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Conditional Probability & Bayes' Theorem · Question 108 of 125

Question 108

MediumOriginal question

P(E)=0.4P(E) = 0.4, P(A∣E)=0.5P(A \mid E) = 0.5 and P(A∣E′)=0.2P(A \mid E') = 0.2. What is P(E∣A)P(E \mid A)?

Options for question 108

Show the answer and explanation
CorrectNot quiteAnswer D

Correct answer: D

Explanation

  1. P(A)=0.4×0.5+0.6×0.2=0.2+0.12=0.32P(A) = 0.4 \times 0.5 + 0.6 \times 0.2 = 0.2 + 0.12 = 0.32.

  2. P(E∣A)=0.20.32=0.625P(E \mid A) = \frac{0.2}{0.32} = 0.625.

Worth noting

0.5 is P(A∣E)P(A \mid E), the conditional the other way round.

Concept
Bayes' theorem from given probabilities
Typical time
90 s
Your time
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The rule this testsBayes' theorem
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Ten questions from Conditional Probability & Bayes' Theorem, at your own pace.

Practise this chapter →

Ten questions from Conditional Probability & Bayes' Theorem, at your own pace.