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Question
If A and B are two independent events such that P (`bar A` ∩ B) = 2/15 and P (A ∩`bar B` ) = 1/6, then find P (B).
Solution
\[\text{ Let }: \]
\[P\left( A \right) = x \]
\[P\left( B \right) = y\]
\[P\left( \bar{A} \cap B \right) = \frac{2}{15}\]
\[ \Rightarrow P\left( \bar{A} \right) \times P\left( B \right) = \frac{2}{15}\]
\[ \Rightarrow \left( 1 - x \right)y = \frac{2}{15} . . . \left( 1 \right)\]
\[P\left( A \cap \bar{B} \right) = \frac{1}{6}\]
\[ \Rightarrow P\left( A \right) \times P\left( B \right) = \frac{1}{6} \]
\[ \Rightarrow \left( 1 - y \right)x = \frac{1}{6} . . . \left( 2 \right)\]
\[\text{ Subtracting} \left( 2 \right) \text{from} \left( 1 \right), \text{ we get } \]
\[x - y = \frac{1}{30}\]
\[x = y + \frac{1}{30}\]
\[\text {Substituting the value of x in } \left( 2 \right),\text{ we get } \]
\[\left( y + \frac{1}{30} \right)\left( 1 - y \right) = \frac{1}{6}\]
\[ \Rightarrow 30 y^2 - 29y + 4 = 0\]
\[ \Rightarrow y = \frac{1}{6}, \frac{4}{5}\]
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