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Probability

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Question : 105 of 127

Marks: +1, -0

Given that

P(A)=\frac{1}{3}, P(B)=\frac{3}{4}

P(A \cup B)=

\frac{11}{12},

the probability,

P\left(\frac{B}{A}\right)

is

$\frac{1}{6}$
$\frac{4}{9}$
$\frac{1}{2}$
$\frac{1}{4}$

Solution:

Concept:

Conditional probability:

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

.

Explanation:

Use the union formula:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

.

So

P(A \cap B) = P(A) + P(B) - P(A \cup B)

.

Substitute:

P(A \cap B) = \frac{1}{3} + \frac{3}{4} - \frac{11}{12}

.

Convert to denominator 12:

\frac{4}{12} + \frac{9}{12} - \frac{11}{12} = \frac{2}{12} = \frac{1}{6}

.

Now

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{1/6}{1/3} = \frac{1}{6} \times 3 = \frac{1}{2}

.

Answer:

\frac{1}{2}

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