2023 May/June · Paper 53 · Q5 (a, b)7 marksConditional probability · Two-way tables & independence
2022
2022 Oct/Nov · Paper 51 · Q5 (a, b, c, d)10 marksConditional probability · Tree diagrams · Two-way tables & independence
2022 Oct/Nov · Paper 52 · Q1 (a, b)5 marksConditional probability · Tree diagrams
2022 Oct/Nov · Paper 52 · Q3 (a)3 marksCombined events
2022 Oct/Nov · Paper 52 · Q5 (a, b)4 marksCombined events · Conditional probability
2022 Oct/Nov · Paper 53 · Q7 (a, b, c)8 marksConditional probability · Probability without replacement
2022 May/June · Paper 51 · Q6 (a, b, c)8 marksConditional probability · Tree diagrams
2022 May/June · Paper 52 · Q7 (a, b, c)9 marksConditional probability · Probability without replacement
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school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Conditional probability · Core · Paper 5 style · 4 marks
(a)Events A and B are such that P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.15. Find P(A | B) and P(B | A).[3]
(b)Find P(A ∪ B).[1]
Show answer
(a) P(A | B) = 0.3, P(B | A) = 0.375
P(A | B) = 0.15/0.5 = 0.3 M1 A1; P(B | A) = 0.15/0.4 = 0.375 A1
(b) 0.75
0.4 + 0.5 − 0.15 = 0.75 B1
Q2Conditional probability · Core · Paper 5 style · 4 marks
(a)In a group of 200 students at a school in Lahore, 120 study Physics, 90 study Chemistry and 50 study both. A student is chosen at random. Given that the student studies Physics, find the probability that the student also studies Chemistry.[2]
(b)Given that the student does not study Chemistry, find the probability that the student studies Physics.[2]
Show answer
(a) 5/12
50/120 M1; = 5/12 A1
(b) 7/11
Not Chemistry: 110 students, of whom 120 − 50 = 70 study Physics M1; 70/110 = 7/11 A1
Q3Conditional probability · Core · Paper 5 style · 4 marks
(a)Two fair six-sided dice are thrown. Given that the total score is 8, find the probability that at least one of the dice shows a 5.[2]
(b)Given that at least one die shows a 5, find the probability that the total score is 8.[2]
Show answer
(a) 2/5
Outcomes with total 8: (2,6), (3,5), (4,4), (5,3), (6,2) M1; 2 of the 5 contain a 5, so 2/5 A1
(b) 2/11
11 outcomes contain at least one 5 M1; 2 of them total 8, so 2/11 A1
Q4Conditional probability · Standard · Paper 5 style · 5 marks
(a)In a city, 2% of people have a certain infection. A screening test gives a positive result for 95% of people who have the infection and for 4% of people who do not. A person is chosen at random and tested. Find the probability that the result is positive.[2]
(b)Given that the result is positive, find the probability that the person has the infection. Give your answer correct to 3 significant figures.[2]
(c)Comment briefly on what your answer to part (b) shows about using this test on its own.[1]
Show answer
(a) 0.0582
0.02 × 0.95 + 0.98 × 0.04 M1; = 0.0582 A1
(b) 0.326
0.019/0.0582 M1; = 0.326 A1
(c) Most people who test positive do not have the infection, so a positive result needs a further test
B1 for a sensible comment referring to the answer to (b) being less than 0.5
Q5Conditional probability · Standard · Paper 5 style · 6 marks
(a)For events A and B, P(A) = 0.3, P(B | A) = 0.6 and P(B | A′) = 0.2. Find P(B).[2]
Q6Conditional probability · Standard · Paper 5 style · 6 marks
(a)A cricket team has probability 0.5 of winning the toss before a match. If it wins the toss, the probability that it wins the match is 0.65; if it loses the toss, the probability that it wins the match is 0.45. Find the probability that the team wins the match.[2]
(b)Given that the team won the match, find the probability that it won the toss. Give your answer correct to 3 significant figures.[2]
(c)Given that the team lost the match, find the probability that it won the toss.[2]
Show answer
(a) 0.55
0.5 × 0.65 + 0.5 × 0.45 M1; = 0.55 A1
(b) 13/22 ≈ 0.591
0.325/0.55 M1; = 13/22 = 0.591 A1
(c) 7/18 ≈ 0.389
0.5 × 0.35 = 0.175 M1; 0.175/0.45 = 7/18 A1
Q7Conditional probability · Standard · Paper 5 style · 5 marks
