 3000 × 0.51
✓ 1
A1 1530
✓ 1
Part (c)
M1 2000 ÷ 0.51 = 3921.6...
✓ 1
A1 3922
✓ 1
Part (d)
M1 income 1530 × 4.20 = 6426
✓ 1
M1 cost 3000 × 1.35 = 4050
✓ 1
A1 €2376.00
✓ 1
Exam tip: Round up when the question asks for the smallest whole number that reaches a target.
104IGMC3-8-09r18-14Combined events, tree diagrams and conditional probabilityPaper CHard8 marks
Calculator
A random note generator produces a sharp, a natural or a flat note. The probability of a sharp note is 2x, of a natural note is x + 0.1 and of a flat note is 0.3. Successive notes are independent.
(a)
Form an equation in x and solve it.
(3)
(b)
Two successive notes are generated. Find the probability that they are of the same type.
(3)
(c)
The generator produces 150 notes. Estimate the number of sharp notes.
Exam tip: Use the fact that all probabilities add to 1 to find the unknown before answering later parts.
105IGMC3-8-11r41-14Single events, relative frequency and expected outcomesPaper CHard6 marks
No calculator
A mountain hut sells three soups. The probabilities that a customer chooses tomato, mushroom or leek are 2x, x + 0.2 and x − 0.1 respectively.
(a)
Find the value of x.
(3)
(b)
Find the probability that a customer does not choose tomato.
(1)
(c)
The hut serves 640 soups in a season. Find the number of leek soups expected.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 2x + (x + 0.2) + (x − 0.1) = 1
✓ 1
M1 4x + 0.1 = 1 (or 4x = 0.9)
✓ 1
A1 x = 0.225
✓ 1
Part (b)
B1 1 − 0.45 = 0.55 (ECF from (a))
✓ 1
Part (c)
M1 640 × (0.225 − 0.1) = 640 × 0.125
✓ 1
A1 80
✓ 1
Exam tip: Form one equation from 'all probabilities add to 1', then check that every individual probability is between 0 and 1.
106IGMC3-8-11r47-09Single events, relative frequency and expected outcomesPaper CHard8 marks
Calculator
A box contains red, white and green marker buoys. One buoy is chosen at random. The probability it is red is 0.32, the probability it is green is x and the probability it is white is 3x.
(a)
Find the value of x.
(3)
(b)
A buoy is chosen, its colour is noted and it is put back. A second buoy is then chosen. Find the probability that both are white. Give your answer correct to 3 significant figures.
(2)
(c)
Find the probability that exactly one of the two buoys is red. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.32 + x + 3x = 1
✓ 1
M1 4x = 0.68
✓ 1
A1 x = 0.17
✓ 1
Part (b)
B1 P(white) = 3 × 0.17 = 0.51
✓ 1
A1 0.51 × 0.51 = 0.2601 = 0.260
✓ 1
Part (c)
M1 0.32 × 0.68
✓ 1
M1 multiplied by 2 (two orders)
✓ 1
A1 0.4352 = 0.435
✓ 1
Common error: Because the first buoy is put back, the probabilities for the second choice are unchanged.
107IGMC-048-01-07Combined events, tree diagrams and conditional probabilityPaper EMedium5 marks
Calculator
The probability that it rains on a given day is 0.3. If it rains, the probability that Amira cycles to work is 0.25. If it does not rain, the probability that Amira cycles to work is 0.8.
Tree diagram with the given probabilities
(a)
Find the probability that Amira cycles to work on a given day.
(3)
(b)
On one day Amira cycles to work. Find the probability that it rained on that day.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: 0.3 × 0.25
✓ 1
M1: 0.7 × 0.8 added to their first product
✓ 1
A1: 0.635
✓ 1
Part (b)
M1: \(\frac{0.3 \times 0.25}{0.635}\)
✓ 1
A1: 0.118 to 3 s.f.
✓ 1
Exam tip: For 'given that' questions, divide the probability of the required route by the total probability of the condition. Here that is P(rain and cycles) ÷ P(cycles).
108IGMC-048-01-08Combined events, tree diagrams and conditional probabilityPaper EMedium6 marks
No calculator
There are 30 students in a class. 18 students study French, 12 students study Spanish and 5 students study both languages.
Venn diagram (not to scale)
(a)
Find the number of students who study neither language.
(2)
(b)
A student who studies French is chosen at random. Find the probability that this student also studies Spanish.
(2)
(c)
Two different students are chosen at random from those who study Spanish. Find the probability that both of them also study French.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: 30 − (18 + 12 − 5), or a correct Venn diagram method using 13, 5 and 7
✓ 1
A1: 5
✓ 1
Part (b)
M1: denominator of 18
✓ 1
A1: \(\frac{5}{18}\)
✓ 1
Part (c)
M1: \(\frac{5}{12} \times \frac{4}{11}\)
✓ 1
A1: \(\frac{20}{132} = \frac{5}{33}\)
✓ 1
Common error: In conditional questions the denominator is the size of the restricted group, not the whole class of 30.
109IGMC-048-01-10Combined events, tree diagrams and conditional probabilityPaper EMedium5 marks
No calculator
A bag contains 3 red, 4 green and 5 yellow sweets. Two sweets are taken at random, without replacement.
(a)
Show that the probability that both sweets are the same colour is \(\frac{19}{66}\).
(3)
(b)
Given that both sweets are the same colour, find the probability that they are both yellow.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: at least one correct product, for example \(\frac{5}{12} \times \frac{4}{11}\)
✓ 1
M1: adds all three, \(\frac{6}{132} + \frac{12}{132} + \frac{20}{132}\)
Exam tip: For a 'show that' answer, every product and the sum must be written out before the given fraction appears.
110IGMC3-8-05d-10Combined events, tree diagrams and conditional probabilityPaper EMedium5 marks
Calculator
An engineer chooses two satellites one after the other from a fleet of 20, without replacement, for a software check. The fleet has 8 weather satellites, 7 communications satellites and 5 navigation satellites.
(a)
Find the probability that both satellites chosen are weather satellites.
(2)
(b)
Find the probability that the two satellites chosen are of different types.
Common error: Without replacement, the second denominator is one smaller, and the numerator also drops by one when the same type is chosen again.
111IGMC3-8-05d-12Single events, relative frequency and expected outcomesPaper EMedium6 marks
Calculator
In a tennis match a player wins a point on her first serve with probability 0.75 if the first serve lands in. The probability that her first serve lands in is 0.6. If it does not, she makes a second serve and wins the point with probability 0.5.
Tree diagram with the given probabilities
(a)
Find the probability that she wins a point when serving.
(3)
(b)
In 200 service points, find the number of points she is expected to win.
(1)
(c)
She wins a particular point. Find the probability that her first serve was in.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.6 × 0.75 = 0.45
✓ 1
M1 0.4 × 0.5 = 0.2
✓ 1
A1 0.65
✓ 1
Part (b)
130
✓ 1
Part (c)
M1 0.45 / 0.65
✓ 1
A1 9/13 = 0.692
✓ 1
Exam tip: Draw the tree in your head: multiply along branches, add the branches that lead to the same outcome, and divide only for "given that".
112IGMC3-8-08r15-09Combined events, tree diagrams and conditional probabilityPaper EMedium6 marks
No calculator
A fisheries researcher tags 100 fish. 60 are cod and 40 are haddock. Of the cod, 24 carry a tracking tag; of the haddock, 6 carry a tracking tag.
(a)
Find the probability that a fish chosen at random is a cod with a tag.
(1)
(b)
Find the probability that a fish chosen at random has a tag.
(1)
(c)
A fish with a tag is chosen. Find the probability that it is a cod.
