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Conditional Probability & Bayes' Theorem · Question 74 of 125

Question 74

HardOriginal question

AA and BB are independent, with P(A∩B)=16P(A \cap B) = \frac{1}{6}, P(A′∩B′)=13P(A' \cap B') = \frac{1}{3} and P(A)>P(B)P(A) > P(B). What is P(A)P(A)?

Options for question 74

Show the answer and explanation
CorrectNot quiteAnswer C

Correct answer: C

Explanation

  1. With p=P(A)p = P(A) and q=P(B)q = P(B): pq=16pq = \frac{1}{6} and (1−p)(1−q)=1−(p+q)+pq=13(1 - p)(1 - q) = 1 - (p + q) + pq = \frac{1}{3}, so p+q=56p + q = \frac{5}{6}.

  2. pp and qq are the roots of t2−56t+16=0t^{2} - \frac{5}{6}t + \frac{1}{6} = 0, namely 12\frac{1}{2} and 13\frac{1}{3}; the larger is P(A)=12P(A) = \frac{1}{2}.

Worth noting

13\frac{1}{3} is the smaller root, P(B)P(B); the question says P(A)>P(B)P(A) > P(B).

Concept
An unknown probability from independence
Typical time
120 s
Your time
—
The rule this testsIndependent events
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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.