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प्रश्न
Mark the correct alternative in the following question:
\[\text{ If A and B are two events such that } P\left( A \right) = \frac{1}{2}, P\left( B \right) = \frac{1}{3}, P\left( A|B \right) = \frac{1}{4}, \text{ then } P\left( A \cap B \right) \text{ equals} \]
पर्याय
\[ \frac{1}{12}\]
\[ \frac{3}{4} \]
\[ \frac{1}{4} \]
\[ \frac{3}{16}\]
उत्तर
\[\text{ We have } , \]
\[P\left( A \right) = \frac{1}{2}, P\left( B \right) = \frac{1}{3} \text{ and } P\left( A|B \right) = \frac{1}{4}\]
\[\text{ As }, P\left( A|B \right) = \frac{1}{4}\]
\[ \Rightarrow \frac{P\left( A \cap B \right)}{P\left( B \right)} = \frac{1}{4}\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{1}{4} \times P\left( B \right)\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{1}{4} \times \frac{1}{3}\]
\[ \Rightarrow P\left( A \cap B \right) = \frac{1}{12}\]
\[\text{ Also } , P\left( A \cup B \right) = P\left( A \right) + P\left( B \right) - P\left( A \cap B \right)\]
\[ = \frac{1}{2} + \frac{1}{3} - \frac{1}{12}\]
\[ = \frac{6 + 4 - 1}{12}\]
\[ = \frac{9}{12}\]
\[ = \frac{3}{4}\]
\[\text{ Now } , \]
\[P\left( \overline {A} \cap \overline {B} \right) = P\left( \overline {A \cup B} \right)\]
\[ = 1 - P\left( \overline {A} \cup B \right)\]
\[ = 1 - \frac{3}{4}\]
\[ = \frac{4 - 3}{4}\]
\[ = \frac{1}{4}\]
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