Question

Mark the correct alternative in the following question:

\[ \text{ If }  P\left( B \right) = \frac{3}{5}, P\left( A|B \right) = \frac{1}{2} \text{ and }  P\left( \overline{A \cup B }\right) = \frac{4}{5}, \text{ then }  P\left( \overline{ A } \cup B \right) + P\left( A \cup B \right) = \]

विकल्प

• \[\frac{1}{5}\]

• \[ \frac{4}{5} \]

• \[  \frac{1}{2} \]

• \[ 1\]

Answer 1

\[\text{ We have } , \]
\[P\left( B \right) = \frac{3}{5}, P\left( A|B \right) = \frac{1}{2} \text{ and } P\left( A \cup B \right) = \frac{4}{5}\]
\[\text{ As } , P\left( A|B \right) = \frac{1}{2}\]
\[ \Rightarrow \frac{P\left( A \cap B \right)}{P\left( B \right)} = \frac{1}{2}\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{1}{2} \times P\left( B \right)\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{1}{2} \times \frac{3}{5}\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{3}{10}\]
\[\text{ As } , P\left( A \cup B \right) = \frac{4}{5}\]
\[ \Rightarrow P\left( A \right) + P\left( B \right) - P\left( A \cap B \right) = \frac{4}{5}\]
\[ \Rightarrow P\left( A \right) + \frac{3}{5} - \frac{3}{10} = \frac{4}{5}\]
\[ \Rightarrow P\left( A \right) + \frac{3}{10} = \frac{4}{5}\]
\[ \Rightarrow P\left( A \right) = \frac{4}{5} - \frac{3}{10}\]
\[ \Rightarrow P\left( A \right) = \frac{5}{10}\]
\[ \Rightarrow P\left( A \right) = \frac{1}{2}\]
\[\text{ Now } , \]
\[P\left(\overline{ A \cup B } \right) + P\left( \overline{ A } \cup B \right) = \left[ 1 - P\left( A \cup B \right) \right] + \left[ 1 - P\left( \text {Only A } \right) \right]\]
\[ = \left[ 1 - \frac{4}{5} \right] + 1 - \left[ P\left( A \right) - P\left( A \cap B \right) \right]\]
\[ = \frac{1}{5} + 1 - \left[ \frac{1}{2} - \frac{3}{10} \right]\]
\[ = \frac{6}{5} - \frac{2}{10}\]
\[ = \frac{10}{10}\]
\[ = 1\]