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Question
If A and B are two events such that P (A) = \[\frac{1}{3},\] P (B) = \[\frac{1}{5}\] and P (A ∪ B) = \[\frac{11}{30}\] , find P (A/B) and P (B/A).
Solution
\[\text{ Given } : \]
\[P\left( A \right) = \frac{1}{3}\]
\[P\left( B \right) = \frac{1}{5}\]
\[P\left( A \cup B \right) = \frac{11}{30}\]
\[\text{ Now } , \]
\[P\left( A \cup B \right) = P\left( A \right) + P\left( B \right) - P\left( A \cap B \right)\]
\[ \Rightarrow \frac{11}{30} = \frac{1}{3} + \frac{1}{5} - P\left( A \cap B \right)\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{1}{3} + \frac{1}{5} - \frac{11}{30} = \frac{10 + 6 - 11}{30} = \frac{5}{30} = \frac{1}{6}\]
\[P\left( A/B \right) = \frac{P\left( A \cap B \right)}{P\left( B \right)} = \frac{\frac{1}{6}}{\frac{1}{5}} = \frac{5}{6}\]
\[P\left( B/A \right) = \frac{P\left( A \cap B \right)}{P\left( A \right)} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{3}{6} = \frac{1}{2}\]
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