Question

If A and B are two events such that\[ P\left( A \right) = \frac{6}{11}, P\left( B \right) = \frac{5}{11} \text{ and } P\left( A \cup B \right) = \frac{7}{11}, \text{ then find } P\left( A \cap B \right), P\left( A|B \right) \text { and } P\left( B|A \right) . \]

Answer 1

 We have , 
\[P\left( A \right) = \frac{6}{11}, P\left( B \right) = \frac{5}{11} \text{ and}  P\left( A \cup B \right) = \frac{7}{11}\]
\[As, P\left( A \cup B \right) = P\left( A \right) + P\left( B \right) - P\left( A \cap B \right)\]
\[ \Rightarrow P\left( A \cap B \right) = P\left( A \right) + P\left( B \right) - P\left( A \cup B \right)\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{6}{11} + \frac{5}{11} - \frac{7}{11}\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{6 + 5 - 7}{11}\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{4}{11}\]
\[\text{ Now } , \]
\[P\left( A|B \right) = \frac{P\left( A \cap B \right)}{P\left( B \right)} = \frac{\left( \frac{4}{11} \right)}{\left( \frac{5}{11} \right)} = \frac{4}{5} \text { and }\]
\[P\left( B|A \right) = \frac{P\left( A \cap B \right)}{P\left( A \right)} = \frac{\left( \frac{4}{11} \right)}{\left( \frac{6}{11} \right)} = \frac{4}{6} = \frac{2}{3}\]