(2)
(d)
Two fish are chosen at random without replacement. Find the probability that both have a tag.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 24/100 = 6/25
✓ 1
Part (b)
B1 30/100 = 3/10
✓ 1
Part (c)
M1 24/30 (or (24/100)/(30/100))
✓ 1
A1 4/5
✓ 1
Part (d)
M1 30/100 × 29/99
✓ 1
A1 29/330
✓ 1
Exam tip: For a conditional probability, the 'given' information shrinks the total you divide by.
113IGMC3-8-08r15-10Combined events, tree diagrams and conditional probabilityPaper EMedium7 marks
Calculator
A weather-sensing robot decides whether to shelter. The probability that rain is forecast is 0.3. If rain is forecast, the probability that the robot shelters is 0.9. If rain is not forecast, the probability that it shelters is 0.2.
Tree diagram with the given probabilities
(a)
Find the probability that rain is forecast and the robot shelters.
(2)
(b)
Find the probability that the robot shelters.
(2)
(c)
The robot shelters. Find the probability that rain was forecast. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.3 × 0.9
✓ 1
A1 0.27
✓ 1
Part (b)
M1 0.27 + 0.7 × 0.2
✓ 1
A1 0.41
✓ 1
Part (c)
M1 conditional probability P(rain and shelter)/P(shelter)
✓ 1
M1 0.27/0.41
✓ 1
A1 0.659
✓ 1
Common error: P(A given B) is not the same as P(B given A); divide by the probability of the event that is given.
114IGMC3-8-08r15-11Combined events, tree diagrams and conditional probabilityPaper EMedium7 marks
No calculator
A box holds 5 silk ties and 3 cotton ties. Two ties are taken at random without replacement.
(a)
Show that the probability that both ties are silk is 5/14.
(2)
(b)
Find the probability that the second tie is silk.
(3)
(c)
Given that the second tie is silk, find the probability that the first tie was also silk.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5/8 × 4/7
✓ 1
A1 20/56 = 5/14
✓ 1
Part (b)
M1 cotton then silk: 3/8 × 5/7 = 15/56
✓ 1
M1 adds 20/56 + 15/56
✓ 1
A1 35/56 = 5/8
✓ 1
Part (c)
M1 (5/14) ÷ (5/8)
✓ 1
A1 4/7
✓ 1
Exam tip: In 'show that' questions, write the unsimplified product before simplifying to the given value.
115IGMC3-8-09r18-04Combined events, tree diagrams and conditional probabilityPaper EMedium6 marks
Calculator
A trainee pilot takes a simulator test. The probability of passing at the first attempt is 0.7. A pilot who fails the first attempt takes the test a second time and passes with probability 0.85. No pilot takes the test more than twice.
Tree diagram with the given probabilities (first attempt, then second attempt)
(a)
Find the probability that a trainee passes at the second attempt.
(2)
(b)
Find the probability that a trainee passes the test.
(2)
(c)
200 trainees take the test. Calculate the number expected to fail both attempts.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.3 × 0.85
✓ 1
A1 0.255
✓ 1
Part (b)
M1 0.7 + their 0.255 (or 1 − 0.3 × 0.15)
✓ 1
A1 0.955 (ECF from (a))
✓ 1
Part (c)
M1 200 × 0.3 × 0.15 (or 200 × 0.045)
✓ 1
A1 9
✓ 1
Common error: The second-attempt probability applies only to pilots who failed the first time, so multiply it by 0.3, not by 1.
116IGMC3-8-09r18-06Combined events, tree diagrams and conditional probabilityPaper EMedium7 marks
No calculator
Nine pilots are available for standby duty: 5 are senior pilots and 4 are junior pilots. Two of the nine are chosen at random to be on standby.
(a)
Show that the probability that both are senior pilots is 5/18.
(2)
(b)
Find the probability that exactly one of the two is a senior pilot.
(3)
(c)
Find the probability that at least one of the two is a junior pilot.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5/9 × 4/8
✓ 1
A1 20/72 = 5/18 shown
✓ 1
Part (b)
M1 5/9 × 4/8 (senior then junior) or 4/9 × 5/8 (junior then senior)
✓ 1
M1 adding both orders: 20/72 + 20/72
✓ 1
A1 5/9
✓ 1
Part (c)
M1 1 − 5/18
✓ 1
A1 13/18
✓ 1
Exam tip: 'At least one' is often quickest as 1 minus the probability of none.
117IGMC3-8-09r18-12Single events, relative frequency and expected outcomesPaper EMedium7 marks
No calculator
A box holds 12 eggs, of which 3 are cracked. Three eggs are taken from the box at random, one after another, without replacement.
(a)
Find the probability that the first two eggs taken are both cracked.
(2)
(b)
Find the probability that none of the three eggs is cracked.
(2)
(c)
Find the probability that exactly one of the three eggs is cracked.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 3/12 × 2/11
✓ 1
A1 1/22
✓ 1
Part (b)
M1 9/12 × 8/11 × 7/10
✓ 1
A1 21/55
✓ 1
Part (c)
M1 one ordering, e.g. 3/12 × 9/11 × 8/10 (= 9/55)
✓ 1
M1 multiplying by 3 for the three possible orderings
✓ 1
A1 27/55
✓ 1
Common error: Without replacement, both numerator and denominator change after each egg is taken.
118IGMC3-8-10r20-07Combined events, tree diagrams and conditional probabilityPaper EMedium6 marks
Calculator
A data centre keeps 12 spare drives in a tray. 4 of the drives are faulty. Two drives are taken from the tray at random, one after the other, without replacement.
(a)
Find the probability that both drives are faulty. Give your answer as a fraction in its simplest form.
(2)
(b)
Find the probability that exactly one of the drives is faulty. Give your answer as a fraction in its simplest form.
(3)
(c)
Find the probability that at least one drive is not faulty. Give your answer as a fraction in its simplest form.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 4/12 × 3/11
✓ 1
A1 1/11
✓ 1
Part (b)
M1 4/12 × 8/11
✓ 1
M1 doubles (or adds 8/12 × 4/11)
✓ 1
A1 16/33
✓ 1
Part (c)
B1 1 − 1/11 = 10/11 (ECF from (a))
✓ 1
Exam tip: Without replacement, the second probability changes: reduce the denominator by 1 and adjust the numerator for the first pick.
119IGMC3-8-10r20-08Single events, relative frequency and expected outcomesPaper EMedium6 marks
Calculator
A survey team scans a field with a metal detector. The probability that buried metal is present in a scanned area is 0.2. If metal is present, the detector signals with probability 0.9. If no metal is present, the detector signals (falsely) with probability 0.1.
Tree diagram with the given probabilities
(a)
Find the probability that the detector signals in a scanned area.
(3)
(b)
The detector signals. Find the probability that metal is present. Give your answer as a fraction in its simplest form.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.2 × 0.9 (= 0.18)
✓ 1
M1 0.8 × 0.1 (= 0.08) and adds the two products
✓ 1
A1 0.26
✓ 1
Part (b)
M1 P(metal and signal) ÷ P(signal)
✓ 1
M1 0.18 ÷ 0.26 (ECF from (a))
✓ 1
A1 9/13 (0.692)
✓ 1
Exam tip: For a conditional probability, divide the probability of both events by the probability of the condition.
120IGMC3-8-10r20-11Single events, relative frequency and expected outcomesPaper EMedium7 marks
Calculator
The table shows 80 artefacts from a dig, sorted by material and period.
Bronze: 14 early, 22 late Ceramic: 26 early, 18 late
One artefact is chosen at random.
(a)
Find the probability that the artefact is bronze.
(1)
(b)
Given that the artefact is ceramic, find the probability that it is late. Give your answer as a fraction in its simplest form.
