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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 15 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
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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.
11IGMC-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.
12IGMC-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.
13IGMC-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.
14IGMC-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.
15IGMC-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.
16IGMC-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.
17IGMC-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.
18IGMC-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
(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.
19IGMC-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.
(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).
20IGMC-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.
(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.
21IGMC-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.
22IGMC-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.
23IGMC-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.
(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.
24IGMC-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.
25IGMC-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.
26IGMC-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)
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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.
27IGMC-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)
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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.
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