Question

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

• A. $P\left(Y\right)=\frac{4}{15}$
• B. $P\left(X^{\prime }|Y\right)=\frac{1}{2}$
• C. $P(X\cap Y)=\frac{1}{5}$
• D. $P(X\cup Y)=\frac{2}{5}$

Answer 1

Answer: (A), (B)

The correct options are
A $P\left(Y\right)=\frac{4}{15}$
B $P\left(X^{\prime }|Y\right)=\frac{1}{2}$

$P\left(X\right)=\frac{1}{3},P\left(X|Y\right)=\frac{1}{2}$ and $P\left(Y|X\right)=\frac{2}{5}$

$P\left(X|Y\right)=\frac{P(X\cap Y)}{P\left(Y\right)}=\frac{1}{2}........\left(i\right)P\left(Y|X\right)=\frac{P(X\cap Y)}{P\left(X\right)}=\frac{2}{5}.......\left(ii\right)\Rightarrow P(X\cap Y)=\frac{2}{5}\times \frac{1}{3}\Rightarrow P(X\cap Y)=\frac{2}{15}$
Option C is incorrect.
Using equation $\left(i\right)$,
$P\left(Y\right)=2\times \frac{2}{15}=\frac{4}{15}$
Option A is correct.
$P(X\cup Y)=P\left(X\right)+P\left(Y\right)-P(X\cap Y)=\frac{1}{3}+\frac{4}{15}-\frac{2}{15}=\frac{7}{15}$
Option D is incorrect.
$P\left(X^{\prime }|Y\right)=\frac{P(X^{\prime }\cap Y)}{P\left(Y\right)}=\frac{P(X\cup Y)-P\left(X\right)}{P\left(Y\right)}=\frac{7/15-1/3}{4/15}=\frac{1}{2}$
Option B is correct.