If X and Y be two events such that $P(X/Y)=\frac{1}{2}$, $P(Y/X)=\frac{1}{3}$ and $P(X \cap Y)=\frac{1}{6}$, then $P(X \cup Y)$ is: (a) $\frac{1}{6}$ (b) $\frac{1}{3}$ (c) $\frac{2}{3}$ (d) $\frac{2}{5}$

If X and Y be two events such that P(X/Y)=(1)/(2), P(Y/X)=(1)/(3) and P(X ∩ Y)=(1)/(6), then P(X ∪ Y) is: (a)

Question 68 of 89

P(X/Y)=\frac{1}{2}

(a)

\frac{1}{6}

(b)

\frac{1}{3}

(c)

\frac{2}{3}

(d)

\frac{2}{5}

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2022MCQ· 1mImportance★★★★★

76% · 68/89 Questions

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Solving for $P(X)=\frac12$ and $P(Y)=\frac13$ from the given conditionals, the addition rule gives $P(X\cup Y)=\frac23$ .

From $P(X/Y)=\dfrac{P(X\cap Y)}{P(Y)}$ : $\dfrac12 = \dfrac{1/6}{P(Y)} \Rightarrow P(Y) = \dfrac{1/6}{1/2} = \dfrac13$ .

From $P(Y/X)=\dfrac{P(X\cap Y)}{P(X)}$ : $\dfrac13 = \dfrac{1/6}{P(X)} \Rightarrow P(X) = \dfrac{1/6}{1/3} = \dfrac12$ .

…