Let X and Y be two events such that P(X) =\frac{1}{3} , \textrm{ } P \left(\right. X \left|\right. Y \left.\right) = \frac{1}{2} \textrm{ }\textrm{ } \textrm{ } and \textrm{ } P \left(\right. Y \left|\right. X \left.\right) \textrm{ } = \frac{2}{5} . \textrm{ }\textrm{ } Then

Engineering

Mathematics

Conditional Probability

Question

Let X and Y be two events such that P(X) = $\frac{1}{3}, P (X | Y) = \frac{1}{2} and P (Y | X) = \frac{2}{5} . Then$

$P (X \cup Y) = \frac{2}{5}$

$P (Y) = \frac{4}{15}$

$P (X \cap Y) = \frac{1}{5}$

P(X'|Y) = $\frac{1}{2}$

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P(X) = $\frac{1}{3}; \frac{P (Y \cap X)}{P (X)} = \frac{2}{5}$ ⇒ $P (X \cap Y) = \frac{2}{15}$

P(X|Y) = $\frac{1}{2} \Rightarrow \frac{2 / 15}{P (Y)} = \frac{1}{2} \Rightarrow P (Y) = \frac{4}{15}$ ⇒ (B)

P(X'|Y) = $\frac{P (X' \cap Y)}{P (Y)} = \frac{P (Y) - P (X \cap Y)}{P (Y)} = \frac{\frac{4}{15} - \frac{2}{15}}{\frac{4}{15}} = \frac{1}{2}$ ⇒ (C)

$P (X \cup Y) = \frac{1}{3} + \frac{4}{15} - \frac{2}{15} = \frac{7}{15} .$