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.
Answer & explanation
Correct answer: option 2
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: $\displaystyle x+y=\frac56$