(2)
(c)
Given that the artefact is early, find the probability that it is ceramic. Give your answer as a fraction in its simplest form.
(2)
(d)
Two different artefacts are chosen at random, without replacement. Find the probability that both are late. Give your answer as a fraction in its simplest form.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 36/80 = 9/20
✓ 1
Part (b)
M1 18/44
✓ 1
A1 9/22
✓ 1
Part (c)
M1 26/40
✓ 1
A1 13/20
✓ 1
Part (d)
M1 40/80 × 39/79
✓ 1
A1 39/158
✓ 1
Exam tip: For 'given that', the total you divide by is only the group named in the condition, not all 80 artefacts.
121IGMC3-8-12r30-06Combined events, tree diagrams and conditional probabilityPaper EMedium7 marks
Calculator
A library trolley holds 9 books: 5 fiction and 4 non-fiction. Two books are taken from the trolley at random, one after the other, without replacement.
(a)
Find the probability that both books are fiction.
(2)
(b)
Find the probability that one book is fiction and the other is non-fiction.
(3)
(c)
Find the probability that at least one of the books is non-fiction.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5/9 × 4/8
✓ 1
A1 5/18
✓ 1
Part (b)
M1 5/9 × 4/8 (fiction then non-fiction)
✓ 1
M1 add the other order, 4/9 × 5/8
✓ 1
A1 5/9
✓ 1
Part (c)
M1 1 − P(both fiction) (ECF from (a))
✓ 1
A1 13/18
✓ 1
Exam tip: 'At least one' is usually quickest as 1 minus the probability of none.
122IGMC3-8-11r32-10Combined events, tree diagrams and conditional probabilityPaper EMedium7 marks
Calculator
A data centre has two types of server. 60% are type X and 40% are type Y. In one year, a type X server fails with probability 0.02 and a type Y server fails with probability 0.05. A server is chosen at random.
Tree diagram with the given probabilities
(a)
Show that the probability that the chosen server fails during the year is 0.032.
(2)
(b)
The data centre has 4000 servers. Estimate the number that fail during the year.
(2)
(c)
The chosen server fails. Find the probability that it is type Y.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.6 × 0.02 and 0.4 × 0.05 (0.012 and 0.02)
✓ 1
A1 0.012 + 0.02 = 0.032 (shown)
✓ 1
Part (b)
M1 4000 × 0.032
✓ 1
A1 128
✓ 1
Part (c)
M1 0.4 × 0.05 = 0.02 as the numerator
✓ 1
M1 divides by 0.032
✓ 1
A1 0.625 (or 5/8)
✓ 1
Common error: For a conditional probability the denominator is the probability of the condition (the server failed), not 1.
123IGMC3-8-11r32-11Combined events, tree diagrams and conditional probabilityPaper EMedium6 marks
No calculator
A tray holds 12 beads: 7 are glass and 5 are stone. Two beads are taken at random, one after the other, without replacement.
(a)
Find the probability that both beads are glass.
(2)
(b)
Find the probability that the second bead is glass.
(2)
(c)
Given that the second bead is glass, find the probability that the first bead was also glass.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 7/12 × 6/11
✓ 1
A1 7/22
✓ 1
Part (b)
M1 7/12 × 6/11 + 5/12 × 7/11
✓ 1
A1 77/132 = 7/12
✓ 1
Part (c)
M1 (7/22) ÷ (7/12) or 42/132 ÷ 77/132 (ECF)
✓ 1
A1 6/11
✓ 1
Exam tip: For "given that" questions, write the probability of both events together over the probability of the given event.
124IGMC3-8-11r39-07Combined events, tree diagrams and conditional probabilityPaper EMedium8 marks
Calculator
In a hospital, 4% of patients have a particular condition. A screening test gives a positive result for 95% of patients who have the condition and for 6% of patients who do not have the condition.
Tree diagram with the given probabilities
(a)
Find the probability that a patient chosen at random gets a positive result.
(3)
(b)
Given that a patient gets a positive result, find the probability that the patient has the condition. Give your answer correct to 3 significant figures.
(3)
(c)
5000 patients are screened. Work out the expected number who do not have the condition but get a positive result.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.04 × 0.95 (= 0.038)
✓ 1
M1 0.96 × 0.06 (= 0.0576)
✓ 1
A1 0.0956
✓ 1
Part (b)
M1 numerator 0.04 × 0.95
✓ 1
M1 divided by 0.0956 (ECF from (a))
✓ 1
A1 0.397
✓ 1
Part (c)
M1 5000 × 0.96 × 0.06
✓ 1
A1 288
✓ 1
Common error: P(positive | condition) and P(condition | positive) are different probabilities: the second one has the whole positive group as its denominator.
125IGMC3-8-11r39-12Combined events, tree diagrams and conditional probabilityPaper EMedium7 marks
Calculator
A hospital has 120 nurses. The table shows how many work each shift pattern and how many of them have worked overtime this month. Day shifts: 50 nurses, 30 worked overtime. Night shifts: 40 nurses, 28 worked overtime. Weekend shifts: 30 nurses, 12 worked overtime.
(a)
A nurse is chosen at random. Find the probability that this nurse works night shifts and has worked overtime. Give your answer as a fraction in its simplest form.
(1)
(b)
A nurse who has worked overtime is chosen at random. Find the probability that this nurse works night shifts. Give your answer as a fraction in its simplest form.
(2)
(c)
Two different nurses are chosen at random. Find the probability that both work weekend shifts. Give your answer correct to 3 significant figures.
(2)
(d)
Two different nurses are chosen at random. Find the probability that both have worked overtime. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 \(\frac{7}{30}\)
✓ 1
Part (b)
M1 28 ÷ 70
✓ 1
A1 \(\frac{2}{5}\)
✓ 1
Part (c)
M1 \(\frac{30}{120}\) × \(\frac{29}{119}\)
✓ 1
A1 0.0609
✓ 1
Part (d)
M1 \(\frac{70}{120}\) × \(\frac{69}{119}\)
✓ 1
A1 0.338
✓ 1
Exam tip: In a two-way table, for 'given that' questions use only the row or column named in the condition as the total.
126IGMC3-8-12r40-07Combined events, tree diagrams and conditional probabilityPaper EMedium6 marks
Calculator
At an airport, 20% of travellers are on business trips and the rest are on leisure trips. A business traveller carries hand luggage only with probability 0.7. A leisure traveller carries hand luggage only with probability 0.4. One traveller is chosen at random.
Tree diagram with the given probabilities
(a)
Find the probability that the traveller is on a business trip and carries hand luggage only.
(2)
(b)
Find the probability that the traveller carries hand luggage only.
(2)
(c)
Given that the traveller carries hand luggage only, find the probability that they are on a business trip. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.2\times0.7\)
✓ 1
A1 0.14
✓ 1
Part (b)
M1 \(0.2\times0.7+0.8\times0.4\) (sum of two branches)
✓ 1
A1 0.46
✓ 1
Part (c)
M1 \(\frac{0.14}{0.46}\) (ECF from (a) and (b))
✓ 1
A1 0.304
✓ 1
Exam tip: In a conditional probability question, the denominator is the probability of the condition, here "hand luggage only", found by adding all the branches that lead to it.
127IGMC3-8-12r40-09Combined events, tree diagrams and conditional probabilityPaper EMedium7 marks
Calculator
A music school has 100 students. Of these, 45 play the piano, 30 play the guitar and 12 play both instruments. One student is chosen at random.
Venn diagram (not to scale)
(a)
Find the probability that the student plays neither instrument.
(2)
(b)
Given that the student plays the piano, find the probability that they also play the guitar. Give your answer as a fraction in its simplest form.
