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IB Mathematics: Applications and Interpretation · Statistics and Probability

Venn/tree/sample space; combined events; conditional; independence practice questions.

36 exam-style questions on Venn/tree/sample space; combined events; conditional; independence — topic 4.6 of Statistics and Probability in the IB Mathematics: Applications and Interpretation syllabus — with answers and explanations. Samples below; the full set is free with an account.

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About this topic

Syllabus code
4.6
Subject
IB Mathematics: Applications and Interpretation
Topic
4 · Statistics and Probability
Questions
36 with worked answers
Question types
12 fill-in-the-gap questions · 12 multiple-choice questions · 5 true/false questions · 4 flashcards · 2 diagram tasks · 1 matching task
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Sample questions on Venn/tree/sample space; combined events; conditional; independence

In a class of 30 students, 18 study Physics, 14 study Chemistry, and 8 study both. Determine how many students study neither subject.

  • 4
  • 6
  • 8
  • 10

Answer: 6

By inclusion–exclusion, students studying at least one = 18 + 14 − 8 = 24, so 30 − 24 = 6 study neither.

A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. Calculate the probability that both are red.

  • 25/64
  • 5/14
  • 5/16
  • 15/56

Answer: 5/14

Without replacement: P(both red) = (5/8) × (4/7) = 20/56 = 5/14.

For events A and B, P(A) = 0.6, P(B) = 0.5 and P(A ∪ B) = 0.8. Determine P(A ∩ B).

  • 0.1
  • 0.2
  • 0.3
  • 0.4

Answer: 0.3

P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.6 + 0.5 − 0.8 = 0.3.

A coin is biased so that P(heads) = 0.7. It is tossed twice. Calculate the probability of getting exactly one head.

  • 0.21
  • 0.42
  • 0.49
  • 0.09

Answer: 0.42

Exactly one head = HT or TH = (0.7)(0.3) + (0.3)(0.7) = 0.21 + 0.21 = 0.42.

Given P(A) = 0.5, P(B) = 0.4 and P(A | B) = 0.5, identify the correct statement about events A and B.

  • Independent; P(A ∩ B) = 0.20
  • Dependent; P(A ∩ B) = 0.20
  • Independent; P(A ∩ B) = 0.25
  • Mutually exclusive; P(A ∩ B) = 0

Answer: Independent; P(A ∩ B) = 0.20

P(A | B) = P(A) = 0.5, so A and B are independent; P(A ∩ B) = P(A | B) × P(B) = 0.5 × 0.4 = 0.20.

A disease affects 1% of a population. A test gives a positive result for 90% of people who have the disease and a negative result for 90% of people who do not. A person tests positive. Identify the numerator of P(disease | positive).

  • 0.90 × 0.01
  • 0.90 × 0.99
  • 0.10 × 0.01
  • 0.99 × 0.10

Answer: 0.90 × 0.01

P(disease | positive) = P(disease and positive) / P(positive); the numerator is P(disease) × P(positive | disease) = 0.01 × 0.90. The product 0.99 × 0.10 is the false-positive term, which belongs only in the denominator.

That was 6 of 36 Venn/tree/sample space; combined events; conditional; independence questions.

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36 Venn/tree/sample space; combined events; conditional; independence questions — free with an account. Spaced repetition, streaks and full exam simulations included.

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