Question

If $A$ and $B$ are two independents events, the probability that both $A$ and $B$ occur is $\frac{1}{8}$ and the probability that neither of them occurs in $\frac{3}{8}$. The probability of the occurrence of $A$ is:

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

Answer 1

Answer: (A), (C)

$\frac{1}{2}$
$\frac{1}{4}$

Let probability of occurrence of A $=x$

Let probability of occurrence of A $=y$

Probability of occurrence of both =$xy$ = $\frac{1}{8}$

Probability that both will not occur = $(1-x)(1-y)$ = $\frac{3}{8}$

Therefore,

$x+y=\frac{3}{4}$

$y=\frac{3}{4}-x$

Substituting in the above equation,

($\frac{3}{4}-x)x=\frac{1}{8}$

$8x^2-6x+1=0$

x = $\frac{1}{4}$ or $\frac{1}{2}$