Edexcel International A Level Maths · S1 · Statistics 1
Probability
28 free, high-quality past-paper-style questions on Probability for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 246 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
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Spinner P is fair and has three equal sectors numbered 1, 2 and 3. Spinner Q is fair and has five equal sectors numbered 1, 2, 3, 4 and 5. Both spinners are spun once and the score is the positive difference between the two numbers (for example, 2 and 5 give a score of 3, and 3 and 3 give a score of 0).
(a)
Find the probability that the score is 0.
(2)
(b)
Find the probability that the score is at least 2.
(2)
(c)
Given that the score is not 0, find the probability that the score is 1.
(3)
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Answer
Mark
Notes
Part (a)
sample space of 3 × 5 = 15 equally likely outcomes, e.g. a grid of differences
M1
3 outcomes give 0 (1,1; 2,2; 3,3), so 3/15 = 1/5
A1
Part (b)
counts outcomes with difference 2, 3 or 4: (1,3), (1,4), (1,5), (2,4), (2,5), (3,1), (3,5)
Exam tip: Draw the 3 × 5 grid of scores first. Every probability in this question is then a count of cells; for part (c) the 12 non-zero cells become the new total.
A guest staying at a hotel is chosen at random. G is the event that the guest uses the gym and W is the event that the guest uses the swimming pool. P(G) = 0.55, P(W) = 0.35 and the probability that the guest uses neither the gym nor the pool is 0.25.
(a)
Find P(G ∪ W).
(1)
(b)
Find P(G ∩ W).
(2)
(c)
Find the probability that the guest uses exactly one of the gym and the pool.
(2)
(d)
Find the probability that the guest uses the pool, given that the guest uses the gym.
Each time Rhea throws a dart, the probability that she hits the bullseye is 0.2, independently of all other throws. She throws three darts.
Figure: tree diagram for the three darts (B = hits the bullseye)
(a)
Find the probability that she does not hit the bullseye with any of the three darts.
(2)
(b)
Find the probability that she hits the bullseye with exactly one of the three darts.
(3)
(c)
Find the probability that she hits the bullseye with at least two of the three darts.
(2)
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Answer
Mark
Notes
Part (a)
multiplies three probabilities of missing, 0.8 × 0.8 × 0.8
M1
0.512
A1
Part (b)
one path, e.g. hit, miss, miss: 0.2 × 0.8 × 0.8 = 0.128
M1
three such paths (the hit can be on any of the three throws), × 3
M1
0.384
A1
Part (c)
1 − 0.512 − 0.384
M1
0.104 ft on their (a) and (b)
A1
Exam tip: A three-stage tree has 8 paths. You do not need to draw all of them: write one path, multiply, then count how many paths have the same probability.
A day is chosen at random. E is the event that a small bakery sells all of its bread by closing time and F is the event that it rains during the day. P(E) = 0.6, P(F) = 0.25 and P(E ∩ F) = 0.15.
(a)
Show that E and F are independent.
(2)
(b)
Find P(E ∪ F).
(2)
(c)
Find P(E | F′) and explain how your answer relates to part (a).
(2)
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Answer
Mark
Notes
Part (a)
P(E) × P(F) = 0.6 × 0.25 = 0.15
M1
= P(E ∩ F), so E and F are independent
A1
Part (b)
0.6 + 0.25 − 0.15
M1
0.7
A1
Part (c)
P(E ∩ F′) ÷ P(F′) = (0.6 − 0.15) ÷ 0.75 = 0.6
B1
this equals P(E): since E and F are independent, so are E and F′, and knowing it did not rain does not change the probability
B1
Model sketch
Figure: Venn diagram of the probabilities
Exam tip: In a “show that” independence question, write both numbers — 0.6 × 0.25 = 0.15 and P(E ∩ F) = 0.15 — and finish with a sentence. A bare “they are equal” is not enough.
Figure 1 shows two spinners, A and B. When a spinner is spun, the pointer is equally likely to stop at any angle, so the probability of each number is proportional to the angle of its sector (the angles are marked in Figure 1). Both spinners are spun once, independently, and the score is the sum of the two numbers shown.
