Let X and Y be two events such that P(X)=13, P(X/Y)=12 and P(Y/X)=25. Then,

A
$P (Y) = \frac{7}{15}$
B
$P (X \cap Y) = \frac{1}{5}$ $P (X \cup Y) = \frac{2}{6}$
C
$P (X \cup Y) = \frac{2}{5}$
D
$P (Y) = \frac{4}{15}$

Solution:

We have,

$P (Y / X) = \frac{2}{5} \Rightarrow \frac{P (X \cap Y)}{P (X)} = \frac{2}{5} \Rightarrow P (X \cap Y) = \frac{2}{5} P (X) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}$

It is given that

$P (X / Y) = \frac{1}{2} \Rightarrow \frac{P (X \cap Y)}{P (Y)} = \frac{1}{2} \Rightarrow P (Y) = 2 P (X \cap Y) = 2 \times \frac{2}{15} = \frac{4}{15}$

So, option $(d)$ is correct.

Thus, we have

$P (X) = \frac{1}{3}, P (Y) = \frac{4}{15}$ and $P (X \cap Y) = \frac{2}{15}$

$\begin{aligned} ∴ P (X \cup Y) = P (X) + P (Y) - P (X \cap Y) \\ = \frac{1}{3} + \frac{4}{15} - \frac{2}{15} = \frac{7}{15} \end{aligned}$

So,

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,

Solution:

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