(a)A card is chosen at random from 20 cards numbered 1 to 20. A is the event 'the number is a multiple of 3', B is the event 'the number is even' and C is the event 'the number is greater than 15'. Find P(A | B) and hence show that A and B are independent.[3]
(b)Determine whether B and C are independent, justifying your answer.[2]
Show answer
(a) P(A | B) = 3/10 = P(A), so independent
Even numbers: 10, multiples of 3 among them: 6, 12, 18 M1; P(A | B) = 3/10 A1; P(A) = 6/20 = 3/10, equal so independent A1
(b) Not independent: P(C | B) = 3/10 but P(C) = 1/4
Even numbers greater than 15: 16, 18, 20, so P(C | B) = 3/10 M1; P(C) = 5/20 = 1/4 ≠ 3/10, not independent A1
Q8Conditional probability · Challenge · Paper 5 style · 7 marks
(a)A factory in Sialkot makes footballs on three machines. Machine A makes 50% of the balls, B makes 30% and C makes 20%. The proportions of faulty balls are 2% from A, 3% from B and 5% from C. Find the probability that a randomly chosen ball is faulty.[2]
(b)A ball is found to be faulty. Find the probability that it was made by machine C. Give your answer correct to 3 significant figures.[2]
(c)Two balls are chosen at random, independently. Given that at least one of them is faulty, find the probability that both are faulty. Give your answer correct to 3 significant figures.[3]
Q9Conditional probability · Challenge · Paper 5 style · 7 marks
(a)Ali travels to college by bus or by bicycle. The probability that it rains on a given morning is 0.3. If it rains, the probability that he takes the bus is 0.8; if it does not rain, the probability that he takes the bus is 0.4. When he takes the bus, the probability that he is late is 0.25. When he cycles, the probability that he is late is 0.2 if it is raining and 0.05 if it is not. Find the probability that Ali is late on a randomly chosen morning.[3]
(b)Given that Ali is late, find the probability that it was raining. Give your answer correct to 3 significant figures.[2]
(c)Given that Ali is late, find the probability that he cycled. Give your answer correct to 3 significant figures.[2]
(c) Not independent, since P(A) × P(B) = 0.16 ≠ 0.1 = P(A ∩ B)
P(A)P(B) = 0.16 M1; ≠ 0.1 so not independent A1
Q11Tree diagrams · Core · Paper 5 style · 4 marks
(a)Sara takes a driving test. The probability that she passes at the first attempt is 0.6. If she fails, she takes the test a second time, and the probability that she passes at the second attempt is 0.75. Draw a tree diagram to show this information.[2]
(b)Find the probability that Sara passes within two attempts.[2]
Show answer
(a) First branch pass 0.6 / fail 0.4; after fail: pass 0.75 / fail 0.25
Correct structure B1; all probabilities correct B1
(b) 0.9
0.6 + 0.4 × 0.75 M1; = 0.9 A1
Q12Tree diagrams · Core · Paper 5 style · 4 marks
(a)A biased coin, for which P(head) = 0.3, is tossed and a fair six-sided die is thrown. Find the probability of obtaining a head and a six.[1]
(b)Find the probability of obtaining either a head or a six, but not both.[3]
(a)In Islamabad in spring, the probability that it rains on a Monday is 0.4. If it rains on Monday, the probability that it rains on Tuesday is 0.6; if it does not rain on Monday, the probability that it rains on Tuesday is 0.25. Find the probability that it rains on Tuesday.[2]
(b)Find the probability that it rains on exactly one of the two days.[2]
Show answer
(a) 0.39
0.4 × 0.6 + 0.6 × 0.25 M1; = 0.39 A1
(b) 0.31
0.4 × 0.4 + 0.6 × 0.25 M1; = 0.31 A1
Q14Tree diagrams · Standard · Paper 5 style · 6 marks
(a)Hamza plays three games of chess against his sister. The probability that he wins the first game is 0.6. After a game he wins, the probability that he wins the next game is 0.7; after a game he does not win, the probability that he wins the next game is 0.3. Find the probability that Hamza wins at least two of the three games.[4]
(b)Given that Hamza wins at least two of the games, find the probability that he won the first game. Give your answer correct to 3 significant figures.[2]
Show answer
(a) 0.558
WWW 0.294, WWL 0.126, WLW 0.054, LWW 0.084 M1 A1 (any two correct); sum of four M1; = 0.558 A1
(b) 0.849
(0.294 + 0.126 + 0.054)/0.558 M1; = 0.849 A1
Q15Tree diagrams · Standard · Paper 5 style · 5 marks