(a)
Write down the probability that spinner A shows 2.
(1)
(b)
Find the probability that the score is 5.
(3)
(c)
Given that the score is 5, find the probability that spinner A shows 1.
(2)
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Answer
Mark
Notes
Part (a)
120/360 = 1/3
B1
Part (b)
identifies the two outcomes giving 5: A = 1 with B = 4, and A = 3 with B = 2
Common error: Treating each number as equally likely because each sector carries one number. Here the probabilities come from the angles: 150°, 120° and 90° out of 360° on spinner A.
Figure 1 shows 20 tokens in a bag. Each token is a circle, a square or a triangle, and each token is either navy or gold: navy tokens are solid and marked N, and gold tokens are hatched and marked G. One token is taken from the bag at random. N is the event that the token is navy and C is the event that the token is a circle.
(a)
Find the probability that the token is a triangle.
(1)
(b)
Find P(N ∪ C).
(2)
(c)
Given that the token is not a circle, find the probability that it is navy.
(2)
(d)
Show that the events N and C are independent.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
6/20 = 3/10
B1
Part (b)
12/20 + 10/20 − 6/20, or counts 16 tokens that are navy or a circle or both
M1
= 16/20 = 4/5
A1
Part (c)
10 tokens are not circles, of which 6 are navy (2 squares, 4 triangles)
M1
6/10 = 3/5
A1
Part (d)
P(N) × P(C) = 12/20 × 10/20 = 0.3
M1
P(N ∩ C) = 6/20 = 0.3, equal, so N and C are independent
A1
Exam tip: Count from the figure into a small two-way table (colour × shape) first and check that the six counts add to 20. Every part is then read from the table instead of recounting the picture.
500 runners completed a marathon. Each runner was recorded as having trained with a running club or not, and as finishing in under 4 hours or not. The results are shown in the frequency tree in Figure 1, in which three of the numbers have been replaced by a, b and c.
(a)
Find the values of a, b and c.
(2)
(b)
One of the runners is chosen at random. Find the probability that the runner finished in under 4 hours.
(2)
(c)
Given that the chosen runner finished in under 4 hours, find the probability that the runner trained with a club. Give your answer to 3 significant figures.
(2)
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Answer
Mark
Notes
Part (a)
a = 500 − 180 = 320 and b = 180 − 117 = 63
B1
c = 320 − 144 = 176 ft on their a
B1
Part (b)
117 + 144 = 261
M1
261/500 = 0.522
A1
Part (c)
117 ÷ 261 ft on their 261
M1
= 13/29 = 0.448
A1
Exam tip: In a frequency tree each pair of branches must add up to the number they come from. For the conditional probability, the “given” group (261 runners) becomes the new total.
A visitor to a library is chosen at random. F is the event that the visitor borrows a fiction book and C is the event that the visitor uses a library computer. The Venn diagram in Figure 1 shows the probabilities of the four regions, where x and y are probabilities. P(F) = 0.45.
(a)
Find the value of x and the value of y.
(3)
(b)
Find the probability that the visitor uses a library computer, given that the visitor does not borrow a fiction book. Give your answer to 3 significant figures.
(2)
(c)
Determine whether the events F and C are independent. Show your working.
(2)
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Answer
Mark
Notes
Part (a)
x = 0.45 − 0.32 = 0.13
B1
uses total probability 1: 0.32 + 0.13 + y + 0.26 = 1
Common error: Using 0.32 as P(F). In a Venn diagram the number inside a circle but outside the overlap is “F only”; P(F) is the whole circle, 0.32 + x.
A drawer contains 12 socks: 5 black, 4 grey and 3 white. In the dark, Tom takes two socks from the drawer at random, one after the other, without replacement.
Figure: tree diagram for the two socks (Bl = black, G = grey, W = white)
(a)
Find the probability that the two socks are the same colour.
(3)
(b)
Find the probability that at least one of the socks is white.
(2)
(c)
Given that the two socks are different colours, find the probability that the first sock is black.
