Cambridge O Level Statistics: Independent events and selection with or without replacement (4040) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover part of topic 11, “Elementary ideas of probability”, in the Cambridge O Level Statistics (4040) syllabus for examination in 2025, 2026 and 2027: independent events, the multiplication rule, and selections made with or without replacement. The course is not tiered. Learn the method first from the study guide or the revision notes. The Cambridge O Level Statistics hub and the printable checklist show the rest of the course, and the free 10-minute diagnostics help you choose what to revise next.

Give answers as exact fractions or as decimals to 3 significant figures unless the question says otherwise.

Questions

1. The events A and B are independent, with P(A) = 0.6 and P(B) = 0.35. Find

(a) P(A ∩ B) [1] (b) P(A ∪ B) [2]

2. A club has 60 members. 24 swim, 15 cycle and 6 do both. Show, by calculation, whether “swims” and “cycles” are independent events. [3]

3. A fair six-sided die is rolled and a fair coin is flipped. Find the probability of

(a) a six and a tail [1] (b) neither a six nor a tail [2]

4. Each battery from a factory is defective with probability 0.03, independently of the others. Four batteries are chosen. Find the probability that, giving both answers to 3 significant figures,

(a) none is defective [1] (b) at least one is defective [2]

5. A bag contains 4 black and 5 white counters. Two counters are picked with replacement. Draw a tree diagram, and find the probability that

(a) both are black [2] (b) they are different colours [2]

6. A bag of 15 sweets holds 6 mint and 9 toffee sweets. Two sweets are taken at random without replacement. Find the probability that

(a) both are mint [1] (b) one of each type is taken [2]

7. A drawer holds 8 socks: 5 grey and 3 navy. Three socks are taken without replacement. Find the probability that

(a) all three are grey [2] (b) exactly two are grey [2] (c) at least one is navy [1]

8. A class has 20 students, of whom 8 are girls. Two students are chosen as monitors. Find the probability that both are girls if

(a) the same student may be chosen twice [2] (b) the two monitors must be different students [2] (c) Explain why your answer to (b) is smaller than your answer to (a). [1]

9. A box holds 10 components, of which 3 are faulty. Components are tested one at a time, without replacement, until a faulty one is found. Find the probability that the first faulty component is found

(a) on the second test [2] (b) on the third test [2] (c) after more than three tests [2]

10. Bag A holds 4 red and 6 white counters. Bag B holds 3 red and 2 white counters. One counter is taken from bag A. Then two counters are taken from bag B without replacement. Find the probability that

(a) all three counters are red [2] (b) exactly one of the three counters is red [3] (c) at least two of the three counters are red [3]

11. A bag contains 10 counters, r of them red and the rest green. Two counters are taken without replacement, and the probability that both are red is 2/15.

(a) Show that r² − r − 12 = 0. [2] (b) Solve the equation and state the value of r. [2] (c) Using this value of r, find the probability that exactly one red counter is taken if the two counters are taken with replacement. [2]

Answers

1. (a) P(A ∩ B) = 0.6 × 0.35 [1] = 0.21. (b) P(A ∪ B) = 0.6 + 0.35 − 0.21 [1] = 0.74 [1]. Examiner insight: the addition rule for “or” needs the overlap subtracted, and a solution that adds 0.6 and 0.35 gives 0.95 and shows the rule has been misremembered.

2. P(swim) = 24/60 = 0.4 and P(cycle) = 15/60 = 0.25 [1]. The product is 0.4 × 0.25 = 0.1 [1]. P(swim and cycle) = 6/60 = 0.1, which equals the product, so the events are independent [1]. Examiner insight: a conclusion needs both values written out and a comparison stated in words. Saying “yes” with no figures, or comparing only one pair of numbers, leaves the argument incomplete.

3. (a) 1/6 × 1/2 [1] = 1/12. (b) 5/6 × 1/2 [1] = 5/12 [1]. Examiner insight: for “neither”, each factor is the chance of the event not happening. Using 1/6 × 1/2 again, or subtracting 1/12 from 1, answers a different question.

4. (a) 0.97⁴ [1] = 0.885. (b) 1 − 0.97⁴ [1] = 1 − 0.8853 = 0.115 [1]. Examiner insight: “at least one” is quickest as 1 − P(none). Rounding the answer to part (a) before subtracting can shift the third significant figure of part (b), so keep the unrounded value in your working.

5. Tree: first pick B (4/9) or W (5/9); second pick B (4/9) or W (5/9) from each point, the same numbers because the counter is replaced [1]. (a) P(BB) = 4/9 × 4/9 = 16/81 [1]. (b) P(BW) + P(WB) = 4/9 × 5/9 + 5/9 × 4/9 [1] = 40/81 [1]. Examiner insight: “different colours” has two paths, black then white and white then black. An answer that includes only one path is a common loss of credit.