(2)
(c)
Determine whether playing the piano and playing the guitar are independent events. Show your working.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(100-(45+30-12)\) or \(100-63\) (students who play at least one)
A1 \(P(\text{piano and guitar})=0.12\), which is not equal to 0.135
✓ 1
R1 so the events are not independent
✓ 1
Exam tip: To test for independence, compare P(A and B) with P(A) × P(B). Quote both values before you conclude.
128IGMC3-8-11r41-06Combined events, tree diagrams and conditional probabilityPaper EMedium5 marks
Calculator
On a mountain pass, the probability of a storm on a given day is 0.2. If there is a storm, the probability that a mast loses its signal is 0.35. If there is no storm, the probability that it loses its signal is 0.05.
Tree diagram with the given probabilities
(a)
Find the probability that there is a storm and the mast loses its signal.
(1)
(b)
Find the probability that the mast loses its signal on a given day.
(2)
(c)
Given that the mast loses its signal, find the probability that there was a storm. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.2 × 0.35 = 0.07
✓ 1
Part (b)
M1 0.2 × 0.35 + 0.8 × 0.05 (= 0.07 + 0.04)
✓ 1
A1 0.11
✓ 1
Part (c)
M1 0.07 ÷ 0.11 (ECF from (a) and (b))
✓ 1
A1 0.636
✓ 1
Exam tip: For 'given that' questions, put the probability of both events on top and the probability of the condition underneath.
129IGMC3-8-11r41-08Combined events, tree diagrams and conditional probabilityPaper EMedium6 marks
Calculator
A box contains 15 food samples: 9 are fermented and 6 are fresh. Two samples are chosen at random, one after the other, without replacement.
(a)
Show that the probability that both samples are fermented is 12/35.
(2)
(b)
Find the probability that at least one of the samples is fermented.
(2)
(c)
Given that at least one sample is fermented, find the probability that both are fermented. Give your answer as a fraction in its simplest form.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 9/15 × 8/14
✓ 1
A1 72/210 = 12/35 shown
✓ 1
Part (b)
M1 1 − P(both fresh) = 1 − 6/15 × 5/14
✓ 1
A1 6/7
✓ 1
Part (c)
M1 (12/35) ÷ (6/7)
✓ 1
A1 2/5 (= 0.4)
✓ 1
Exam tip: 'At least one' is usually quickest as 1 minus the probability of none.
130IGMC3-8-11r41-11Combined events, tree diagrams and conditional probabilityPaper EMedium6 marks
Calculator
On a two-day climb, the probability that day 1 is clear is 0.6. If day 1 is clear, the probability that day 2 is clear is 0.7. If day 1 is not clear, the probability that day 2 is clear is 0.4.
Tree diagram with the given probabilities
(a)
Find the probability that both days are clear.
(1)
(b)
Find the probability that exactly one of the two days is clear.
(3)
(c)
Given that exactly one day is clear, find the probability that it was day 1 that was clear. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.6 × 0.7 = 0.42
✓ 1
Part (b)
M1 0.6 × 0.3 (clear then not clear)
✓ 1
M1 0.4 × 0.4 (not clear then clear)
✓ 1
A1 0.18 + 0.16 = 0.34
✓ 1
Part (c)
M1 0.18 ÷ 0.34 (ECF from (b))
✓ 1
A1 0.529
✓ 1
Exam tip: On a tree diagram, the second-stage probabilities depend on the branch before them, so they need not be the same on each branch.
131IGMC3-8-11r47-10Single events, relative frequency and expected outcomesPaper EMedium6 marks
Calculator
A diver surveys coral reefs. The probability that a reef is bleached is 0.3. If a reef is bleached, the probability that reef sharks are present is 0.2. If a reef is not bleached, the probability that reef sharks are present is 0.7.
Tree diagram with the given probabilities
(a)
Find the probability that reef sharks are present on a reef chosen at random.
(3)
(b)
A reef is found to have no reef sharks. Find the probability that it is bleached. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.3 × 0.2 = 0.06
✓ 1
M1 0.7 × 0.7 = 0.49 added to 0.06
✓ 1
A1 0.55
✓ 1
Part (b)
M1 P(bleached and no sharks) = 0.3 × 0.8 = 0.24
✓ 1
M1 divided by 1 − 0.55 = 0.45
✓ 1
A1 0.5333... = 0.533 (or 8/15)
✓ 1
Exam tip: For conditional probability, divide the probability of both events happening by the probability of the condition.
132IGMC3-8-11r47-11Combined events, tree diagrams and conditional probabilityPaper EMedium7 marks
No calculator
A bag holds 9 anchor bolts: 5 are galvanised and 4 are plain. Two bolts are taken from the bag at random, one after the other, without replacement.
(a)
Find the probability that both bolts are galvanised.
(2)
(b)
Find the probability that both bolts are of the same type.
(3)
(c)
Given that both bolts are of the same type, find the probability that both are galvanised.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5/9 × 4/8
✓ 1
A1 5/18
✓ 1
Part (b)
M1 4/9 × 3/8 = 1/6
✓ 1
M1 added to 5/18
✓ 1
A1 4/9
✓ 1
Part (c)
M1 (5/18) ÷ (4/9)
✓ 1
A1 5/8 (= 0.625)
✓ 1
Common error: Without replacement, the second fraction has one fewer item on the top (if the same type) and one fewer on the bottom.
133IGMC3-8-11r47-13Combined events, tree diagrams and conditional probabilityPaper EMedium6 marks
Calculator
Of 80 divers at a training centre, 44 are certified for deep dives (D), 36 are certified for wreck dives (W) and 14 hold both certificates.
Venn diagram (not to scale)
(a)
Work out how many divers hold neither certificate.
(2)
(b)
A diver is chosen at random from those certified for wreck dives. Find the probability that this diver is also certified for deep dives. Give your answer as a fraction in its simplest form.
(2)
(c)
Two divers are chosen at random from all 80, without replacement. Find the probability that neither holds a certificate. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 44 + 36 − 14 = 66
✓ 1
A1 14
✓ 1
Part (b)
M1 14/36
✓ 1
A1 7/18
✓ 1
Part (c)
M1 14/80 × 13/79
✓ 1
A1 182/6320 = 0.0288
✓ 1
Common error: In (b) the sample space is the 36 wreck-certified divers, not all 80.
134IGMC-048-01-12Combined events, tree diagrams and conditional probabilityPaper EHard7 marks
No calculator
A bag contains n red counters and 4 blue counters. Two counters are taken at random, without replacement. The probability that both counters are red is \(\frac{1}{3}\).
(a)
Show that \(n^2 - 5n - 6 = 0\).
(3)
(b)
Solve the equation to find the value of n.
(2)
(c)
Given that the two counters taken are the same colour, find the probability that they are both blue.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(\frac{n}{n+4} \times \frac{n-1}{n+3}\)
✓ 1
M1: \(3n(n-1) = (n+4)(n+3)\)
✓ 1
A1: \(3n^2 - 3n = n^2 + 7n + 12\) leading to \(2n^2 - 10n - 12 = 0\), and so \(n^2 - 5n - 6 = 0\)
✓ 1
Part (b)
M1: \((n - 6)(n + 1) = 0\)
✓ 1
A1: \(n = 6\) only, with \(n = -1\) rejected
✓ 1
Part (c)
M1: \(\frac{12/90}{30/90 + 12/90}\)
✓ 1
A1: \(\frac{2}{7}\)
✓ 1
Exam tip: Always reject solutions that make no sense in context. A number of counters cannot be negative.
135IGMC-048-01-13Combined events, tree diagrams and conditional probabilityPaper EHard7 marks
Calculator
2% of the people in a town have a certain condition. A test gives a positive result for 95% of people who have the condition. The test gives a positive result for 10% of people who do not have the condition.