(3)
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Answer
Mark
Notes
Part (a)
at least one correct product with denominators 12 and 11, e.g. 5/12 × 4/11
M1
adds the three same-colour paths: 20/132 + 12/132 + 6/132
M1
= 38/132 = 19/66
A1
Part (b)
1 − P(neither white) = 1 − 9/12 × 8/11
M1
= 1 − 72/132 = 5/11
A1
Part (c)
P(different) = 1 − 19/66 = 47/66 ft
B1
P(first black and different) = 5/12 × 7/11 = 35/132
M1
(35/132) ÷ (94/132) = 35/94
A1
Common error: Using 4/12 for the second grey sock after a grey has been taken. Without replacement, both the number of grey socks and the total go down by one: 3/11.
400 commuters were asked how they usually travel to work and whether they are satisfied with their journey. The results are shown below.
Satisfied
Not satisfied
Car
98
72
Train
96
54
Bicycle
62
18
One of these commuters is chosen at random. T is the event that the commuter travels by train and S is the event that the commuter is satisfied.
(a)
Find the probability that the commuter travels by bicycle.
(1)
(b)
Find the probability that the commuter is satisfied, given that the commuter travels by car. Give your answer to 3 significant figures.
(2)
(c)
Find P(T | S′).
(2)
(d)
Determine, showing your working, whether the events T and S are independent.
(3)
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Answer
Mark
Notes
Part (a)
80/400 = 0.2
B1
Part (b)
98 ÷ (98 + 72) = 98/170
M1
= 49/85 = 0.576
A1
Part (c)
n(S′) = 72 + 54 + 18 = 144
B1
54/144 = 0.375
B1
Part (d)
P(T) = 150/400 = 0.375 and P(S) = 256/400 = 0.64
B1
P(T) × P(S) = 0.24 and P(T ∩ S) = 96/400 = 0.24
M1
equal, so T and S are independent
A1
Exam tip: Add a total row and a total column to the table before answering anything. Every probability is then a cell, row total or column total divided by the right total.
A student at a college is chosen at random. A, B and C are the events that the student studies Art, Biology and Chemistry respectively. Because of the timetable, no student studies both Biology and Chemistry. The events A and B are independent. P(A) = 0.5, P(B) = 0.3, P(C) = 0.25 and P(A ∩ C) = x. The probability that the student studies none of the three subjects is 0.2.
(a)
Find P(A ∩ B).
(1)
(b)
Show that x = 0.1.
(3)
(c)
Draw a Venn diagram showing the probabilities for the events A, B and C.
(2)
(d)
Find the probability that the student studies Art, given that the student studies Biology or Chemistry.
(2)
(e)
Determine whether A and C are independent.
(2)
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Answer
Mark
Notes
Part (a)
0.5 × 0.3 = 0.15
B1
Part (b)
P(A ∪ B ∪ C) = 1 − 0.2 = 0.8
B1
B and C do not overlap, so P(A ∪ B ∪ C) = 0.5 + (0.3 − 0.15) + (0.25 − x)
M1
0.9 − x = 0.8, so x = 0.1
A1
Part (c)
circles B and C drawn not overlapping, each overlapping A
B1
values: A only 0.25, A ∩ B 0.15, B only 0.15, A ∩ C 0.1, C only 0.15, outside 0.2
Figure: Venn diagram for A, B and C (B and C do not overlap)
Exam tip: Two pieces of information decide the shape of the Venn diagram before any numbers go in: “no student studies both” means the B and C circles do not overlap, and independence gives the A ∩ B region by multiplying.
An online shop sends 70% of its orders by standard delivery and the rest by express delivery. An order sent by standard delivery arrives late with probability 0.15; an order sent by express delivery arrives late with probability 0.04. An order is chosen at random.
Figure: tree diagram for the delivery of an order
(a)
Find the probability that the order arrives late.
(3)
(b)
Given that the order arrives late, find the probability that it was sent by express delivery. Give your answer to 3 significant figures.
(3)
(c)
Two orders are chosen at random, independently. Find the probability that exactly one of them arrives late. Give your answer to 3 significant figures.
