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Conditional Probability

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Question 11

The events A and B are such that P(A)=15\displaystyle P(A) = \frac{1}{5}P(A)=51​, P(A∩B)=110\displaystyle P(A \cap B) = \frac{1}{10}P(A∩B)=101​, P(A∣B)=12\displaystyle P(A|B) = \frac{1}{2}P(A∣B)=21​.

a.

State the formula for P(A∣B)P(A|B)P(A∣B) in terms of P(A∩B)P(A \cap B)P(A∩B) and P(B)P(B)P(B)

[1]
b.

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

[3]
c.

Find P(A∪B)P(A \cup B)P(A∪B)

[3]
d.

Find P(B∣A′)P(B|A')P(B∣A′)

[3]
Markscheme

Conditional Probability Questions

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

368 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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