A and B are two independent events. The probability that both events A and B occur is $\frac{1}{6}$ 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.

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

Since A and B are independent:

$P(A \text{ and } B) = P(A) \times P(B)$

$xy = \frac{1}{6}$

---

For the probability that neither occurs, since A and B are independent, A' and B' are also independent:

$P(A' \text{ and } B') = P(A') \times P(B')$

Where $P(A') = 1 - x$ and $P(B') = 1 - y$:

$(1-x)(1-y) = \frac{1}{3}$

---

Expanding the equation:

$(1-x)(1-y) = \frac{1}{3}$

$1 - x - y + xy = \frac{1}{3}$

---

Substituting $xy = \frac{1}{6}$:

$1 - x - y + \frac{1}{6} = \frac{1}{3}$

$\frac{7}{6} - (x+y) = \frac{1}{3}$

$\frac{7}{6} - (x+y) = \frac{2}{6}$

$x + y = \frac{7}{6} - \frac{2}{6}$

$x + y = \frac{5}{6}$

Therefore, the value of x + y is $\frac{5}{6}$.