AQA A-Level Mathematics: M: Probability (7357) – 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 Section M: Probability (M1 to M3) of the AQA A-level Mathematics (7357) specification, version 1.3, for A-level exams from June 2018 onwards. Section M is listed under Paper 3. A calculator is required in every 7357 paper. Questions 8, 10 and 11 are extended problems. Give answers to 3 significant figures unless told otherwise.

Learn the methods first in the Probability study guide and the Probability revision notes. The 7357 course hub and the printable checklist show where this topic fits, and the free 10-minute diagnostics help you find weak spots.

Questions

1. Events A and B are mutually exclusive, with P(A) = 0.3 and P(B) = 0.5.

(a) Find P(A ∪ B). [1] (b) Find P(A′ ∩ B′). [1]

2. For events C and D, P(C) = 0.6, P(D) = 0.25 and P(C ∩ D) = 0.15.

(a) Determine whether C and D are independent. Show your working. [2] (b) Find P(C ∪ D). [1]

3. A college records how 250 students travel to college.

	Bus	Car	Walk	Total
Year 12	48	22	60	130
Year 13	42	38	40	120
Total	90	60	100	250

One student is chosen at random.

(a) Find the probability that the student walks. [1] (b) Given that the student travels by car, find the probability that the student is in Year 13. [2] (c) Determine whether the events “travels by bus” and “is in Year 12” are independent. [3]

4. For events A and B, P(A) = 0.45, P(B) = 0.3 and P(A | B) = 0.5.

(a) Find P(A ∩ B). [1] (b) Find P(A ∪ B). [2] (c) Find P(B | A). [1] (d) Find P(B′ | A′). [2]

5. A box contains 7 green and 5 yellow counters. Two counters are taken at random, without replacement.

(a) Show that the probability that both counters are the same colour is 31/66. [3] (b) Given that both counters are the same colour, find the probability that they are both green. [2]

6. Events A, B and C are such that A and B are independent, and C is mutually exclusive with both A and B. P(A) = 0.5, P(B) = 0.4 and P(C) = 0.15.

(a) Find P(A ∩ B). [1] (b) Find the probability that none of A, B and C happens. [2] (c) Find P(C | B′). [2]

7. A machine makes bolts. Each bolt is defective with probability 0.04. Bolts are packed in boxes of 10.

(a) Find the probability that a box contains no defective bolts. [2] (b) State an assumption you made in part (a), other than the probability being 0.04. [1] (c) Defective bolts tend to be made in runs, when the machine drifts out of adjustment. Explain the likely effect of this on your answer to part (a). [2]

8. On any day, P(rain) = 0.3 at a bakery. It sells out of bread with probability 0.2 on a rainy day and 0.65 on a dry day.

(a) Find the probability that the bakery sells out on a given day. [2] (b) Given that the bakery sells out, find the probability that it rained. [2] (c) Assuming days are independent, find the probability that the bakery sells out on both of two consecutive days. [2] (d) Comment on the assumption in part (c). [1]

9. The mass, X grams, of honey in a jar is modelled by X ~ N(500, 8²). A jar is rejected if X < 488 or X > 515.

(a) Find the probability that a jar is rejected. [2] (b) Three jars are chosen independently. Find the probability that at least one is rejected. [2] (c) Given that a jar is rejected, find the probability that it is underweight. [2]

10. A spinner gives a score X with the probability distribution below.

x	1	2	3	4
P(X = x)	0.1	0.3	k	2k

(a) Find the value of k. [2]

The spinner is spun twice, and the spins are independent. E is the event “the total score is 5” and F is the event “the first score is 2”.

(b) Find P(E). [3] (c) Find P(F | E). [2] (d) Determine whether E and F are independent. [2]

11. An archer shoots three arrows. On the first arrow she hits the gold with probability 0.3.

Model A: every arrow hits the gold with probability 0.3, independently. Model B: after a gold, the next arrow hits the gold with probability 0.4; after a miss, the next arrow hits the gold with probability 0.25.

(a) Using Model A, find the probability that exactly one arrow hits the gold. [2] (b) Using Model B, find the probability that exactly one arrow hits the gold. [4] (c) Using Model B, find the probability that the first arrow hit the gold, given that exactly one arrow hit the gold. [2] (d) Give one reason why Model B may be more realistic than Model A. [1]

Answers

1. (a) P(A ∪ B) = 0.3 + 0.5 = 0.8 [1] (b) P(A′ ∩ B′) = 1 − 0.8 = 0.2 [1] Examiner insight: A follow-through mark is usually allowed in (b) if P(A ∪ B) in (a) is wrong.

2. (a) P(C) × P(D) = 0.6 × 0.25 = 0.15 [1] This equals P(C ∩ D) = 0.15, so C and D are independent [1] (b) P(C ∪ D) = 0.6 + 0.25 − 0.15 = 0.7 [1] Examiner insight: The conclusion mark needs both numbers shown and compared; “they are independent” with no product earns nothing.

3. (a) P(walks) = 100/250 = 0.4 [1] (b) Car column total is 60, of whom 38 are Year 13 [1] P(Year 13 | car) = 38/60 = 19/30 = 0.633 (3 s.f.) [1] (c) P(bus) = 90/250 = 0.36 and P(Year 12) = 130/250 = 0.52 [1] P(bus) × P(Year 12) = 0.36 × 0.52 = 0.1872, while P(bus ∩ Year 12) = 48/250 = 0.192 [1] 0.1872 ≠ 0.192, so the events are not independent [1] Examiner insight: In (b) a denominator of 250 or 120 loses the method mark at once; “given car” means divide by the car total.

