Question

The probability of the occurrence of two independent events $A$ and $B$ is $\frac{1}{6}.$ The probability that neither occurs is $\frac{1}{3}.$ Then the probability of occurrence of only A is

• A. $\frac{1}{3}$ only
• B. $\frac{1}{3}$ or $\frac{1}{2}$
• C. $\frac{1}{2}$ only
• D. None of these

Answer 1

The correct option is B $\frac{1}{3}$ or $\frac{1}{2}$

$P(A\cap B)=P\left(A\right)\times P\left(B\right)$ since the events are independent
Given $P\left(A\right)\times P\left(B\right)=\frac{1}{6}$ ... (1)
$\Rightarrow xy=\frac{1}{6}$ (say)
Since A and B are independent , $\overline{A}$ and $\overline{B}$ are also independent.
$\therefore$ $P\left(\overline{A}\right)\times P\left(\overline{B}\right)=\frac{1}{3}$
$(1-x\left)\right(1-y)=\frac{1}{3}$
1 $-x-y+xy=\frac{1}{3}$
Hence $x+y=1+\frac{1}{6}-\frac{1}{3}=\frac{5}{6}$
$\therefore$ either $x=\frac{1}{2}$ and
$y=\frac{1}{3}$ or$x=\frac{1}{3}$ and $y=\frac{1}{2}$