Tree diagram with the given probabilities
(a)
Find the probability that a person chosen at random tests positive.
(3)
(b)
A person tests positive. Find the probability that this person has the condition.
(2)
(c)
Two people are chosen at random and tested independently. Find the probability that at least one of them tests positive.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: 0.02 × 0.95 (= 0.019)
✓ 1
M1: 0.98 × 0.10 (= 0.098) added to their first product
✓ 1
A1: 0.117
✓ 1
Part (b)
M1: \(\frac{0.019}{0.117}\) or \(\frac{0.02 \times 0.95}{\text{their (part 1)}}\)
✓ 1
A1: 0.162 (0.1623...) or \(\frac{19}{117}\)
✓ 1
Part (c)
M1: 1 − (1 − 0.117)2, or 2 × 0.117 × 0.883 + 0.1172, using their answer to part 1
✓ 1
A1: 0.220 (0.2203...)
✓ 1
Common error: Giving 0.95 as the probability that a person who tests positive has the condition confuses P(positive | condition) with P(condition | positive). Because the condition is rare, most positive results come from people without it, so the second probability is much smaller.
136IGMC-048-01-14Combined events, tree diagrams and conditional probabilityPaper EHard7 marks
No calculator
A box contains 10 light bulbs, of which 3 are faulty. Bulbs are taken at random, one at a time, without replacement, and tested. Testing stops as soon as a faulty bulb is found.
(a)
Find the probability that the first faulty bulb is found on the third test.
(2)
(b)
Find the probability that more than three tests are needed.
(2)
(c)
Given that a faulty bulb is found within the first three tests, find the probability that it was found on the first test.
M1: P(found within three tests) = \(1 - \frac{7}{24} = \frac{17}{24}\)
✓ 1
M1: \(\frac{3}{10} \div \frac{17}{24}\)
✓ 1
A1: \(\frac{36}{85}\)
✓ 1
Exam tip: 'More than three tests' means the first three bulbs are all working, so only one product is needed.
137IGMC-048-00-12Single events, relative frequency and expected outcomesPaper EHard6 marks
No calculator
A box contains only black beads and white beads. There are 12 more white beads than black beads. A bead is taken at random. The probability that it is black is \(\frac{2}{7}\).
(a)
Find the total number of beads in the box.
(3)
(b)
k black beads and k white beads are added to the box. The probability of taking a black bead is now \(\frac{3}{8}\). Find k.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(\frac{b}{2b+12} = \frac{2}{7}\)
✓ 1
M1: \(7b = 4b + 24\) leading to \(b = 8\)
✓ 1
A1: 28
✓ 1
Part (b)
M1: \(\frac{8+k}{28+2k} = \frac{3}{8}\)
✓ 1
M1: \(64 + 8k = 84 + 6k\)
✓ 1
A1: \(k = 10\)
✓ 1
Common error: Adding k of each colour increases the total by 2k, not by k.
138IGMC-048-00-13Single events, relative frequency and expected outcomesPaper EHard6 marks
Calculator
A biased spinner can land on red, yellow or green only. The probability of yellow is twice the probability of red. The probability of green is 0.1 more than the probability of yellow.
(a)
Find the probability that the spinner lands on red.
(3)
(b)
Pedro spins the spinner 50 times and it lands on red 15 times. He says this shows the probability in part (a) is wrong. Comment on his statement.
(1)
(c)
The spinner is spun 450 times. How many more times is it expected to land on green than on red?
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(r + 2r + (2r + 0.1) = 1\)
✓ 1
M1: \(5r = 0.9\)
✓ 1
A1: 0.18
✓ 1
Part (b)
B1: relative frequency 0.3 differs from 0.18, but 50 spins is too few to be sure; more spins are needed oe
✓ 1
Part (c)
M1: \(450 \times (0.46 - 0.18)\) or \(207 - 81\)
✓ 1
A1: 126
✓ 1
Exam tip: Write each probability in terms of one unknown, then use the fact that the probabilities of all outcomes add up to 1.
139IGMC-048-00-14Single events, relative frequency and expected outcomesPaper EHard6 marks
Calculator
A bag contains 40 counters. Each counter is red or blue. Lin takes a counter at random, records its colour and replaces it. She does this 250 times and records red 165 times.
(a)
Estimate the number of red counters in the bag. Give your answer as a whole number.
(2)
(b)
Lin adds 10 more red counters to the bag. Use your answer to the previous part to estimate the probability of taking a red counter now.
(2)
(c)
Lin now takes a counter at random, records its colour and replaces it. She does this 300 times. Estimate the number of times she takes a red counter.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1: \(\frac{165}{250} \times 40\) or 26.4 seen
✓ 1
A1: 26
✓ 1
Part (b)
M1: \(\frac{26+10}{40+10}\), ft their whole-number answer from the previous part
✓ 1
A1: 0.72 oe, e.g. \(\frac{18}{25}\)
✓ 1
Part (c)
M1: \(300 \times 0.72\), ft their probability from the previous part
✓ 1
A1: 216
✓ 1
Common error: The number of counters must be a whole number, so round 26.4 to 26 before using it. Also add the 10 extra counters to the total as well as to the red counters: the new total is 50, not 40.
140IGMC3-8-05d-11Combined events, tree diagrams and conditional probabilityPaper EHard6 marks
Calculator
A disease affects 2% of a population. A screening test gives a positive result for 95% of people who have the disease and for 4% of people who do not have the disease. A person is chosen at random and tested.
Tree diagram with the given probabilities
(a)
Find the probability that the person has the disease and tests positive.
(2)
(b)
Find the probability that the person tests positive.
(2)
(c)
The person tests positive. Find the probability that the person has the disease. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.02 × 0.95
✓ 1
A1 0.019
✓ 1
Part (b)
M1 0.98 × 0.04 = 0.0392
✓ 1
A1 0.019 + 0.0392 = 0.0582
✓ 1
Part (c)
M1 0.019 / 0.0582
✓ 1
A1 0.326
✓ 1
Common error: A positive test does not mean the person has the disease; conditional probability divides by the probability of a positive result, not by 1.
141IGMC3-8-05d-13Combined events, tree diagrams and conditional probabilityPaper EHard7 marks
Calculator
In a given year, a river floods with probability 0.08, independently of other years.
(a)
Find the smallest number of years n for which the probability of at least one flood in n years exceeds 0.5.
(3)
(b)
Find the probability of no flood in 3 years. Give your answer to 3 significant figures.
(2)
(c)
Find the probability of at least one flood in 3 years. Use the unrounded value from (b). Give your answer to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 1 − 0.92n > 0.5, so 0.92n < 0.5
✓ 1
M1 evidence of testing values, e.g. 0.928 = 0.513 and 0.929 = 0.472
✓ 1
A1 n = 9
✓ 1
Part (b)
M1 0.923
✓ 1
A1 0.779
✓ 1
Part (c)
M1 1 − 0.923
✓ 1
A1 0.221
✓ 1
Exam tip: "At least one" problems are solved through the complement: 1 minus the probability of none.
142IGMC3-8-05d-14Combined events, tree diagrams and conditional probabilityPaper EHard8 marks
Calculator
A storage rack on a space station holds 5 red-capped vials and n blue-capped vials. Two vials are taken at random, one after the other, without replacement. The probability that both are red is 2/9.
(a)
Show that n2 + 9n − 70 = 0.
(3)
(b)
Solve the equation to find n.
