OxfordAQA International AS and A-level Mathematics: Probability laws, conditional probability and independence (9660) – 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 are aligned to the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, section S1.1 (Further probability). The spec states this content is part of International AS, Unit PSM1. Read the study guide first, or use the revision notes for a quick recap. See also the course hub, the printable checklist and the free 10-minute diagnostics.

Show your working. Give exact fractions or decimals unless a question says otherwise.

Questions

1. An integer is chosen at random from 1 to 40 inclusive. Event A is “the number is a multiple of 4” and event B is “the number is a square number”.

(a) List the outcomes in A ∩ B and find P(A ∩ B). [2] (b) Find P(A ∪ B). [1] (c) Find P(A’). [1]

2. Events A and B are mutually exclusive, with P(A) = 0.52 and P(B) = 0.3.

(a) Find P(A ∪ B). [1] (b) Find the probability that neither A nor B occurs. [1] (c) Explain, with a calculation, why A and B are not independent. [2]

3. For events A and B, P(A) = 0.6, P(B) = 0.35 and P(A ∩ B) = 0.21.

(a) Show that A and B are independent. [2] (b) Find P(A ∪ B). [2]

4. A school surveyed 200 pupils about how they travel to school.

	Bus	Walk	Car
Year 10	38	29	36
Year 11	42	21	34

A pupil is chosen at random from the 200.

(a) Find P(walks). [1] (b) Find P(Year 11 and bus). [1] (c) Find P(bus | Year 10). [2] (d) Find P(Year 11 | car). [2] (e) Determine whether “walks” and “Year 11” are independent. [3]

5. A and B are independent events with P(A) = 0.65 and P(A ∪ B) = 0.93. Find P(B). [4]

6. A drawer holds 6 black socks and 4 white socks. Three socks are taken at random without replacement.

(a) Find the probability that all three are black. [2] (b) Find the probability that at least one is white. [2] (c) Find the probability that the first two socks taken are the same colour. [3]

7. A parcel passes through three independent stages. The probabilities that each stage is on time are 0.9, 0.85 and 0.8 respectively.

(a) Find the probability that all three stages are on time. [2] (b) Find the probability that at least one stage is late. [1] (c) Find the probability that exactly one stage is late. [3]

8. For events A and B, P(A) = 0.45, P(B) = 0.5 and P(A | B) = 0.4.

(a) Find P(A ∩ B). [1] (b) Find P(A ∪ B). [2] (c) Find P(B | A), giving your answer as a fraction. [2]

9. A factory has two machines. Machine X makes 60% of the items and machine Y makes 40%. Of the items from X, 3% are defective. Of the items from Y, 5% are defective. One item is chosen at random.

(a) Find the probability that it comes from Y and is defective. [1] (b) Find the probability that it is defective. [2] (c) Given that the item is defective, find the probability that it came from X. [3]

10. P(A) = 0.3 and P(B) = 0.4.

(a) Find P(A ∪ B) if A and B are mutually exclusive. [1] (b) Find P(A ∪ B) if A and B are independent. [2] (c) Explain why two events with non-zero probabilities cannot be both mutually exclusive and independent. [2]

11. In a group of 80 people at a leisure centre, 45 use the pool, 38 use the sauna and 15 use neither.

(a) Find how many people use both. [2] (b) One person who uses the sauna is chosen at random. Find the probability that they also use the pool. [2] (c) Determine whether using the pool and using the sauna are independent. [3] (d) Two different people are chosen at random from the 80. Find the probability that neither uses the pool or the sauna. [3]

Answers

1. (a) A ∩ B = {4, 16, 36} [1], so P(A ∩ B) = 3/40 [1] (b) A has 10 outcomes and B = {1, 4, 9, 16, 25, 36} has 6, so P(A ∪ B) = (10 + 6 − 3)/40 = 13/40 [1] (c) P(A’) = 1 − 10/40 = 3/4 [1] Examiner insight: Listing the outcomes in the overlap shows the count behind your fraction, which makes it much easier to see where a slip occurred.

2. (a) P(A ∪ B) = 0.52 + 0.3 = 0.82 [1] (b) 1 − 0.82 = 0.18 [1] (c) Mutually exclusive means P(A ∩ B) = 0 [1], but P(A) × P(B) = 0.52 × 0.3 = 0.156, which is not 0, so the events are not independent [1] Examiner insight: An explanation that quotes the independence condition and evaluates both sides is stronger than the bare statement that exclusive events cannot be independent.

3. (a) P(A) × P(B) = 0.6 × 0.35 = 0.21 [1], which equals P(A ∩ B), so the events are independent [1] (b) P(A ∪ B) = 0.6 + 0.35 − 0.21 [1] = 0.74 [1] Examiner insight: On a “show that” question, state the comparison and the conclusion in words as well as showing the arithmetic.