(3)
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Answer
Mark
Notes
Part (a)
tree diagram with first branches 0.7, 0.3 and second branches (late) 0.15, 0.04
B1
0.7 × 0.15 + 0.3 × 0.04
M1
= 0.105 + 0.012 = 0.117
A1
Part (b)
numerator 0.3 × 0.04 = 0.012
M1
divided by their 0.117
M1
= 0.10256… = 0.103
A1
Part (c)
P(not late) = 1 − 0.117 = 0.883 ft
B1
2 × 0.117 × 0.883
M1
= 0.206622 = 0.207
A1
Common error: Answering (b) with 0.04. That is P(late | express); the question gives the lateness and asks for the delivery type, so divide one path by the total probability of being late.
A fair red die and a fair blue die, each numbered 1 to 6, are rolled together. A is the event that the red die shows 4, B is the event that the total of the two scores is 7, and C is the event that the total is at least 9.
(a)
Find P(C).
(2)
(b)
Show that A and B are independent.
(3)
(c)
State, with a reason, whether B and C are mutually exclusive.
(1)
(d)
Find P(A | C).
(2)
(e)
Hence state whether A and C are independent, giving a reason.
yes: a total cannot be both 7 and at least 9, so P(B ∩ C) = 0
B1
Part (d)
A ∩ C: red 4 with blue 5 or 6, 2 outcomes
M1
2/10 = 1/5
A1
Part (e)
P(A | C) = 1/5 ≠ 1/6 = P(A), so not independent ft on their (d)
B1
Exam tip: With two dice, draw the 6 × 6 grid of totals. Conditional probabilities are then counts inside the given event: for P(A | C) the 10 cells of C are the whole sample space.
An email arriving at an office is chosen at random. L is the event that the email contains a link and J is the event that the email is sent to the junk folder. P(L) = 0.4 and P(J | L) = 0.35. Of the emails that do not contain a link, a proportion p are sent to the junk folder. P(J) = 0.29.
Figure: tree diagram for an email
(a)
Find P(L ∩ J).
(1)
(b)
Find the value of p.
(3)
(c)
Find the probability that an email sent to the junk folder contains a link. Give your answer to 3 significant figures.
(2)
(d)
Find the probability that an email neither contains a link nor is sent to the junk folder.
(2)
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Answer
Mark
Notes
Part (a)
0.4 × 0.35 = 0.14
B1
Part (b)
P(J) = P(L ∩ J) + P(L′) × p
M1
0.14 + 0.6p = 0.29
A1
p = 0.25
A1
Part (c)
P(L | J) = 0.14 ÷ 0.29
M1
= 14/29 = 0.483
A1
Part (d)
0.6 × (1 − p) or 1 − (0.4 + 0.29 − 0.14)
M1
= 0.45 ft on their p
A1
Exam tip: Sketch the tree even when the question does not ask for one: L / L′ first, then J / J′. The unknown p sits on a branch, and P(J) is the sum of the two paths that end in J.
A repair shop records the fault type of each of the last 200 phones it repaired. Every phone was made by brand X or brand Y. The stacked bar chart in Figure 1 shows the numbers. One of these 200 phones is chosen at random.
(a)
Find the probability that the phone is a brand Y phone with a battery fault.
(1)
(b)
Find the probability that the phone has a screen fault, given that it is a brand X phone.
(2)
(c)
Find the probability that the phone is a brand Y phone, given that it does not have a screen fault.
(2)
(d)
Show that the event “the phone is brand X” is independent of the event “the phone has an other fault”, but is not independent of the event “the phone has a screen fault”.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
40/200 = 1/5
B1
Part (b)
brand X total = 50 + 30 + 20 = 100
M1
50/100 = 1/2
A1
Part (c)
200 − 90 = 110 phones without a screen fault, of which 40 + 20 = 60 are brand Y
0.5 × 0.45 = 0.225 ≠ 50/200 = 0.25 = P(X ∩ screen), so not independent
A1
Exam tip: Read each block of a stacked bar as a difference of two readings: the brand Y block for screen faults runs from 50 up to 90, so it represents 40 phones, not 90.
A gardener plants seeds of one variety of flower. Each seed either germinates (G) or does not; a seed that does not germinate cannot flower. A seed that germinates either flowers (F) or does not. The tree diagram in Figure 1 shows the probabilities, where a and b are probabilities.
