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Let $A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur together is $1/6$ and the probability that neither of them occurs is $1/3$. The probability of occurrence of $A$ is
$0$ or $1$
$\frac{1}{2}$ or $\frac{1}{3}$
$\frac{1}{2}$ or $\frac{1}{4}$
$\frac{1}{3}$ or $\frac{1}{4}$
Solution
(b) $P(A \cap B) = \frac{1}{6}$ and $P({A^c} \cap {B^c}) = \frac{1}{3}$
Now $P{(A \cup B)^c} = P({A^c} \cap {B^c}) = \frac{1}{3}$
$⇒$ $1 – P(A \cup B) = \frac{1}{3}$$ \Rightarrow \,P(A \cup B) = \frac{2}{3}$
But $P(A \cup B) = P(A) + P(B) – P(A \cap B)$
$ \Rightarrow P(A)\, + P(B) = \frac{5}{6}$$…..(i)$
$\because$ $A$ and $B$ are independent events
$\therefore$ $P(A \cap B)\, = \,P(A)\,P(B)$ $⇒$ $P(A)\,P(B) = \frac{1}{6}$
${[P(A)\, – \,P(B)\,]^2} = {[P(A) + P(B)]^2} – 4P(A)\,P(B)$
$ = \frac{{25}}{{36}} – \frac{4}{6} = \frac{1}{{36}}$
$ \Rightarrow $ $P(A) – P(B) = \pm \,\frac{1}{6}$……$(ii)$
Solving $(i)$ and $(ii)$, we get $P(A) = \frac{1}{2}$ or $\frac{1}{3}.$