6. (a) 6/15 × 5/14 [1] = 30/210 = 1/7. (b) 6/15 × 9/14 + 9/15 × 6/14 [1] = 54/210 + 54/210 = 108/210 = 18/35 [1]. Examiner insight: write the two orders separately before adding. If the second-stage denominator is 15 rather than 14, the working shows a replacement model was used where none was stated.

7. (a) 5/8 × 4/7 × 3/6 [1] = 60/336 = 5/28 [1]. (b) Orders GGN, GNG, NGG each have probability 5/8 × 4/7 × 3/6 = 60/336 [1], so the total is 3 × 60/336 = 180/336 = 15/28 [1]. (c) 1 − 5/28 = 23/28 [1]. Examiner insight: “at least one navy” is the complement of “all grey”, which was found in part (a). Using that result is quicker than adding three separate cases, and a correctly identified complement keeps the working short.

8. (a) 8/20 × 8/20 = 64/400 [1] = 4/25 = 0.16 [1]. (b) 8/20 × 7/19 [1] = 56/380 = 14/95 ≈ 0.147 [1]. (c) Without replacement, once one girl has been chosen there are fewer girls and fewer students left, so the second choice is less likely to be a girl: 7/19 is less than 8/20 [1]. Examiner insight: an explanation scores when it refers to the changed numbers in the second pick. A statement such as “it is different” without saying why does not explain anything.

9. (a) Non-faulty then faulty: 7/10 × 3/9 [1] = 21/90 = 7/30 [1]. (b) Non-faulty, non-faulty, faulty: 7/10 × 6/9 × 3/8 [1] = 126/720 = 7/40 [1]. (c) More than three tests means the first three are all non-faulty: 7/10 × 6/9 × 5/8 [1] = 210/720 = 7/24 [1]. Examiner insight: each path must show the right number of “not yet faulty” factors before the faulty one. A useful check is that the probabilities of finding the first faulty component on test 1, 2, 3 and after three tests add to 1: 3/10 + 7/30 + 7/40 + 7/24 = 1.

10. (a) P(red from A) = 4/10 and P(both red from B) = 3/5 × 2/4 = 3/10 [1]. The stages are independent, so P = 4/10 × 3/10 = 12/100 = 3/25 [1]. (b) From bag B, P(no red) = 2/5 × 1/4 = 1/10 and P(exactly one red) = 3/5 × 2/4 + 2/5 × 3/4 = 6/20 + 6/20 = 3/5 [1]. Exactly one red overall is red from A with no red from B, or white from A with one red from B: 4/10 × 1/10 + 6/10 × 3/5 [1] = 4/100 + 36/100 = 2/5 [1]. (c) At least two red is: all three red, or red from A with exactly one red from B, or white from A with two red from B. Using 3/25 from part (a): red from A with one red from B = 4/10 × 3/5 = 6/25 [1]; white from A with two red from B = 6/10 × 3/10 = 9/50 [1]. Total = 6/50 + 12/50 + 9/50 = 27/50 [1]. So P = 27/50 = 0.54. Examiner insight: in a multi-stage question, first find the probabilities for each source separately, then combine. A list of every case that fits the wording, with no overlap, is easier to mark and easier to check than one long expression.

11. (a) P(both red) = r/10 × (r − 1)/9 [1] = 2/15, so r(r − 1) = 90 × 2/15 = 12, giving r² − r − 12 = 0 [1]. (b) (r − 4)(r + 3) = 0 [1], so r = 4 or r = −3, and r must be positive, so r = 4 [1]. (c) P(red) = 4/10 = 2/5 and P(green) = 3/5 on each pick; P(exactly one red) = 2 × 2/5 × 3/5 [1] = 12/25 [1]. Examiner insight: in a “show that” question every step from the first expression to the final equation must appear. When solving, rejecting the negative root with a reason is part of the answer, since a number of counters cannot be negative.

Where marks are usually lost

• Using the same denominator for the second pick when the question says the counter is not replaced.
• Reducing the denominator but not the count of the colour just removed.
• Giving only one of the two orders for “one of each” or “different colours”.
• Adding along a branch when the branch probabilities should be multiplied.
• Claiming independence without testing it, or testing it with figures that are not written down.
• Treating “mutually exclusive” as the same thing as “independent”.
• Computing “at least one” by adding many cases and missing one, when 1 − P(none) is shorter.
• Rounding early, then giving a final answer that differs in the third significant figure.
• Giving a “show that” answer that jumps from the setup to the result.
• Writing an explanation that restates the answer instead of pointing to what changed between the first and second pick.

Next steps

• Go back to the revision notes for the rules in one place.
• Read the study guide for the full worked examples.
• Visit the Cambridge O Level Statistics hub and the checklist.
• Try the free 10-minute diagnostics to find your next topic.
• Book a free trial class if you want help from a teacher.

Official syllabus

These questions follow the Cambridge O Level Statistics 4040 syllabus for examination in 2025, 2026 and 2027, published by Cambridge University Press & Assessment (Cambridge International), topic 11 “Elementary ideas of probability”.