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Question
Mark the correct alternative in the following question:
\[\text{ If } P\left( B \right) = \frac{3}{5}, P\left( A|B \right) = \frac{1}{2} \text{ and } P\left( A \cup B \right) = \frac{4}{5}, \text{ then } P\left( B|\overline{ A } \right) = \]
Options
\[\frac{1}{5}\]
\[ \frac{3}{10}\]
\[\frac{1}{2} \]
\[ \frac{3}{5}\]
Solution
We have,
\[P\left( B \right) = \frac{3}{5}, P\left( A|B \right) = \frac{1}{2} \text{ and } P\left( A \cup B \right) = \frac{4}{5}\]
\[\text{ As } , P\left( A|B \right) = \frac{1}{2}\]
\[ \Rightarrow \frac{P\left( A \cap B \right)}{P\left( B \right)} = \frac{1}{2}\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{1}{2} \times P\left( B \right)\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{1}{2} \times \frac{3}{5}\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{3}{10}\]
\[\text{ And } , P\left( B \cap \overline{ A }\right) = P\left( B \right) - P\left( A \cap B \right)\]
\[ = \frac{3}{5} - \frac{3}{10}\]
\[ = \frac{3}{10}\]
\[\text{ Also } , P\left( A \cup B \right) = P\left( A \right) + P\left( B \right) - P\left( A \cap B \right)\]
\[ \Rightarrow \frac{4}{5} = P\left( A \right) + \frac{3}{5} - \frac{3}{10}\]
\[ \Rightarrow \frac{4}{5} = P\left( A \right) + \frac{3}{10}\]
\[ \Rightarrow P\left( A \right) = \frac{4}{5} - \frac{3}{10}\]
\[ \Rightarrow P\left( A \right) = \frac{5}{10}\]
\[ \Rightarrow P\left( A \right) = \frac{1}{2}\]
\[\text{ So } , P\left( \overline{A} \right) = 1 - P\left( A \right) = 1 - \frac{1}{2} = \frac{1}{2}\]
\[\text{ Now } , \]
\[P\left( B| \overline{A} \right) = \frac{P\left( B \cap \overline{A} \right)}{P\left( \overline{A} \right)}\]
\[ = \frac{\left( \frac{3}{10} \right)}{\left( \frac{1}{2} \right)}\]
\[ = \frac{3 \times 2}{10}\]
\[ = \frac{3}{5}\]
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