4. (a) 29 + 21 = 50 walk, so P(walks) = 50/200 = 1/4 [1] (b) P(Year 11 and bus) = 42/200 = 21/100 [1] (c) There are 38 + 29 + 36 = 103 Year 10 pupils [1], so P(bus | Year 10) = 38/103 [1] (d) There are 36 + 34 = 70 car users [1], so P(Year 11 | car) = 34/70 = 17/35 [1] (e) P(Year 11) = 97/200 and P(walks) = 1/4, so the product is 97/800 = 0.12125 [1]. P(walks and Year 11) = 21/200 = 0.105 [1]. These are not equal, so the events are not independent [1] Examiner insight: In part (c) and part (d), the denominator is the total for the condition, so state that total before dividing, because a wrong denominator is the most common slip.

5. Let P(B) = p. Independence gives P(A ∩ B) = 0.65p [1]. Then 0.93 = 0.65 + p − 0.65p [1], so 0.28 = 0.35p [1], giving P(B) = 0.8 [1] Examiner insight: Setting up the equation with a named unknown shows method clearly; substituting back to check that 0.65 + 0.8 − 0.52 = 0.93 guards against arithmetic slips.

6. (a) 6/10 × 5/9 × 4/8 [1] = 120/720 = 1/6 [1] (b) P(at least one white) = 1 − P(all black) [1] = 1 − 1/6 = 5/6 [1] (c) Both black: 6/10 × 5/9 = 30/90 [1]. Both white: 4/10 × 3/9 = 12/90 [1]. Total = 42/90 = 7/15 [1] Examiner insight: Without replacement, show both the reduced numerator and the reduced denominator on the second and third picks, so that your method is easy for a reader to follow.

7. (a) 0.9 × 0.85 × 0.8 [1] = 0.612 [1] (b) 1 − 0.612 = 0.388 [1] (c) Stage 1 late only: 0.1 × 0.85 × 0.8 = 0.068; stage 2 late only: 0.9 × 0.15 × 0.8 = 0.108; stage 3 late only: 0.9 × 0.85 × 0.2 = 0.153 [2]. Sum = 0.329 [1] Examiner insight: For “exactly one”, list the three separate cases before adding them; a single product with no cases shown is easy to get wrong and hard to check.

8. (a) P(A ∩ B) = P(B) × P(A | B) = 0.5 × 0.4 = 0.2 [1] (b) P(A ∪ B) = 0.45 + 0.5 − 0.2 [1] = 0.75 [1] (c) P(B | A) = 0.2 / 0.45 [1] = 4/9 [1] Examiner insight: Part (c) needs the other conditional probability, so quote P(A ∩ B) / P(A) and check that you have divided by P(A), not P(B).

9. (a) 0.4 × 0.05 = 0.02 [1] (b) P(X and defective) = 0.6 × 0.03 = 0.018 [1]. P(defective) = 0.018 + 0.02 = 0.038 [1] (c) P(X | defective) = P(X and defective) / P(defective) [1] = 0.018 / 0.038 [1] = 9/19 (about 0.474) [1] Examiner insight: Write the conditional probability as a fraction of two named probabilities before you substitute; the denominator is the total from part (b), not the whole population.

10. (a) P(A ∪ B) = 0.3 + 0.4 = 0.7 [1] (b) P(A ∩ B) = 0.3 × 0.4 = 0.12 [1], so P(A ∪ B) = 0.3 + 0.4 − 0.12 = 0.58 [1] (c) If they are mutually exclusive, P(A ∩ B) = 0 [1]. If they are also independent, P(A ∩ B) = P(A) × P(B), which is non-zero when both probabilities are non-zero, so the two statements contradict each other [1] Examiner insight: A good explanation uses the two defining equations side by side; a general comment about “one affecting the other” is less precise.

11. (a) 80 − 15 = 65 use the pool or the sauna [1]. Using both: 45 + 38 − 65 = 18 [1] (b) P(pool | sauna) = 18/38 [1] = 9/19 [1] (c) P(pool) × P(sauna) = 45/80 × 38/80 = 0.2672 (4 d.p.) [1]. P(pool and sauna) = 18/80 = 0.225 [1]. These differ, so the events are not independent [1] (d) P(neither) = 15/80 × 14/79 [1], with the second fraction reduced because the first person is not replaced [1]. This gives 210/6320 = 21/632 [1] Examiner insight: For a multi-part problem, carry exact values from earlier parts into later ones, and say which earlier result you are using.

Where marks are usually lost

• Using the wrong total in a conditional probability, especially “given” questions on a table.
• Adding P(A) and P(B) for events that overlap, without subtracting P(A ∩ B).
• Declaring independence or dependence with no numerical comparison.
• Mixing up mutually exclusive with independent.
• Not reducing the denominator on later picks when there is no replacement.
• Missing cases in “exactly one” questions, or forgetting to add them.
• Calculating P(A | B) when the question asked for P(B | A).
• Rounding partway through, which can change the final answer.
• Stopping at the intersection or union without answering the actual question.

Next steps

• Recap the laws in the revision notes.
• Re-read the worked examples in the study guide.
• Visit the OxfordAQA Mathematics hub and the printable checklist.
• Take one of the free 10-minute diagnostics to find your weakest topics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA. Section S1.1 (Further probability), Unit PSM1, International AS exams from May/June 2018 and International A-level exams from May/June 2019.