AQA A-Level Mathematics Paper 3, June 2025: Question 19

10 marks · Medium difficulty · Multi-step Problem

Given probabilities relating to three events X, Y, and Z, where X and Z are mutually exclusive, find P(Z), complete a Venn diagram, calculate compound and conditional probabilities, and show whether two events are independent.

Practise this question

Question

Question 19 presents information about three events X, Y, and Z: P(X) = P(Y) = 0.38, P(X ∪ Z) = 0.5, P[(X ∪ Y)'] = 0.35, and X and Z are mutually exclusive. Part (a) asks to find P(Z) for 1 mark. Part (b) asks to complete a Venn diagram with three intersecting circles for X, Y, and Z where X and Z do not overlap; 0.27 is filled in the region X only, and 0.3 is in the background outside all circles. Part (c) asks for P[(X ∩ Y) ∪ (Y ∩ Z)] for 2 marks. Part (d) asks for P(Y' | Z') for 2 marks. Part (e) asks to show that the events Y and Z are not independent of each other for 2 marks.

Mark scheme

Show the mark scheme Mark scheme for Question 19: 19(a) gives 0.12 (B1). 19(b) awards M1 for obtaining at least one correct value, M1 for at least two, and A1 for all correct: X ∩ Y = 0.11, Y only = 0.2, Y ∩ Z = 0.07, Z only = 0.05. 19(c) gives M1 for 0.11 + 0.07 and A1F for 0.18. 19(d) gives M1 for stating conditional probability formula or finding numerator 0.57 or denominator 0.88, and A1 for 57/88 or awfw [0.647, 0.65]. 19(e) gives M1 for comparing P(Y) × P(Z) = 0.0456 with P(Y ∩ Z) = 0.07 (or equivalent conditional probability comparison) and R1 for completing a reasoned argument concluding they are not independent.

How to answer it

Probability, Sets & Venn Diagrams

WHAT THIS QUESTION TESTS

AQA A-Level Mathematics (Paper 3 / Statistics):

  • Applying the addition law for mutually exclusive events: P(A ∪ B) = P(A) + P(B).
  • Systematically completing a 3-event Venn diagram using set notation, complements, and regions.
  • Identifying composite set regions from Venn diagrams: P[(X ∩ Y) ∪ (Y ∩ Z)].
  • Calculating conditional probabilities: P(A | B) = P(A ∩ B) / P(B).
  • Statistically proving non-independence using P(A ∩ B) ≠ P(A) × P(B) or conditional probability comparisons.

Part (a) — Finding P(Z)

[1 Mark]

📐 Calculations

  1. Identify rule: X and Z are mutually exclusive, so P(X ∩ Z) = 0.
  2. Use the addition rule: P(X ∪ Z) = P(X) + P(Z).
  3. Substitute values: 0.5 = 0.38 + P(Z).
  4. P(Z) = 0.5 − 0.38 = 0.12

✅ Correct Answer

P(Z) = 0.12

Mark scheme: [B1] for correctly stating 0.12.

💡 Key Knowledge

When two events are mutually exclusive, they cannot happen at the same time:

  • P(A ∩ B) = 0
  • P(A ∪ B) = P(A) + P(B)

❌ Common Errors

Assuming X and Z are independent instead of mutually exclusive, incorrectly using P(X ∪ Z) = P(X) + P(Z) − P(X)P(Z).

Part (b) — Complete the Venn Diagram

[3 Marks]

📐 Step-by-Step Region Calculations

  1. Region (X ∩ Y):
    Total P(X) = 0.38. The region for X only is given as 0.27. Since X and Z do not overlap:
    P(X ∩ Y) = 0.38 − 0.27 = 0.11
  2. Region (Z only):
    We are given P[(X ∪ Y)′] = 0.35. Everything outside X and Y consists of "Z only" and the "outside region".
    From the diagram, the outside region = 0.3.
    P(Z only) = 0.35 − 0.30 = 0.05
  3. Region (Y ∩ Z):
    From part (a), total P(Z) = 0.12.
    P(Y ∩ Z) = P(Z) − P(Z only) = 0.12 − 0.05 = 0.07
  4. Region (Y only):
    Given P(Y) = 0.38. Y contains (X ∩ Y), Y only, and (Y ∩ Z).
    P(Y only) = 0.38 − 0.11 − 0.07 = 0.20
Completed Venn Diagram Values (Left to Right):
[X only: 0.27]  |  [X ∩ Y: 0.11]  |  [Y only: 0.2]  |  [Y ∩ Z: 0.07]  |  [Z only: 0.05]  |  [Outside: 0.3]

✅ Correct Values

  • X ∩ Y = 0.11
  • Y only = 0.2 (or 0.20)
  • Y ∩ Z = 0.07
  • Z only = 0.05
Mark scheme:
[M1] At least one of 0.11, 0.2, 0.07, 0.05 in correct position.
[M1] At least two of 0.11, 0.2, 0.07, 0.05 in correct position.
[A1] All four regions completely correct.

