$A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur is $1/6$ and the probability that neither of them occur is $1/3$. Find the probability of the occurrence of $A$.

$\frac{1}{6}$ or $\frac{1}{3}$

The correct answer is Option (2) → $\frac{1}{6}$ or $\frac{1}{3}$ ##

Let $P(A) = x$ and $P(B) = y$.

According to the question:

$P(A \cap B) = \frac{1}{6} \text{ and } P(A' \cap B') = \frac{1}{3}$

$P(A)P(B) = \frac{1}{6} \text{ and } P(A')P(B') = \frac{1}{3}$

$xy = \frac{1}{6} \text{ and } (1-x)(1-y) = \frac{1}{3}$

On solving we get $x = \frac{1}{2}$ or $\frac{1}{3}$.