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

Answer 1

Answer: (A), (B)

$P\left(\frac{X^{\prime }}{Y}\right)=\frac{1}{2}$

$P\left(X\right)=\frac{1}{3}...\left(i\right)P\left(X|Y\right)=\frac{P(X\cap Y)}{P\left(Y\right)}=\frac{1}{2}...\left(ii\right)P\left(Y|X\right)=\frac{P(X\cap Y)}{P\left(X\right)}=\frac{2}{5}...\left(iii\right)\text{By solving eqns. (i), (ii) and (iii) . . .}P(X\cap Y)=\frac{2}{15}P\left(Y\right)=\frac{4}{15}P\left(X^{\prime }|Y\right)=\frac{P\left(Y\right)-P(X\cap Y)}{P\left(Y\right)}=\frac{\frac{4}{15}-\frac{2}{15}}{\frac{4}{15}}=\frac{1}{2}P(X\cup Y)=\frac{1}{3}+\frac{4}{15}-\frac{2}{15}=\frac{7}{15}=\frac{7}{15}$