(a)
Write down the value of a and the value of b.
(2)
(b)
Find the probability that a seed flowers.
(1)
(c)
Two seeds are planted and behave independently. Find the probability that at least one of them flowers.
(2)
(d)
Given that a seed does not flower, find the probability that it germinated. Give your answer to 3 significant figures.
(3)
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Answer
Mark
Notes
Part (a)
a = 0.4
B1
b = 0.15
B1
Part (b)
0.85 × 0.6 = 0.51
B1
Part (c)
1 − (1 − 0.51)2 = 1 − 0.492 ft on their (b)
M1
= 0.7599
A1
Part (d)
P(G ∩ F′) = 0.85 × 0.4 = 0.34
M1
divides by P(does not flower) = 0.34 + 0.15 = 0.49
M1
0.34 ÷ 0.49 = 34/49 = 0.694
A1
Common error: Forgetting the seeds that never germinate in part (d). “Does not flower” has two paths: germinates but does not flower (0.34) and does not germinate at all (0.15).
A student at a school is chosen at random. Each student travels to school by exactly one of walking, bus or cycling. In Figure 1, the unit square is divided into three columns whose widths are the probabilities of each travel mode. In each column, the height of the shaded part is the conditional probability that a student using that mode arrives before 08:30, given that mode. E is the event that the chosen student arrives before 08:30.
(a)
Use Figure 1 to write down the probability that the student cycles and the probability that the student arrives before 08:30 given that the student travels by bus.
(2)
(b)
Find P(E).
(2)
(c)
Given that the student does not arrive before 08:30, find the probability that the student travels by bus.
(3)
(d)
State, with a reason based on Figure 1, whether E is independent of the event “the student travels by bus”.
not independent: P(E | bus) = 0.6, but P(E) = 0.72 (the shaded heights are not all equal)
B1
Exam tip: In this kind of diagram each shaded rectangle is a path of a tree diagram: width × height = P(mode) × P(E | mode). Adding the shaded areas gives P(E).
Figure 1 shows a system of one-way tunnels in a laboratory maze. A mouse is released at S. At every junction (S, A, B and C) the mouse chooses one of the tunnels leading away from that junction at random, each being equally likely, independently of earlier choices. The mouse moves only in the direction of the arrows and stops when it reaches one of the exits X, Y or Z.
(a)
Find the probability that the mouse passes through junction C.
(2)
(b)
Show that the probability that the mouse leaves by exit X is 5/12.
(2)
(c)
Given that the mouse leaves by exit Z, find the probability that it passed through junction C.
(3)
(d)
Two mice are released one after the other and behave independently. Find the probability that they leave by different exits.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
two routes to C: S→A→C and S→B→C, each ½ × ½
M1
¼ + ¼ = 1/2
A1
Part (b)
direct route S→A→X: ½ × ½ = ¼, and via C: ½ × ⅓ = 1/6
M1
¼ + 1/6 = 3/12 + 2/12 = 5/12
A1
Part (c)
P(Z) = P(S→B→Z) + P(via C, then Z) = ¼ + ½ × ⅓ = 5/12
Exam tip: Turn the maze into a tree: at each junction the branches are the arrows leaving it, each with probability 1 ÷ (number of arrows out). Then list every route to the exit you want and add the route probabilities.
Figure 1 shows a signal system between P and Q. The signal passes from P to Q if component A works and at least one of components B and C works. The number under each component is the probability that it works. The components work or fail independently of each other.
(a)
Find the probability that both B and C fail.
(1)
(b)
Show that the probability that the signal passes from P to Q is 0.893.
(2)
(c)
Given that the signal passes, find the probability that both B and C work. Give your answer to 3 significant figures.