(a)A fair spinner has three sections labelled 1, 2 and 3. It is spun once and the number shown is the number of balls that are then taken, one at a time with replacement, from a bag containing 3 red balls and 2 blue balls. Find the probability that all the balls taken are red.[3]
(b)Given that all the balls taken are red, find the probability that the spinner showed 1. Give your answer correct to 3 significant figures.[2]
Q16Tree diagrams · Standard · Paper 5 style · 6 marks
(a)In a factory, 8% of items are defective. Each item is checked by a first inspector, who rejects a defective item with probability 0.9. Any defective item that she misses is checked by a second inspector, who rejects it with probability 0.7. Non-defective items are never rejected. Find the probability that a randomly chosen item is rejected.[3]
(b)Given that an item is accepted, find the probability that it is defective. Give your answer correct to 3 significant figures.[3]
Q17Tree diagrams · Challenge · Paper 5 style · 7 marks
(a)Ahmed and Bilal take turns to throw a fair six-sided die, starting with Ahmed. The first person to throw a 6 wins. If neither has thrown a 6 after each has had three throws, the game is a draw. Find the probability that Ahmed wins. Give your answer correct to 3 significant figures.[3]
(b)Find the probability that the game is a draw. Give your answer correct to 3 significant figures.[1]
(c)The rule is changed so that the game continues until someone throws a 6. Show that the probability that Ahmed wins is 6/11.[3]
Show answer
(a) 0.363
Ahmed wins on his 1st, 2nd or 3rd throw: 1/6 + (5/6)²(1/6) + (5/6)⁴(1/6) M1 A1; = 2821/7776 = 0.363 A1
(b) 0.335
(5/6)⁶ = 0.335 B1
(c) 6/11
P = (1/6)(1 + 25/36 + (25/36)² + …) M1; geometric series with r = 25/36: (1/6)/(1 − 25/36) M1; = (1/6)(36/11) = 6/11 A1
Q18Tree diagrams · Challenge · Paper 5 style · 6 marks
(a)Bag A contains 4 red and 6 green balls. Bag B contains 5 red and 3 green balls. A fair coin is tossed: if it shows heads, two balls are taken at random from bag A without replacement; if tails, two balls are taken from bag B without replacement. Find the probability that the two balls are the same colour.[4]
(b)Given that the two balls are the same colour, find the probability that they came from bag A.[2]
Q19Tree diagrams · Challenge · Paper 5 style · 7 marks
(a)Zainab plays two games of squash. The probability that she wins the first game is p. If she wins the first game, the probability that she wins the second is p + 0.3; if she loses the first game, the probability that she wins the second is ½p. The probability that she wins both games is 0.18. Find p.[3]
(b)Find the probability that she wins exactly one of the two games.[2]
(c)Given that she wins exactly one game, find the probability that it was the first game.[2]
Q20Combined events · Core · Paper 5 style · 4 marks
(a)Events A and B have P(A) = 0.35 and P(B) = 0.45. Given that A and B are mutually exclusive, find P(A ∪ B) and the probability that neither A nor B occurs.[2]
(b)Given instead that A and B are independent, find P(A ∪ B).[2]
(b) Independent, since P(A) × P(B) = 0.2 = P(A ∩ B)
0.5 × 0.4 = 0.2 M1; equal so independent A1
(c) 0.5
P(A ∩ B′) = 0.3 M1; 0.3/0.6 = 0.5 A1
Q22Combined events · Standard · Paper 5 style · 5 marks
(a)Three friends, Asad, Bushra and Chen, each try independently to solve a puzzle. The probabilities that they solve it are ½, ⅓ and ¼ respectively. Find the probability that the puzzle is solved by at least one of them.[2]
(b)Find the probability that exactly one of them solves the puzzle.[3]
Q23Combined events · Challenge · Paper 5 style · 8 marks
(a)A fair six-sided die is thrown twice. Event A is 'the total is 7', event B is 'the first throw is a 4' and event C is 'the product of the two scores is even'. Show that A and B are independent.[3]
(b)Find P(C) and P(A ∩ C), and determine whether A and C are independent.[4]
(c)Find P(A ∪ C).[1]
Show answer
(a) P(A ∩ B) = 1/36 = P(A)P(B)
P(A) = 6/36 = 1/6, P(B) = 1/6 B1; P(A ∩ B) = P(4 then 3) = 1/36 B1; = (1/6)(1/6), so independent B1