🧠 Exam Technique: Quick Check

Always sum all regions in your completed Venn diagram to check that they equal 1.00:

0.27 + 0.11 + 0.20 + 0.07 + 0.05 + 0.30 = 1.00

If they do not add up to 1, re-check your subtractions!

Part (c) — Find P[(X ∩ Y) ∪ (Y ∩ Z)]

[2 Marks]

📐 Calculations

  1. Identify the two distinct intersection regions:
    P(X ∩ Y) = 0.11
    P(Y ∩ Z) = 0.07
  2. Since X and Z cannot overlap, (X ∩ Y) and (Y ∩ Z) are mutually disjoint.
  3. Add the two probabilities:
    0.11 + 0.07 = 0.18

✅ Correct Answer

0.18

Mark scheme:
[M1] States (their 0.11) + (their 0.07).
[A1F] Follow-through answer: 0.18 (provided final answer is positive and < 0.38).

❌ Common Errors

Subtracting an overlap between (X ∩ Y) and (Y ∩ Z). Since X ∩ Z = ∅, the intersection of all three (X ∩ Y ∩ Z) is 0, so no double counting occurs.

Part (d) — Find P(Y′ | Z′)

[2 Marks]

📐 Calculations

  1. Recall the conditional probability definition:
    P(Y′ | Z′) = P(Y′ ∩ Z′) / P(Z′)
  2. Find denominator P(Z′):
    P(Z′) = 1 − P(Z) = 1 − 0.12 = 0.88
  3. Find numerator P(Y′ ∩ Z′) — regions outside both Y and Z:
    This equals X only + outside region:
    0.27 + 0.30 = 0.57
  4. Compute ratio:
    0.57 / 0.88 = 57/88 ≈ 0.648 (3 s.f.)

✅ Correct Answer

57/88 or 0.648 (accept any answer in range [0.647, 0.65])

Mark scheme:
[M1] Valid formula seen, or 0.57 in numerator, or 1 − their P(Z) = 0.88 in denominator.
[A1] 57/88 or decimal in [0.647, 0.65].

🧠 Exam Technique: Reading Complements

To find P(Y′ ∩ Z′) quickly on the Venn diagram, simply cover up circles Y and Z entirely. Whatever numbers remain visible are in Y′ ∩ Z′ (here, only 0.27 and 0.3 ).

❌ Common Errors

Inverting the condition: calculating P(Z′ | Y′) instead of P(Y′ | Z′), which divides by P(Y′) = 0.62 instead of P(Z′) = 0.88.

Part (e) — Show that Y and Z are Not Independent

[2 Marks]

📐 Calculations (Standard Method)

  1. Calculate the product of individual probabilities:
    P(Y) × P(Z) = 0.38 × 0.12 = 0.0456
  2. Obtain the intersection from the Venn diagram:
    P(Y ∩ Z) = 0.07
  3. Compare the two values:
    0.0456 ≠ 0.07
    P(Y ∩ Z) ≠ P(Y) × P(Z)
  4. Conclude clearly: Events Y and Z are not independent.

✅ Acceptable Alternatives

  • Method 2: P(Y | Z) = 0.07 / 0.12 ≈ 0.583 ≠ P(Y) [0.38]
  • Method 3: P(Z | Y) = 0.07 / 0.38 ≈ 0.184 ≠ P(Z) [0.12]
Mark scheme:
[M1] States their P(Z) × 0.38, or compares P(Y | Z) with 0.38, or compares P(Z | Y) with their P(Z).
[R1] Reasoned argument with fully correct calculated numerical values and clear conclusion that events are not independent.

🧠 Exam Technique: Securing the Reasoning Mark [R1]

To secure the final communication mark in "Show that..." questions:

  • Never just write numbers: write the formula / statement first ( P(Y) × P(Z) ).
  • Explicitly show both numerical values alongside each other ( 0.0456 ≠ 0.07 ).
  • Always write the final conclusion sentence: "Since P(Y ∩ Z) ≠ P(Y) × P(Z), the events Y and Z are not independent."

Topics

Statistics · M: Probability

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.