(3)
(d)
Given that the signal does not pass, find the probability that component A failed. Give your answer to 3 significant figures.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
0.2 × 0.3 = 0.06
B1
Part (b)
P(at least one of B, C works) = 1 − 0.06 = 0.94
M1
0.95 × 0.94 = 0.893
A1
Part (c)
P(A, B and C all work) = 0.95 × 0.8 × 0.7 = 0.532
M1
divides by 0.893
M1
= 0.59574… = 0.596
A1
Part (d)
P(signal does not pass) = 1 − 0.893 = 0.107
B1
if A fails the signal cannot pass, so P(A fails ∩ no signal) = 0.05
M1
0.05 ÷ 0.107 = 0.46728… = 0.467
A1
Exam tip: For components side by side (in parallel), work with failure: P(at least one works) = 1 − P(all fail). For components one after another (in series), multiply the probabilities of working.
Figure 1 shows seven cards, each with a number on it. Two of the cards are taken at random, one after the other, without replacement.
(a)
Find the probability that both cards show odd numbers.
(2)
(b)
Find the probability that the sum of the two numbers is even.
(2)
(c)
Given that the sum of the two numbers is even, find the probability that both cards show odd numbers.
(2)
(d)
Explain why the events “the sum is 7” and “both cards show odd numbers” are mutually exclusive.
(1)
(e)
Find the probability that the second card shows an odd number.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
4 odd cards (1, 3, 5, 9): 4/7 × 3/6
M1
= 2/7
A1
Part (b)
even sum needs both odd or both even: 2/7 + 3/7 × 2/6
M1
= 2/7 + 1/7 = 3/7
A1
Part (c)
(2/7) ÷ (3/7) ft on their (a) and (b)
M1
= 2/3
A1
Part (d)
the sum of two odd numbers is even, so it can never be 7; the events cannot happen together
B1
Part (e)
odd, odd + even, odd: 4/7 × 3/6 + 3/7 × 4/6
M1
= 12/42 + 12/42 = 4/7
A1
Exam tip: Part (e) is the same as the probability that the first card is odd. Without replacement, every position in the order is equally likely to hold any particular card — but you still need the two-path working to show it.
A box contains 15 chocolates that all look the same. 3 of the chocolates contain nuts and the other 12 do not. Sam takes 3 chocolates from the box at random, one after another, without replacement.
(a)
Find the probability that none of Sam’s chocolates contains nuts.
(2)
(b)
Show that the probability that exactly one of Sam’s chocolates contains nuts is 198/455.
(3)
(c)
Find the probability that at least two of Sam’s chocolates contain nuts.
(2)
(d)
Given that at least one of Sam’s chocolates contains nuts, find the probability that exactly one does.
(3)
(e)
If instead each chocolate were replaced before the next was taken, the probability of no nuts would be (12/15)3. Find this probability and explain why it is greater than your answer to part (a).
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
12/15 × 11/14 × 10/13
M1
= 44/91
A1
Part (b)
one order, e.g. nut first: 3/15 × 12/14 × 11/13 = 396/2730
M1
three orders, each with the same probability, so × 3
M1
1188/2730 = 198/455
A1
Part (c)
1 − 44/91 − 198/455
M1
= 37/455
A1
Part (d)
P(at least one) = 1 − 44/91 = 47/91
B1
(198/455) ÷ (47/91)
M1
= 198/235
A1
Part (e)
0.83 = 0.512
B1
without replacement, each nut-free chocolate taken lowers the proportion of nut-free chocolates left (11/14, then 10/13, both less than 12/15), so the product is smaller
B1
Exam tip: For “exactly one” without replacement, the three orders (nut first, second or third) contain the same numbers in different places, so they have equal probabilities. Work out one and multiply by 3, but say why.
Mira plays three points of table tennis against a practice partner. The probability that she wins the first point is 0.6. For each later point, the probability that she wins it is 0.7 if she won the previous point and 0.4 if she lost the previous point. W1 and W2 are the events that she wins the first point and the second point respectively.
Figure: tree diagram for the three points (W = wins the point, L = loses it)
(a)
Find the probability that she wins all three points.
(2)
(b)
Find the probability that she wins exactly two of the three points.
(3)
(c)
Find P(W2).
(2)
(d)
Show that W1 and W2 are not independent.
(2)
(e)
Given that she wins exactly two of the three points, find the probability that she won the first point. Give your answer to 3 significant figures.
