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Conditional Probability

Conditional Probability

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Question 148

The events A and B are such that P(A)=320\displaystyle P(A) = \frac{3}{20}P(A)=203​, P(A∩B)=120\displaystyle P(A \cap B) = \frac{1}{20}P(A∩B)=201​, P(A∣B)=514\displaystyle P(A|B) = \frac{5}{14}P(A∣B)=145​.

a.

Find P(B)P(B)P(B)

[3]
b.

Find P(A∪B)P(A \cup B)P(A∪B)

[3]
c.

Find P(B∣A′)P(B|A')P(B∣A′)

[3]
d.

Hence find P(A′∣B′)P(A'|B')P(A′∣B′)

[3]
Markscheme

Conditional Probability Questions

  1. A Level
  2. /Maths
  3. /Conditional Probability

330 exam-style questions on Edexcel A Level Maths Conditional Probability, covering 2.1 Set Notation, 2.2 Conditional Probability, 2.3 Conditional Probabilities in Venn Diagrams, 2.4 Probability Formulae, and 2.5 Tree Diagrams. Each one has a worked solution and a mark scheme showing where the marks go.

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