OxfordAQA International GCSE Mathematics: Independent events, tree diagrams and conditional probability (9260) – 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 OxfordAQA International GCSE Mathematics (9260), specification Version 5.1 (for exams May/June 2018 onwards), section 3.4.3 Probability, statements S17 and S18. Every question is Extension only. Start with the study guide or the revision notes if the method is unfamiliar. Course pages: the OxfordAQA IGCSE Mathematics hub and the checklist. You can gauge your level with the free 10-minute diagnostics.

Leave probabilities as fractions or exact decimals unless a question asks you to round. Show every step.

Questions

1. (Extension only) A and B are independent events. P(A) = 0.3 and P(B) = 0.8. Work out P(A ∩ B). [2]

2. (Extension only) Decide whether each pair of events is independent. Show your working.

(a) P(C) = 0.5, P(D) = 0.36 and P(C ∩ D) = 0.18. [2] (b) P(E) = 0.6, P(F) = 0.5 and P(E ∩ F) = 0.25. [2]

3. (Extension only) A cyclist meets two sets of traffic lights. The lights work independently. The probability of a green light is 0.7 at the first set and 0.6 at the second.

(a) Find the probability that both lights are green. [2] (b) Find the probability that exactly one light is green. [3]

4. (Extension only) A bag holds 4 black and 6 white counters. A counter is taken at random, replaced, and a second counter is taken.

(a) Write down the probability of black and the probability of white at each stage. [1] (b) Find the probability that both counters are black. [2] (c) Work out the probability of getting one counter of each colour. [2]

5. (Extension only) A box holds 8 raffle tickets: 3 are winning tickets and 5 are not. Two tickets are drawn one after the other and not returned.

(a) Find the probability that both are winning tickets. [2] (b) Find the probability that exactly one is a winning ticket. [3]

6. (Extension only) A component passes three independent checks with probabilities 0.95, 0.9 and 0.98.

(a) Find the probability that the component passes all three checks. [2] (b) Find the probability that it fails at least one check. [2]

7. (Extension only) The table shows the 80 members of a leisure club.

	Adult	Junior
Gym	14	6
Swimming	12	18
Yoga	20	10

One member is picked at random.

(a) Work out the probability of picking a junior. [1] (b) Given that the member is a junior, find the probability that they swim. [2] (c) Given that the member does yoga, find the probability that they are a junior. [2]

8. (Extension only) The probability of frost on a morning is 0.25. If there is frost, a bus is late with probability 0.6. If there is no frost, the bus is late with probability 0.15.

(a) Find the probability that the bus is late. [3] (b) Given that the bus is late, find the probability that there was frost. [3]

9. (Extension only) A bag holds 4 red and 5 blue counters. Two counters are removed one at a time and not put back.

(a) Work out the probability that the second counter removed is red. [3] (b) Given that the second counter is red, find the probability that the first counter was red. [2]

10. (Extension only) 2% of the devices made in a factory are faulty. A test flags a faulty device with probability 0.95. The test also flags a device that is not faulty with probability 0.04.

(a) Draw a tree diagram for a device being faulty or not faulty and then flagged or not flagged. Write a probability beside each branch. [2] (b) Find the probability that a device chosen at random is flagged. [3] (c) Given that a device is flagged, find the probability that it is faulty. Give your answer to 3 significant figures. [3]

11. (Extension only) A and B are independent events. P(A) = 0.4 and the probability that neither A nor B happens is 0.3.

(a) Find P(B). [4] (b) Find the probability that exactly one of A and B happens. [2]

12. (Extension only) A bag holds 6 mint and 4 toffee sweets. Three sweets are removed one at a time and eaten.

(a) Find the probability that all three are mint. [2] (b) Work out the probability that the three include a toffee. [1] (c) Find the probability that exactly one is toffee. [3] (d) Given that at least one is toffee, find the probability that exactly one is toffee. [2]

Answers

1. The events are independent, so multiply: P(A) × P(B) [1] = 0.3 × 0.8 = 0.24 [1] Examiner insight: Show the multiplication and the rule you are using; a bare decimal leaves nothing to reward if the arithmetic slips.

2. (a) P(C) × P(D) = 0.5 × 0.36 = 0.18 [1]. This equals P(C ∩ D), so C and D are independent [1] (b) P(E) × P(F) = 0.6 × 0.5 = 0.3 [1]. This is not 0.25, so E and F are not independent [1] Examiner insight: A conclusion should come with the product and a clear comparison; saying only “yes” or “no” does not show that you tested the rule.

3. (a) P(both green) = 0.7 × 0.6 [1] = 0.42 [1] (b) P(green then not green) = 0.7 × 0.4 = 0.28 [1]; P(not green then green) = 0.3 × 0.6 = 0.18 [1]; total = 0.28 + 0.18 = 0.46 [1] Examiner insight: “Exactly one” has two separate cases, and answers that include only one case are a common way to lose marks.