P(W1) × P(W2) = 0.6 × 0.58 = 0.348 ≠ 0.42, so not independent ft on their (c)
B1
Part (e)
paths in “exactly two” starting with a win: 0.126 + 0.072 = 0.198
M1
divided by their 0.31
M1
= 0.6387… = 0.639
A1
Common error: Treating the points as independent and using 0.6 for every point. The second and third sets of branches depend on the result of the point before, so each path must follow the tree.
A gym has 150 members. Each member was asked whether they attend yoga (Y), Pilates (P) or spin (S) classes. 12 attend all three; 20 attend yoga and Pilates but not spin; no member attends both Pilates and spin without also attending yoga; x attend yoga and spin but not Pilates; 30 attend yoga only; y attend Pilates only; 25 attend spin only; 16 attend none of the three. In total, 50 members attend Pilates.
Figure: Venn diagram of the numbers of members (150 in total)
(a)
Find the values of x and y.
(3)
(b)
A member is chosen at random. Find the probability that the member attends exactly two of the three classes.
(2)
(c)
Find the probability that a randomly chosen member who attends yoga also attends spin.
(2)
(d)
Find the probability that a randomly chosen member attends yoga, given that the member attends Pilates or spin.
(2)
(e)
Determine whether the events “attends Pilates” and “attends spin” are independent.
P(P ∩ S) = 12/150 = 0.08, which is different, so not independent
A1
Exam tip: Write a 0 in any region the question says is empty (here P ∩ S without Y). It stops you leaving a gap and makes the eight-region total easy to check.
A workshop has two machines, X and Y, which break down independently of each other. In any week, the probability that X breaks down is p and the probability that Y breaks down is p + 0.2. The probability that at least one of the machines breaks down in a week is 0.76.
(a)
Show that p2 − 1.8p + 0.56 = 0.
(3)
(b)
Solve the equation to find p, explaining why one solution must be rejected.
(2)
(c)
Find the probability that exactly one of the machines breaks down in a given week.
(2)
(d)
Given that at least one machine breaks down in a particular week, find the probability that both do.
(2)
(e)
Explain why the events “X breaks down” and “Y breaks down” cannot be mutually exclusive.
(1)
(f)
Assuming that different weeks are independent, find the probability that at least one machine breaks down in each of two consecutive weeks.
A bag contains n red counters and 6 blue counters, where n is a positive integer. Two counters are taken from the bag at random, one after the other, without replacement. Figure 1 shows an incomplete tree diagram, in which p and q are probabilities. R and B stand for red and blue.
(a)
Write down expressions, in terms of n, for p and for q.
(2)
(b)
Show that the probability that the two counters are different colours is 12n/((n + 6)(n + 5)).
(2)
(c)
The probability that the two counters are different colours is 1/2. There are more red counters than blue counters in the bag. Find the value of n.
(3)
(d)
Find the probability that both counters are red.
(2)
(e)
Given that the two counters are the same colour, find the probability that they are both blue.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
p = 6/(n + 5)
B1
q = 5/(n + 5)
B1
Part (b)
adds the two mixed paths: n/(n + 6) × 6/(n + 5) + 6/(n + 6) × n/(n + 5)
Common error: Writing the second-draw denominators as n + 6. After one counter has been removed the bag holds n + 5 counters, whichever colour was taken first.
A visitor to a festival is chosen at random. M, F and W are the events that the visitor goes to the music stage, the food market and a craft workshop respectively. The Venn diagram in Figure 1 shows the probabilities of the regions, where p and q are probabilities. The events M and W are independent.
(a)
Show that p + q = 0.3.
(2)
(b)
Using the independence of M and W, show that p = 0.14, and find the value of q.
(4)
(c)
Find the probability that the visitor goes to exactly two of the three.
(2)
(d)
Find P(F | W).
