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Question
If A and B are two events such that \[ P\left( A \right) = \frac{1}{3}, P\left( B \right) = \frac{1}{4} \text{ and } P\left( A \cup B \right) = \frac{5}{12}, \text{ then find } P\left( A|B \right) \text{ and } P\left( B|A \right) . \]
Solution
We have ,
\[P\left( A \right) = \frac{1}{3}, P\left( B \right) = \frac{1}{4} \text{ and } P\left( A \cup B \right) = \frac{5}{12}\]
\[As, P\left( A \cup B \right) = P\left( A \right) + P\left( B \right) - P\left( A \cap B \right)\]
\[ \Rightarrow P\left( A \cap B \right) = P\left( A \right) + P\left( B \right) - P\left( A \cup B \right)\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{1}{3} + \frac{1}{4} - \frac{5}{12}\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{2}{12}\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{1}{6}\]
\[\text { Now } , \]
\[P\left( A|B \right) = \frac{P\left( A \cap B \right)}{P\left( B \right)} = \frac{\left( \frac{1}{6} \right)}{\left( \frac{1}{4} \right)} = \frac{4}{6} = \frac{2}{3} \text{ and } \]
\[P\left( B|A \right) = \frac{P\left( A \cap B \right)}{P\left( A \right)} = \frac{\left( \frac{1}{6} \right)}{\left( \frac{1}{3} \right)} = \frac{3}{6} = \frac{1}{2}\]
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