Question

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

• A. $\frac{4}{5}$
• B. $\frac{2}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{3}$

Answer 1

Answer: (D)

Given $P \left( A \cap B \right) = \frac{1}{6}$
$\Rightarrow P\left(A\right)P\left(B\right) =\frac{1}{6}\quad\ldots\left(1\right)$
and $P\left(A^{c} \cap B^{c}\right) =\frac{1}{3}$
$\Rightarrow P\left(A^{c}\right)P\left(B^{c}\right) =\frac{1}{3}$
$\Rightarrow \left(1-P \left(A \right)\right) \left(1-P \left(B \right)\right)= \frac{1}{3}\quad\ldots\left(2\right)$
Eliminating $P\left(B\right)$ between $\left(1\right)$ and $\left(2\right)$, we obtain
$\left(1-P\left(A\right)\right) \left(1-\frac{1}{6P\left(A\right)}\right) = \frac{1}{3}$
$\Rightarrow \left(1 - t\right) \left(6t - 1\right) = 2t$, where $t = P\left(A\right)$
$\Rightarrow 6t^{2}-5t+ 1=0$
$\Rightarrow t = \frac{1}{2}$ or $\frac{1}{3}$
$\Rightarrow P\left(A\right) = \frac{1}{2}$ or $\frac{1}{3}$