A and B are two independent events. The probability that both events A and B occur is $\frac{1}{6}$ neither of them occur is and the probability that neither of them occur is $\frac{1}{3}$. If $P(A) = x, P(B) = y$ then the value of $x+y$ is.

$\frac{5}{6}$

The correct answer is Option (2) → $\frac{5}{6}$ **

Given:

$P(A\cap B)=\frac16$

$P(\text{neither A nor B})=\frac13$

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

Independence implies:

$xy=\frac16$

Neither event occurs:

$P(A'=1-x,\; B'=1-y)$

Since $A$ and $B$ are independent, $A'$ and $B'$ are also independent:

$(1-x)(1-y)=\frac13$

Now solve the system:

1) $xy=\frac16$

2) $(1-x)(1-y)=\frac13$

Expand (2):

$1 - x - y + xy = \frac13$

Substitute $xy=\frac16$:

$1 - x - y + \frac16 = \frac13$

$\frac{7}{6} - x - y = \frac13$

$-x - y = \frac13 - \frac76 = -\frac56$

$x + y = \frac56$

Answer: $\displaystyle x+y=\frac56$