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Question
Mark the correct alternative in the following question:
\[\text{ If A and B are two events such that } P\left( A \right) = 0 . 4, P\left( B \right) = 0 . 3 \text{ and } P\left( A \cup B \right) = 0 . 5, \text{ then } P\left( B \cap A \right) \text{ equals } \]
Options
\[\frac{2}{3}\]
\[ \frac{1}{2}\]
\[ \frac{3}{10}\]
\[ \frac{1}{5}\]
Solution
\[\text{ We have, } \]
\[P\left( A \right) = 0 . 4, P\left( B \right) = 0 . 3 \text{ and } P\left( A \cup B \right) = 0 . 5\]
\[\text{ As }, P\left( A \cup B \right) = 0 . 5\]
\[ \Rightarrow P\left( A \right) + P\left( B \right) - P\left( A \cap B \right) = 0 . 5\]
\[ \Rightarrow 0 . 4 + 0 . 3 - P\left( A \cap B \right) = 0 . 5\]
\[ \Rightarrow 0 . 7 - P\left( A \cap B \right) = 0 . 5\]
\[ \Rightarrow P\left( A \cap B \right) = 0 . 7 - 0 . 5\]
\[ \Rightarrow P\left( A \cap B \right) = 0 . 2\]
\[\text{ Now} , \]
\[P\left(\overline{ B } \cap A \right) = P\left( A \right) - P\left( A \cap B \right)\]
\[ = 0 . 4 - 0 . 2\]
\[ = 0 . 2\]
\[ = \frac{2}{10}\]
\[ = \frac{1}{5}\]
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