(b) P(C) = 3/4; P(A ∩ C) = 1/6; not independent
P(C) = 1 − (1/2)(1/2) = 3/4 B1; a total of 7 needs one odd and one even score, so its product is always even: P(A ∩ C) = 1/6 M1 A1; (1/6)(3/4) = 1/8 ≠ 1/6, not independent A1
(c) 3/4
Since A ⊂ C, P(A ∪ C) = P(C) = 3/4 B1
Q24Probability without replacement · Core · Paper 5 style · 4 marks
(a)A bag contains 5 red and 3 blue counters. Two counters are taken at random without replacement. Find the probability that both counters are red.[2]
(b)Find the probability that the counters are of different colours.[2]
Show answer
(a) 5/14
(5/8)(4/7) M1; = 5/14 A1
(b) 15/28
2 × (5/8)(3/7) M1; = 15/28 A1
Q25Probability without replacement · Standard · Paper 5 style · 7 marks
(a)A box contains 12 chocolates: 5 caramel, 4 nut and 3 plain. Fatima takes three chocolates at random without replacement. Find the probability that the three chocolates are all of different types.[3]
(b)Find the probability that at least one of the three chocolates is caramel.[2]
(c)Find the probability that exactly two of the three chocolates are nut.[2]
Q26Probability without replacement · Standard · Paper 5 style · 6 marks
(a)A shelf holds 6 Urdu books and 4 English books. A student picks books at random, one at a time without replacement, until she picks an English book. Find the probability that the first English book is the third book picked.[2]
(b)Find the probability that she picks at most two books.[2]
(c)Find the probability that she needs to pick more than four books.[2]
Show answer
(a) 1/6
(6/10)(5/9)(4/8) M1; = 1/6 A1
(b) 2/3
4/10 + (6/10)(4/9) M1; = 2/3 A1
(c) 1/14
All first four Urdu: (6/10)(5/9)(4/8)(3/7) M1; = 1/14 A1
Q27Probability without replacement · Challenge · Paper 5 style · 7 marks
(a)A bag contains n red sweets and 4 green sweets. Two sweets are taken at random without replacement. The probability that both are red is ⅓. Show that n² − 5n − 6 = 0 and hence find n.[4]
(b)Find the probability that the two sweets are of different colours.[2]
(c)A third sweet is now taken from the bag. Given that the first two sweets were of different colours, find the probability that the third sweet is red.[1]
After one red and one green are removed: 5 red out of 8 B1
Q28Two-way tables & independence · Core · Paper 5 style · 4 marks
(a)A café records the drink chosen by 150 customers. Of the 75 customers under 30, 30 chose tea and 45 chose coffee. Of the 75 customers aged 30 or over, 50 chose tea and 25 chose coffee. A customer is chosen at random. Find the probability that the customer chose tea.[1]
(b)Find the probability that the customer chose tea, given that the customer is under 30.[1]
(c)State, with a reason, whether the events 'chose tea' and 'under 30' are independent.[2]
Show answer
(a) 8/15
80/150 = 8/15 B1
(b) 2/5
30/75 = 2/5 B1
(c) Not independent, since P(tea | under 30) = 2/5 ≠ 8/15 = P(tea)
Compare 2/5 with 8/15 M1; not equal so not independent A1
Q29Two-way tables & independence · Standard · Paper 5 style · 5 marks
(a)In a college in Peshawar there are 200 first-year students: 120 girls and 80 boys. Of the girls, 72 study science and the rest study arts. Of the boys, 48 study science and the rest study arts. Copy and complete a two-way table for this information.[1]
(b)A student is chosen at random. Show that the events 'the student is a girl' and 'the student studies science' are independent.[3]
(c)Find the probability that the student is a boy, given that the student studies arts.[1]
Show answer
(a) Girls: science 72, arts 48; boys: science 48, arts 32; totals: science 120, arts 80
Q30Two-way tables & independence · Challenge · Paper 5 style · 7 marks
(a)The members of a gym in Karachi are classified as follows. Male, morning: 24; male, evening: x; female, morning: 36; female, evening: 54. The events 'a randomly chosen member is male' and 'a randomly chosen member attends in the morning' are independent. Find x.[4]
(b)Find the probability that a randomly chosen member who attends in the evening is female.[1]
(c)Two members are chosen at random without replacement. Find the probability that both attend in the morning.[2]