(3)
(c)
Using your value of n, find the probability that the two vials are different colours.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5/(n+5) × 4/(n+4) = 2/9
✓ 1
M1 20 × 9 = 2(n+5)(n+4), so (n+5)(n+4) = 90
✓ 1
A1 n2 + 9n + 20 = 90, so n2 + 9n − 70 = 0
✓ 1
Part (b)
M1 (n + 14)(n − 5) = 0
✓ 1
A1 n = 5 or n = −14
✓ 1
A1 n = 5 as n cannot be negative
✓ 1
Part (c)
M1 2 × 5/10 × 5/9
✓ 1
A1 5/9 = 0.556
✓ 1
Common error: Reject any solution of the quadratic that gives a negative number of vials.
143IGMC3-8-08r15-12Combined events, tree diagrams and conditional probabilityPaper EHard8 marks
No calculator
A bag contains 5 green counters and n yellow counters. Two counters are taken at random without replacement. The probability that both counters are green is 2/11.
(a)
Show that n2 + 9n − 90 = 0.
(3)
(b)
Solve the equation to find the value of n.
(3)
(c)
Find the probability that exactly one of the two counters is green.
M1 factorises (n + 15)(n − 6) = 0 or uses the quadratic formula
✓ 1
A1 n = 6
✓ 1
R1 rejects n = −15 as the number of counters cannot be negative
✓ 1
Part (c)
M1 2 × 5/11 × 6/10
✓ 1
A1 6/11
✓ 1
Common error: After solving a quadratic from a context, always check that each root makes sense.
144IGMC3-8-08r15-14Combined events, tree diagrams and conditional probabilityPaper EHard6 marks
Calculator
In a clothes shop, the probability that a customer tries on a jacket is 0.4. If a customer tries on a jacket, the probability that they buy it is p. If a customer does not try on a jacket, the probability that they buy one is 0.05. The overall probability that a customer buys a jacket is 0.23.
Tree diagram with the given probabilities
(a)
Find the value of p.
(3)
(b)
A customer buys a jacket. Find the probability that the customer tried one on. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.4 × p
✓ 1
M1 0.4p + 0.6 × 0.05 = 0.23
✓ 1
A1 p = 0.5
✓ 1
Part (b)
M1 conditional probability P(tried and bought)/P(bought)
✓ 1
A1 0.2/0.23
✓ 1
A1 0.870
✓ 1
Common error: The overall probability combines both branches of the tree; do not use only the 'tried on' branch.
145IGMC3-8-09r18-07Single events, relative frequency and expected outcomesPaper EHard8 marks
Calculator
At a music school, 60% of the students learn the piano. Of the students who learn the piano, 30% also learn the violin. Of the students who do not learn the piano, 35% learn the violin. A student is chosen at random.
Tree diagram with the given probabilities
(a)
Find the probability that the student learns both the piano and the violin.
(2)
(b)
Find the probability that the student learns the violin.
(2)
(c)
Given that the student learns the violin, find the probability that the student also learns the piano. Give your answer as a fraction.
(2)
(d)
There are 500 students at the school. Estimate the number who learn the violin but do not learn the piano.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.6 × 0.3
✓ 1
A1 0.18
✓ 1
Part (b)
M1 0.6 × 0.3 + 0.4 × 0.35
✓ 1
A1 0.32
✓ 1
Part (c)
M1 0.18/0.32 (their (a) ÷ their (b))
✓ 1
A1 9/16 (= 0.5625)
✓ 1
Part (d)
M1 500 × 0.4 × 0.35
✓ 1
A1 70
✓ 1
Common error: In 'given that' questions the denominator is the probability of the condition (violin), not the whole group of students.
146IGMC3-8-09r18-13Combined events, tree diagrams and conditional probabilityPaper EHard8 marks
No calculator
A baggage hall belt holds 5 red-tagged suitcases and n blue-tagged suitcases. Two suitcases are taken from the belt at random, one after the other, without replacement.
(a)
Show that the probability that both suitcases are red-tagged is \(\frac{20}{(n+5)(n+4)}\).
(2)
(b)
The probability that both suitcases are red-tagged is 5/14. Form and solve an equation to find the value of n.
(4)
(c)
Using your value of n, find the probability that the second suitcase taken is blue-tagged.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5/(n + 5) × 4/(n + 4)
✓ 1
A1 20/((n + 5)(n + 4)) shown
✓ 1
Part (b)
M1 20/((n + 5)(n + 4)) = 5/14
✓ 1
M1 n² + 9n + 20 = 56, so n² + 9n − 36 = 0
✓ 1
M1 (n + 12)(n − 3) = 0 (or correct formula)
✓ 1
A1 n = 3 (n = −12 rejected)
✓ 1
Part (c)
M1 5/8 × 3/7 + 3/8 × 2/7
✓ 1
A1 3/8
✓ 1
Exam tip: Reject solutions that are not possible in context: a number of suitcases cannot be negative.
147IGMC3-8-10r20-12Combined events, tree diagrams and conditional probabilityPaper EHard9 marks
Calculator
A cyclist carries 9 spare inner tubes in a bag: 5 are road tubes and 4 are mountain tubes. Three tubes are taken from the bag at random, one after another, without replacement. Give each answer as a fraction in its simplest form.
(a)
Find the probability that all three tubes are the same type.
(4)
(b)
Find the probability that exactly two of the three tubes are road tubes.
(3)
(c)
The first tube taken is a road tube. Find the probability that all three tubes are the same type.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5/9 × 4/8 × 3/7 (= 60/504)
✓ 1
M1 4/9 × 3/8 × 2/7 (= 24/504)
✓ 1
M1 adds the two cases
✓ 1
A1 84/504 = 1/6
✓ 1
Part (b)
M1 5/9 × 4/8 × 4/7 (= 80/504) for one order
✓ 1
M1 multiplies by 3 for the three orders
✓ 1
A1 240/504 = 10/21
✓ 1
Part (c)
M1 4/8 × 3/7
✓ 1
A1 3/14
✓ 1
Exam tip: After a conditional statement such as 'the first tube is a road tube', redo the counts with that tube already removed from the bag.
148IGMC3-8-10r20-13Combined events, tree diagrams and conditional probabilityPaper EHard8 marks
Calculator
The probability that a day is hot is 0.25. On a hot day, the probability that the cooling system of a data centre fails is p. On a day that is not hot, the probability that it fails is 0.02. The overall probability that the system fails on a given day is 0.04.
Tree diagram with the given probabilities
(a)
Find the value of p.
(3)
(b)
The system fails on a given day. Find the probability that the day was hot. Give your answer as a fraction in its simplest form.
(2)
(c)
Over two independent days, find the probability that the system fails on exactly one of the days.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.25p + 0.75 × 0.02 = 0.04
✓ 1
M1 0.25p = 0.025
✓ 1
A1 p = 0.1
✓ 1
Part (b)
M1 0.25 × 0.1 ÷ 0.04
✓ 1
A1 5/8
✓ 1
Part (c)
M1 0.04 × 0.96
✓ 1
M1 multiplies by 2 for the two orders
✓ 1
A1 0.0768
✓ 1
Exam tip: Write the probabilities on a tree with hot and not hot first; the overall probability is the sum along the branches that end in failure.
149IGMC3-8-10r20-14Combined events, tree diagrams and conditional probabilityPaper EHard7 marks
Calculator
A cyclist enters two independent races. In each race the probability that she wins is p. The probability that she wins exactly one of the two races is 0.42.
(a)
Show that p2 − p + 0.21 = 0.
(2)
(b)
Solve the equation in part (a) to find the two possible values of p.
(3)
(c)
The cyclist is more likely to win a race than lose it. Find the probability that she wins at least one of the two races.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 2p(1 − p) = 0.42
✓ 1
A1 2p − 2p2 = 0.42, rearranged and divided by −2
✓ 1
Part (b)
M1 attempt to factorise or use the quadratic formula
✓ 1
A1 p = 0.3
✓ 1
A1 p = 0.7
✓ 1
Part (c)
M1 chooses p = 0.7 and uses 1 − (1 − 0.7)2
✓ 1
A1 0.91
✓ 1
Exam tip: A quadratic probability equation often has two roots; use the information in the question to reject or choose between them.
