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

Answer 1

Answer: (B), (C)

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

Given that:

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

Now, we know that:

$P\left(\frac{Y}{X}\right)P\left(X\right)=P(X\cap Y)$

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

$P\left(Y\right)=\frac{P(X\cap Y)}{P\left(\frac{X}{Y}\right)}=\frac{\frac{2}{15}}{\frac{1}{2}}=\frac{4}{15}$

Now,

$P(X\cup Y)=P\left(X\right)+P\left(Y\right)-P(X\cap Y)$

$=\frac{1}{3}+\frac{4}{15}-\frac{2}{15}$

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

$P\left(\frac{X^{\prime }}{Y}\right)=\frac{P(X^{\prime }\cap Y)}{P\left(Y\right)}$

$=\frac{P\left(Y\right)-P(X\cap Y)}{P\left(Y\right)}$

$=1-P\left(\frac{X}{Y}\right)$

$=1-\frac{1}{2}=\frac{1}{2}$

Hence, the correct options are $B,C$.