Discussion Forum

Let X and Y be two events such that  $P(X|Y)=\frac{1}{2},P(Y|X)=\frac{1}{3}$ and  $P(X \cap Y) \frac{1}{6}$ , which of the following is/are correct?

Answer:

Explanation:

 Concept involved 

(i) Conditional probability i.e

$P(A/B)=\frac{P(A\cap B)}{P(B)}$

  (ii) $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

(iii) independent event, then

 $P(A \cap B)=P(A).P(B)$

  sol. Here, $P(X/Y)=\frac{1}{2},P\left(\frac{Y}{X}\right)=\frac{1}{3}$

 and   $P(X \cap Y)=6$

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

 $\Rightarrow$  $\frac{1}{2}= \frac{1/6}{P(Y)}$

 $\Rightarrow$  $P(Y)=\frac{1}{3}$............(i)

    $P(\frac{Y}{X})=\frac{1}{3}$

  $\Rightarrow$   $\frac{ P(X\cap Y)}{P(X)}=\frac{1}{3}$

$\Rightarrow$  $\frac{1}{6}=\frac{1}{3}P(X)$

$\therefore$  $P(X)=\frac{1}{2}$...........(ii)

 $p(X \cup Y)=P(X)+P(Y)-P(X \cap Y)$

 =$\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}$..........(iii)

   $P(X \cap Y)=\frac{1}{6}$

 and    $P(X) P(Y)=\frac{1}{2} \frac{1}{3}=\frac{1}{6}$

 $\Rightarrow$  $P(X \cap Y)=P(X) .P(Y)$

 $\Rightarrow$ Independent events ...........(iv)

   $P(X^{c} \cap Y)=P(Y)-P(X \cap Y)$

  =$\frac{1}{3}-\frac{1}{6}=\frac{1}{6}$ ..........(iv)