150IGMC3-8-12r30-12Combined events, tree diagrams and conditional probabilityPaper EHard8 marks
Calculator
At a wind farm, 70% of days are windy. On a windy day the probability that a turbine's output is above its target is 0.9. On a day that is not windy the probability is 0.2.
Tree diagram with the given probabilities
(a)
Show that the probability that the output is above target on a randomly chosen day is 0.69.
(3)
(b)
Estimate the number of days in a 200-day period on which the output is above target.
(2)
(c)
On one day the output is above target. Find the probability that the day was windy. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.7 × 0.9
✓ 1
M1 0.3 × 0.2
✓ 1
A1 0.63 + 0.06 = 0.69
✓ 1
Part (b)
M1 200 × 0.69
✓ 1
A1 138
✓ 1
Part (c)
M1 0.63 ÷ 0.69
✓ 1
M1 numerator is P(windy and above target)
✓ 1
A1 0.913
✓ 1
Exam tip: For a conditional probability, put P(both events) on top and the probability of the given event underneath.
151IGMC3-8-12r30-13Combined events, tree diagrams and conditional probabilityPaper EHard9 marks
Calculator
In each race a swimmer wins with probability 0.4, independently of other races. The swimmer takes part in three races.
(a)
Find the probability that the swimmer wins all three races.
(1)
(b)
Find the probability that the swimmer wins exactly one race.
(3)
(c)
Find the probability that the swimmer wins at least two races.
(2)
(d)
Given that the swimmer wins at least one race, find the probability that the swimmer wins exactly one race. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.43 = 0.064
✓ 1
Part (b)
M1 0.4 × 0.6 × 0.6
✓ 1
M1 multiply by 3 for the three possible races
✓ 1
A1 0.432
✓ 1
Part (c)
M1 3 × 0.4 × 0.4 × 0.6 = 0.288
✓ 1
A1 0.288 + 0.064 = 0.352
✓ 1
Part (d)
M1 P(at least one) = 1 − 0.63 = 0.784
✓ 1
M1 0.432 ÷ 0.784
✓ 1
A1 0.551
✓ 1
Common error: In 'given that' questions, the denominator is the probability of the condition, not 1.
152IGMC3-8-12r30-14Combined events, tree diagrams and conditional probabilityPaper EHard8 marks
Calculator
A store at a wind farm holds 5 spare blades of type A and n spare blades of type B. Two blades are taken at random without replacement. The probability that both are type A is 2/21.
(a)
Show that n2 + 9n − 190 = 0.
(3)
(b)
Solve the equation to find n.
(3)
(c)
Find the probability that the two blades taken are of different types. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 5/(n + 5) × 4/(n + 4) = 2/21
✓ 1
M1 20 × 21 = 2(n + 5)(n + 4), that is 210 = n2 + 9n + 20
✓ 1
A1 rearranged to n2 + 9n − 190 = 0 with no errors
✓ 1
Part (b)
M1 (n + 19)(n − 10) = 0 or correct formula
✓ 1
A1 n = 10 or −19
✓ 1
A1 n = 10 since the number of blades must be positive
✓ 1
Part (c)
M1 2 × 5/15 × 10/14 (ECF from (b))
✓ 1
A1 0.476
✓ 1
Exam tip: Reject any root that is not a possible whole number of objects.
153IGMC3-8-11r32-12Combined events, tree diagrams and conditional probabilityPaper EHard8 marks
Calculator
A data centre buys hard drives from three suppliers. 50% come from supplier 1, 30% from supplier 2 and 20% from supplier 3. The probability that a drive is defective is 0.01 for supplier 1, 0.02 for supplier 2 and 0.05 for supplier 3. A drive is chosen at random.
Tree diagram with the given probabilities
(a)
Show that the probability that the drive is defective is 0.021.
(2)
(b)
The drive is defective. Find the probability that it came from supplier 3. Give your answer correct to 3 significant figures.
(3)
(c)
Two drives are chosen at random and independently. Find the probability that exactly one is defective. Give your answer correct to 3 significant figures.
Common error: Do not use 0.021 as the numerator for the conditional probability; the numerator is the single branch for the supplier in question.
154IGMC3-8-11r32-13Combined events, tree diagrams and conditional probabilityPaper EHard8 marks
No calculator
A cycling team has n riders, of whom 4 are sprinters. Two riders are chosen at random, without replacement. The probability that both are sprinters is 1/6.
(a)
Show that n2 − n − 72 = 0.
(3)
(b)
Solve the equation to find n.
(3)
(c)
Find the probability that exactly one of the two chosen riders is a sprinter.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 P(first is a sprinter) = 4/n and P(second is a sprinter) = 3/(n − 1)
✓ 1
M1 4/n × 3/(n − 1) = 1/6
✓ 1
A1 72 = n(n − 1), so n2 − n − 72 = 0 (shown)
✓ 1
Part (b)
M1 factorises as (n − 9)(n + 8) = 0 or uses the formula
✓ 1
A1 n = 9
✓ 1
R1 rejects n = −8 because the number of riders must be positive
✓ 1
Part (c)
M1 4/9 × 5/8 × 2 (ECF from (b))
✓ 1
A1 5/9
✓ 1
Exam tip: Express each branch of the tree using the unknown, then form and solve an equation.
155IGMC3-8-11r32-14Combined events, tree diagrams and conditional probabilityPaper EHard8 marks
Calculator
A drive in a data centre fails in its first year with probability 0.03. A drive that survives the first year fails in its second year with probability 0.05. 2000 new drives are installed.
Tree diagram with the given probabilities
(a)
Find the probability that a drive survives both years.
(2)
(b)
Find the probability that a drive fails within the two years.
(1)
(c)
Estimate how many of the 2000 drives fail during the second year.
(2)
(d)
A drive is found to have failed within the two years. Find the probability that it failed in the second year. Give your answer correct to 3 significant figures.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.97 × 0.95
✓ 1
A1 0.9215
✓ 1
Part (b)
B1 1 − 0.9215 = 0.0785 (ECF)
✓ 1
Part (c)
M1 2000 × 0.97 × 0.05
✓ 1
A1 97
✓ 1
Part (d)
M1 0.97 × 0.05 = 0.0485 as the numerator
✓ 1
M1 divides by 0.0785 (ECF)
✓ 1
A1 0.618
✓ 1
Common error: The second-year failure probability applies only to drives that survived year one, so it must be multiplied by 0.97.
156IGMC3-8-11r39-08Combined events, tree diagrams and conditional probabilityPaper EHard9 marks
No calculator
An apiary has 9 hives, 4 of which are infested with mites. An inspector chooses 3 different hives at random, one after another, without replacement. Give each answer as a fraction in its simplest form.
(a)
Find the probability that all 3 hives chosen are infested.
(2)
(b)
Find the probability that exactly one of the 3 hives chosen is infested.
(3)
(c)
Given that at least one of the 3 hives is infested, find the probability that exactly one is infested.
M1 P(at least one) = 1 − \(\frac{5}{42}\) = \(\frac{37}{42}\)
✓ 1
M1 \(\frac{10}{21}\) ÷ \(\frac{37}{42}\) (ECF from (b))
✓ 1
A1 \(\frac{20}{37}\)
✓ 1
Exam tip: For 'at least one' questions, subtracting P(none) from 1 is usually far quicker than adding several cases.
