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# A and B are two independent events. The probability that both A and B occur is $\large\frac{1}{6}$ and the probability that neither of them occurs is $\large\frac{1}{3}$. Then the probability of the two events are respectively

$\begin {array} {1 1} (A)\;\large\frac{1}{2},\large\frac{1}{3} & \quad (B)\;\large\frac{1}{5},\large\frac{1}{6} \\ (C)\;\large\frac{1}{2},\large\frac{1}{6} & \quad (D)\;\large\frac{2}{3},\large\frac{1}{4} \end {array}$

$P(A \cap B ) = \large\frac{1}{6}$
$P(\overline A \cap \overline B ) = \large\frac{1}{3}$
$\Rightarrow P(A)\: P(B) = \large\frac{1}{6}$
$P(\overline{A \cup B} ) = \large\frac{1}{3}$
$P(A \cup B ) = 1-\large\frac{1}{3}$
$= \large\frac{2}{3}$
$\therefore P(A)+P(B) - P( A \cap B ) = \large\frac{2}{3}$
$\Rightarrow P(A) + P(B) = \large\frac{2}{3}+ \large\frac{1}{6}$
$= \large\frac{5}{6}$
Solving (1) and (2) $P(A) = \large\frac{1}{2}$
$P(B) = \large\frac{1}{3}$
$P(A) = \large\frac{1}{3}\: P(B) = \large\frac{1}{2}$
Ans : (A)