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If A and B are two events such that P(A)=12P(A) = \frac{1}{2}, P(B)=13P(B) = \frac{1}{3}, P(BA)=14P(B|A) = \frac{1}{4}, then P(AB)P(A|B) is:

Solution

Correct Option: 3

Given P(A)=12P(A) = \frac{1}{2}, P(B)=13P(B) = \frac{1}{3}, and P(BA)=14P(B|A) = \frac{1}{4}

To find P(AB)P(A|B), we first need P(AB)P(A \cap B).

Using the conditional probability formula:

P(BA)=P(AB)P(A)P(B|A) = \frac{P(A \cap B)}{P(A)}

14=P(AB)12\frac{1}{4} = \frac{P(A \cap B)}{\frac{1}{2}}

P(AB)=14×12P(A \cap B) = \frac{1}{4} \times \frac{1}{2}

P(AB)=18P(A \cap B) = \frac{1}{8}


Now applying the conditional probability formula:

P(AB)=P(AB)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}

P(AB)=1813P(A|B) = \frac{\frac{1}{8}}{\frac{1}{3}}

P(AB)=18×31P(A|B) = \frac{1}{8} \times \frac{3}{1}

P(AB)=38P(A|B) = \frac{3}{8}


Therefore, P(AB)=38P(A|B) = \frac{3}{8}

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