4. (a) P(A ∩ B) = P(B) × P(A | B) = 0.3 × 0.5 = 0.15 [1] (b) P(A ∪ B) = 0.45 + 0.3 − 0.15 [1] = 0.6 [1] (c) P(B | A) = 0.15 / 0.45 = 1/3 [1] (d) P(A′ ∩ B′) = 1 − 0.6 = 0.4 and P(A′) = 0.55 [1] P(B′ | A′) = 0.4 / 0.55 = 8/11 = 0.727 (3 s.f.) [1] Examiner insight: Part (d) needs P(A′ ∩ B′) on top, not P(B′) × P(A′); multiplying assumes independence, which (a) shows is false here.

5. (a) P(G, G) = 7/12 × 6/11 = 42/132 [1] P(Y, Y) = 5/12 × 4/11 = 20/132 [1] P(same) = 42/132 + 20/132 = 62/132 = 31/66 [1] (b) P(both green | same) = (42/132) / (62/132) [1] = 21/31 [1] Examiner insight: On a “show that”, give the intermediate value 62/132 before 31/66; jumping straight to the given answer risks the final mark.

6. (a) P(A ∩ B) = 0.5 × 0.4 = 0.2 [1] (b) P(A ∪ B) = 0.5 + 0.4 − 0.2 = 0.7, and C does not overlap A or B [1] P(none) = 1 − 0.7 − 0.15 = 0.15 [1] (c) C lies wholly outside B, so P(C ∩ B′) = 0.15, and P(B′) = 0.6 [1] P(C | B′) = 0.15 / 0.6 = 0.25 [1] Examiner insight: A labelled Venn sketch with C as a separate circle lets method marks be given even if one region value is wrong.

7. (a) P(no defectives) = 0.96¹⁰ [1] = 0.665 (3 s.f.) [1] (b) Bolts are defective independently of each other [1] (c) Defectives cluster in a few boxes rather than spreading across many [1] so more boxes contain none, and the probability is likely to be higher than 0.665 [1] Examiner insight: The second mark in (c) needs a direction (higher or lower); “the answer would change” is not enough.

8. (a) P(sells out) = 0.3 × 0.2 + 0.7 × 0.65 [1] = 0.06 + 0.455 = 0.515 [1] (b) P(rain | sells out) = 0.06 / 0.515 [1] = 0.117 (3 s.f.) [1] (c) P(both days) = 0.515² [1] = 0.265 (3 s.f.) [1] (d) Rainy days tend to follow rainy days, so sales on consecutive days are probably not independent [1] Examiner insight: In (b) the numerator is the single path “rain and sells out” (0.06), not P(sells out | rain) = 0.2; using 0.2 / 0.515 is a common method-mark loss.

9. (a) P(X < 488) = 0.06681 and P(X > 515) = 0.03040 [1] P(rejected) = 0.06681 + 0.03040 = 0.0972 (3 s.f.) [1] (b) P(at least one rejected) = 1 − (1 − 0.09720)³ [1] = 0.264 (3 s.f.) [1] (c) P(underweight | rejected) = 0.06681 / 0.09720 [1] = 0.687 (3 s.f.) [1] Examiner insight: Write each statement, e.g. P(X < 488), before its calculator value; a wrong bare decimal earns no method mark.

10. (a) 0.1 + 0.3 + k + 2k = 1 [1] 3k = 0.6, so k = 0.2 [1] (b) Totals of 5 come from (1, 4), (4, 1), (2, 3), (3, 2) [1] P(E) = 2(0.1 × 0.4) + 2(0.3 × 0.2) = 0.08 + 0.12 [1] = 0.2 [1] (c) P(F ∩ E) = P(2, 3) = 0.3 × 0.2 = 0.06 [1] P(F | E) = 0.06 / 0.2 = 0.3 [1] (d) P(F) = 0.3 = P(F | E) [1], so E and F are independent [1] Examiner insight: List the ordered pairs in (b); a missed pair loses the accuracy mark, but listed working allows follow-through in (c) and (d).

11. (a) P(exactly one) = 3 × 0.3 × 0.7² [1] = 0.441 [1] (b) P(G, M, M) = 0.3 × 0.6 × 0.75 = 0.135 [1] P(M, G, M) = 0.7 × 0.25 × 0.6 = 0.105 [1] P(M, M, G) = 0.7 × 0.75 × 0.25 = 0.13125 [1] Total = 0.371 (3 s.f.) [1] (c) P(first gold | exactly one) = 0.135 / 0.37125 [1] = 0.364 (3 s.f.) [1] (d) Confidence after a gold or frustration after a miss means arrows are unlikely to be independent [1] Examiner insight: Reusing 0.3 for later arrows in Model B treats it as Model A and loses every path mark.

Where marks are usually lost

• Stating “independent” or “not independent” without both P(A) × P(B) and P(A ∩ B) written as numbers.
• Using the grand total instead of the row or column total for a conditional probability from a table.
• Forgetting that the second draw has one fewer item when sampling without replacement.
• Reversing a conditional on a tree.
• Writing model critiques that name no assumption and give no direction of effect.
• Rounding to 3 s.f. part-way through and carrying the rounded value forward.

Next steps

• Recap with the Probability revision notes.
• Re-read the Probability study guide.
• See all 7357 topics on the course hub and tick them off on the printable checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams June 2018 onwards, published by AQA. Section 3.14, M: Probability.