4. (a) Each stage: P(black) = 4/10 = 2/5 and P(white) = 6/10 = 3/5 [1] (b) P(black, black) = 2/5 × 2/5 [1] = 4/25 [1] (c) P(black, white) = 2/5 × 3/5 = 6/25 and P(white, black) = 3/5 × 2/5 = 6/25 [1]; P(different) = 6/25 + 6/25 = 12/25 [1] Examiner insight: With replacement, both stages carry the same probabilities, so a different set of numbers at stage two suggests you have treated it as without replacement.

5. (a) P(win, win) = 3/8 × 2/7 [1] = 6/56 = 3/28 [1] (b) P(win, lose) = 3/8 × 5/7 = 15/56 [1]; P(lose, win) = 5/8 × 3/7 = 15/56 [1]; total = 30/56 = 15/28 [1] Examiner insight: After the first pick the total drops to 7, and the second-stage numerator also changes if the first ticket is of the same type.

6. (a) P(passes all three) = 0.95 × 0.9 × 0.98 [1] = 0.8379 [1] (b) “Fails at least one” is the complement of “passes all” [1], so 1 − 0.8379 = 0.1621 [1] Examiner insight: Using the complement saves listing seven paths; if you do list them, write each one so a slip can still be followed.

7. (a) Juniors = 6 + 18 + 10 = 34, so P(junior) = 34/80 = 17/40 [1] (b) Given junior, the total is 34 [1], so P(swim | junior) = 18/34 = 9/17 [1] (c) Given yoga, the total is 30 [1], so P(junior | yoga) = 10/30 = 1/3 [1] Examiner insight: Reading the denominator from the “given” row or column is the key step; dividing by 80 in parts (b) or (c) is the usual error.

8. (a) Frost and late: 0.25 × 0.6 = 0.15 [1]; no frost and late: 0.75 × 0.15 = 0.1125 [1]; P(late) = 0.15 + 0.1125 = 0.2625 [1] (b) P(frost | late) = P(frost and late) ÷ P(late) [1] = 0.15 ÷ 0.2625 [1] = 4/7 (0.571 to 3 significant figures) [1] Examiner insight: The denominator in part (b) is the answer to part (a), so check part (a) carefully; show each path so your method can be followed.

9. (a) P(red, red) = 4/9 × 3/8 = 12/72 [1]; P(blue, red) = 5/9 × 4/8 = 20/72 [1]; P(second red) = 32/72 = 4/9 [1] (b) P(first red | second red) = (12/72) ÷ (32/72) [1] = 12/32 = 3/8 [1] Examiner insight: “Second counter” does not mean “after the first is known”; both first-stage colours must be included in part (a).

10. (a) First stage: faulty 0.02, not faulty 0.98 [1]. Second stage after faulty: flagged 0.95, not flagged 0.05; after not faulty: flagged 0.04, not flagged 0.96 [1] (b) Faulty and flagged: 0.02 × 0.95 = 0.019 [1]; not faulty and flagged: 0.98 × 0.04 = 0.0392 [1]; P(flagged) = 0.019 + 0.0392 = 0.0582 [1] (c) P(faulty | flagged) = 0.019 ÷ 0.0582 [1] = 0.32646… [1] = 0.326 (3 significant figures) [1] Examiner insight: Label each branch and check that each pair adds to 1; if the question specifies 3 significant figures, round only on the final line.

11. (a) Let P(B) = b. Independence means P(not A and not B) = P(not A) × P(not B) [1]; P(not A) = 0.6 [1]; 0.6 × (1 − b) = 0.3 [1]; 1 − b = 0.5, so b = 0.5 [1] (b) P(A only) = 0.4 × 0.5 = 0.2 and P(B only) = 0.6 × 0.5 = 0.3 [1]; total = 0.5 [1] Examiner insight: Turn the words “neither A nor B” into a product of complements before you try to solve; unclear working here is hard to credit.

12. (a) P(all mint) = 6/10 × 5/9 × 4/8 [1] = 120/720 = 1/6 [1] (b) P(at least one toffee) = 1 − 1/6 = 5/6 [1] (c) One path, toffee then mint then mint: 4/10 × 6/9 × 5/8 = 120/720 [1]; there are three equally likely orders for the single toffee [1]; 3 × 120/720 = 360/720 = 1/2 [1] (d) P(exactly one | at least one) = (1/2) ÷ (5/6) [1] = 6/10 = 3/5 [1] Examiner insight: Exactly one toffee can occur in three positions, so state the number of orders you are counting.

Where marks are usually lost

• Using the original total at the second stage when counters or tickets are not replaced.
• Adding probabilities along a branch instead of multiplying.
• Multiplying across separate paths instead of adding them.
• Counting only one of the cases in “exactly one” questions.
• Answering with the wrong conditional direction: P(A | B) instead of P(B | A).
• Using the grand total as the denominator when a conditional probability needs a reduced total.
• Stating “independent” without showing the product comparison.
• Rounding part-way through a calculation, then missing the required accuracy at the end.
• Leaving branches that do not sum to 1 and not noticing.
• Listing every path for “at least one” when 1 − P(none) was shorter and safer.

Next steps

• Re-read the method in the study guide where you lost marks.
• Condense it with the revision notes.
• Course hub: OxfordAQA IGCSE Mathematics.
• Tick off statements on the checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), specification Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. These questions follow section 3.4.3 Probability, statements S17 and S18.