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Question
If A and B are two events such that
\[ P\left( A \right) = \frac{1}{2}, P\left( B \right) = \frac{1}{3} \text{ and } P\left( A \cap B \right) = \frac{1}{4}, \text{ then find } P\left( A|B \right), P\left( B|A \right), P\left( \overline{ A }|B \right) \text{ and } P\left( \overline{ A }|\overline{ B } \right) .\]
Solution
We have ,
\[P\left( A \right) = \frac{1}{2}, P\left( B \right) = \frac{1}{3} \text{ and } P\left( A \cap B \right) = \frac{1}{4}\]
\[\text{ Also } , P\left( B \right) = 1 - P\left( B \right) = 1 - \frac{1}{3} = \frac{2}{3}\]
\[\text{ As } , P\left( A \cup B \right) = P\left( A \right) + P\left( B \right) - P\left( A \cap B \right)\]
\[ = \frac{1}{2} + \frac{1}{3} - \frac{1}{4}\]
\[ = \frac{6 + 4 - 3}{12}\]
\[ \Rightarrow P\left( A \cup B \right) = \frac{7}{12}\]
\[\text{ Also } , P\left(\overline{ A } \cap B \right) = P\left( B \right) - P\left( A \cap B \right)\]
\[ \Rightarrow P\left( \overline{ A } \cap B \right) = \frac{1}{3} - \frac{1}{4}\]
\[ \Rightarrow P\left( \overline{ A } \cap B \right) = \frac{4 - 3}{12}\]
\[ \Rightarrow P\left(\overline{ A } \cap B \right) = \frac{1}{12}\]
\[\text{ And } , P\left( \overline{ A } \cap \overline{ B } \right) = P\left( \overline{ A \cup B } \right)\]
\[ = 1 - P\left( A \cup B \right)\]
\[ = 1 - \frac{7}{12}\]
\[ = \frac{5}{12}\]
\[ \text{ Now } , \]
\[P\left( A|B \right) = \frac{P\left( A \cap B \right)}{P\left( B \right)} = \frac{\left( \frac{1}{4} \right)}{\left( \frac{1}{3} \right)} = \frac{3}{4}, \]
\[P\left( B|A \right) = \frac{P\left( A \cap B \right)}{P\left( A \right)} = \frac{\left( \frac{1}{4} \right)}{\left( \frac{1}{2} \right)} = \frac{2}{4} = \frac{1}{2}, \]
\[P\left( \overline{ A }|B \right) = \frac{P\left( \overline{ A} \cap B \right)}{P\left( B \right)} = \frac{\left( \frac{1}{12} \right)}{\left( \frac{1}{3} \right)} = \frac{3}{12} = \frac{1}{4} \text{ and } \]
\[P\left( \overline{ A }|\overline{ B } \right) = \frac{P\left( \overline{ A } \cap \overline{ B} \right)}{P\left( B \right)} = \frac{\left( \frac{5}{12} \right)}{\left( \frac{2}{3} \right)} = \frac{15}{24} = \frac{5}{8}\]
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