(2)
(e)
Determine whether the events F and W are independent.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
the probabilities in all eight regions add to 1
M1
0.20 + 0.12 + 0.10 + 0.04 + 0.06 + 0.18 = 0.70, so p + q = 0.3
A1
Part (b)
P(M) = 0.36 + p and P(W) = q + p + 0.10 = 0.40
B1
P(M ∩ W) = p + 0.06 = (0.36 + p) × 0.40
M1
0.6p = 0.084, so p = 0.14
A1
q = 0.3 − 0.14 = 0.16
A1
Part (c)
0.10 + p + 0.04
M1
= 0.28 ft on their p
A1
Part (d)
(0.04 + 0.06) ÷ 0.40
M1
= 0.25
A1
Part (e)
P(F) = 0.12 + 0.10 + 0.04 + 0.06 = 0.32
B1
P(F | W) = 0.25 ≠ 0.32 = P(F) (or 0.32 × 0.40 = 0.128 ≠ 0.10), so not independent
A1
Exam tip: Write P(M), P(W) and P(M ∩ W) as sums of regions before using independence. Here P(W) turns out not to depend on p and q separately, which makes the equation linear.
In a quiz, a contestant is asked up to three questions. The contestant stops as soon as a question is answered wrongly, and also stops after the third question. Figure 1 shows the tree diagram, where C means a correct answer and W means a wrong answer. The probability that a contestant answers exactly two questions correctly is 0.18.
(a)
Find the probability that a contestant answers exactly one question correctly.
(2)
(b)
Show that x = 0.625.
(3)
(c)
Find the probability that a contestant answers all three questions correctly.
(1)
(d)
Given that a contestant answers at least one question correctly, find the probability that the contestant answers all three correctly.
(2)
(e)
Two contestants take part, independently. Find the probability that they answer the same number of questions correctly.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
correct then wrong: 0.8 × 0.4
M1
= 0.32
A1
Part (b)
exactly two correct means C, C, W: 0.8 × 0.6 × (1 − x)
M1
0.48(1 − x) = 0.18
M1
1 − x = 0.375, so x = 0.625
A1
Part (c)
0.48 × 0.625 = 0.3
B1
Part (d)
P(at least one correct) = 0.8
B1
0.3 ÷ 0.8 = 0.375
A1
Part (e)
probabilities of 0, 1, 2, 3 correct are 0.2, 0.32, 0.18, 0.3 ft
B1
0.22 + 0.322 + 0.182 + 0.32
M1
= 0.04 + 0.1024 + 0.0324 + 0.09 = 0.2648
A1
Exam tip: Check that the four end points of a stopping tree add to 1: here 0.2 + 0.32 + 0.18 + 0.3 = 1. Part (e) then needs all four, so a mistake in one shows up immediately.
A customer at a garden centre is chosen at random. A is the event that the customer buys plants and B is the event that the customer buys garden tools. The Venn diagram in Figure 1 shows the probabilities of the four regions, where x and y are probabilities. The events A and B are independent, and the probability that a customer buys neither plants nor tools is greater than 0.25.
(a)
Show that x2 − 0.5x + 0.06 = 0.
(3)
(b)
Find the value of x and the value of y, explaining your choice.
(3)
(c)
Find P(B | A′) and comment on your answer.
(2)
(d)
A third event C, with P(C) = 0.1, is mutually exclusive with both A and B. Find P(A ∪ B ∪ C) and P(C | B′).
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
P(A) = 0.3 + x and P(B) = 0.2 + x
B1
independence: (0.3 + x)(0.2 + x) = x
M1
0.06 + 0.5x + x2 = x, so x2 − 0.5x + 0.06 = 0
A1
Part (b)
(x − 0.2)(x − 0.3) = 0, so x = 0.2 or 0.3
M1
y = 1 − 0.5 − x gives y = 0.3 or 0.2
M1
y > 0.25, so x = 0.2, y = 0.3
A1
Part (c)
0.2 ÷ (1 − 0.5) = 0.4
B1
equals P(B) = 0.4, as expected because A and B are independent (so A′ and B are too)
B1
Part (d)
P(A ∪ B) + P(C) = 0.7 + 0.1 = 0.8
B1
0.1 ÷ (1 − 0.4) = 1/6
B1
Common error: Writing the independence condition as 0.3 × 0.2 = x. The numbers in the diagram are “A only” and “B only”; independence uses the whole of each circle, P(A) = 0.3 + x.
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