157IGMC3-8-11r39-09Single events, relative frequency and expected outcomesPaper EHard8 marks
Calculator
The probability that driver A starts a Formula 1 race on pole position is 0.4. If A starts on pole, the probability that A wins the race is p. If A does not start on pole, the probability that A wins is 0.15. The probability that A wins a randomly chosen race is 0.31.
Tree diagram with the given probabilities
(a)
Find the value of p.
(3)
(b)
Given that A wins a race, find the probability that A started on pole. Give your answer correct to 3 significant figures.
(3)
(c)
In 50 races, find the expected number in which A starts on pole but does not win.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.4p + 0.6 × 0.15 = 0.31
✓ 1
M1 0.4p = 0.22
✓ 1
A1 p = 0.55
✓ 1
Part (b)
M1 0.4 × 0.55 = 0.22
✓ 1
M1 0.22 ÷ 0.31
✓ 1
A1 0.710
✓ 1
Part (c)
M1 50 × 0.4 × (1 − 0.55) (ECF from (a))
✓ 1
A1 9
✓ 1
Exam tip: Form an equation from the total probability first; the unknown appears only on one branch of the tree.
158IGMC3-8-11r39-13Combined events, tree diagrams and conditional probabilityPaper EHard8 marks
Calculator
A beekeeper inspects a hive each day looking for queen cells. On any day the probability of finding a queen cell is 0.2, independently of other days.
(a)
Find the probability that the first queen cell is found on day 3.
(2)
(b)
The beekeeper wants the probability of finding at least one queen cell in n days to exceed 0.95. Find the smallest value of n.
(3)
(c)
Given that no queen cell is found on days 1 and 2, find the probability that the first queen cell is found on day 5.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 0.8 × 0.8 × 0.2
✓ 1
A1 0.128
✓ 1
Part (b)
M1 1 − 0.8n > 0.95 or 0.8n < 0.05
✓ 1
M1 trial of values or use of logarithms (0.813 = 0.055 and 0.814 = 0.044)
✓ 1
A1 n = 14
✓ 1
Part (c)
M1 P(none on days 1 and 2) = 0.82 = 0.64
✓ 1
M1 P(first found on day 5) = 0.84 × 0.2 = 0.08192, divided by 0.64
✓ 1
A1 0.128
✓ 1
Exam tip: For 'at least one' over several independent trials, write the complement 1 − (probability of none)ⁿ and solve the inequality.
159IGMC3-8-12r40-06Combined events, tree diagrams and conditional probabilityPaper EHard6 marks
No calculator
A crate holds 10 apples: 6 Gala and 4 Fuji. A farmer takes two apples at random, one after the other, without putting the first back.
(a)
Find the probability that both apples are Gala. Give your answer as a fraction in its simplest form.
(2)
(b)
Find the probability that the two apples are different varieties. Give your answer as a fraction in its simplest form.
(2)
(c)
Given that the two apples are different varieties, find the probability that the first apple was Gala.
M1 \(\frac{6}{10}\times\frac49\div\frac{8}{15}\) (conditional probability: favourable ÷ their answer to (b))
✓ 1
A1 \(\frac12\)
✓ 1
Common error: Keeping the denominators at 10 and 10. When the first apple is not replaced, the second probability has denominator 9.
160IGMC3-8-12r40-10Combined events, tree diagrams and conditional probabilityPaper EHard6 marks
Calculator
A farmer tries to start an old tractor, with at most three attempts. The probability that it starts on the first attempt is 0.7. If it fails on the first attempt, the probability that it starts on the second attempt is 0.5. If it fails on the first two attempts, the probability that it starts on the third attempt is 0.3. The farmer stops as soon as the tractor starts, or after three failed attempts.
Tree diagram with the given probabilities (attempts 1, 2 and 3)
(a)
Find the probability that the tractor starts on the second attempt.
(2)
(b)
Find the probability that the tractor starts within the three attempts.
(2)
(c)
Given that the tractor starts within the three attempts, find the probability that it started on the first attempt. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 \(0.3\times0.5\)
✓ 1
A1 0.15
✓ 1
Part (b)
M1 \(1-0.3\times0.5\times0.7\) or \(0.7+0.15+0.3\times0.5\times0.3\)
✓ 1
A1 0.895
✓ 1
Part (c)
M1 \(\frac{0.7}{0.895}\) (ECF from (b))
✓ 1
A1 0.782
✓ 1
Common error: Using 0.5 and 0.3 as the probabilities of failing. They are the probabilities of starting at those attempts, so the failure probabilities are 0.5 and 0.7.
161IGMC3-8-11r41-12Combined events, tree diagrams and conditional probabilityPaper EHard11 marks
Calculator
A signal passes through three relay stations in series. Each relay works independently with probability 0.9. The signal arrives only if all the relays it passes through work.
(a)
Find the probability that all three relays work.
(1)
(b)
Find the probability that exactly one of the three relays fails.
(3)
(c)
A backup rule says that the network is usable if at least two of the three relays work. Find the probability that the network is usable.
(2)
(d)
A longer chain has n relays in series, each working with probability 0.9. Find the largest value of n for which the probability that all n relays work is greater than 0.5.
(3)
(e)
For the chain with this value of n, find the probability that exactly one relay fails. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
B1 0.93 = 0.729
✓ 1
Part (b)
M1 0.9 × 0.9 × 0.1
✓ 1
M1 multiplies by 3 for the three possible positions of the failed relay
✓ 1
A1 0.243
✓ 1
Part (c)
M1 adds P(all work) and P(exactly two work)
✓ 1
A1 0.729 + 0.243 = 0.972 (ECF from (a) and (b))
✓ 1
Part (d)
M1 0.9n > 0.5
✓ 1
M1 tests values (0.96 = 0.531 and 0.97 = 0.478) or uses logarithms (n < 6.58)
✓ 1
A1 n = 6
✓ 1
Part (e)
M1 6 × 0.95 × 0.1 (using n = 6 from (d))
✓ 1
A1 0.354
✓ 1
Exam tip: For 'exactly one' among several identical trials, count the number of positions the single outcome can take and multiply by that number.
162IGMC3-8-11r47-12Combined events, tree diagrams and conditional probabilityPaper EHard7 marks
Calculator
Three depots A, B and C handle 50%, 30% and 20% of a firm's parcels respectively. The probability that a parcel is late is 0.04 for depot A, 0.10 for depot B and 0.15 for depot C.
Tree diagram with the given probabilities
(a)
Find the probability that a parcel chosen at random is late.
(3)
(b)
A late parcel is chosen. Find the probability that it was handled by depot B.
(2)
(c)
Two parcels are chosen at random. Assume that whether one is late is independent of the other. Find the probability that exactly one is late. Give your answer correct to 3 significant figures.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
M1 any one correct product from 0.5 × 0.04 = 0.02, 0.3 × 0.10 = 0.03, 0.2 × 0.15 = 0.03
✓ 1
M1 sum of all three products
✓ 1
A1 0.08
✓ 1
Part (b)
M1 0.03 ÷ 0.08
✓ 1
A1 0.375
✓ 1
Part (c)
M1 2 × 0.08 × 0.92
✓ 1
A1 0.1472 = 0.147
✓ 1
Exam tip: Write the three branch products down separately before adding; it makes the later conditional step much quicker.
163IGMC3-8-11r47-14Combined events, tree diagrams and conditional probabilityPaper EHard8 marks
No calculator
A bag contains 5 red beads and n blue beads. Two beads are taken at random, without replacement. The probability that both are red is 5/18.
(a)
Show that n2 + 9n − 52 = 0.
(3)
(b)
Solve the equation to find the value of n.
(3)
(c)
Find the probability that the two beads are one red and one blue. Give your answer as a fraction in its simplest form.
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