Question

Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $\frac{11}{25}$ and the probability of none of them occurring is $\frac{2}{25}$. If $P\left(T\right)$ denotes the probability of occurrence of the event $T$, then

• A. $P\left(E\right)=\frac{4}{5}$ , $P\left(F\right)=\frac{3}{5}$
• B. $P\left(E\right)=\frac{1}{5},P\left(F\right)=\frac{2}{5}$
• C. $P\left(E\right)=\frac{2}{5}$ , $P\left(F\right)=\frac{1}{5}$
• D. $P\left(E\right)=\frac{3}{5},P\left(F\right)=\frac{4}{5}$

Answer 1

Answer: (A), (D)

$P\left(E\right)=\frac{4}{5}$ , $P\left(F\right)=\frac{3}{5}$
$P\left(E\right)=\frac{3}{5},P\left(F\right)=\frac{4}{5}$

Given, $P(E,F^{\prime })$ or $P(F,E^{\prime })=\frac{11}{25}$
$\Rightarrow P\left(E\right)P\left(F^{\prime }\right)+P\left(F\right)P\left(E^{\prime }\right)=\frac{11}{25}$

Let $P\left(E\right)=x,P\left(F\right)=y$
$\Rightarrow x(1-y)+y(1-x)=\frac{11}{25}$ ......(1)

$P(E^{\prime },F^{\prime })=\frac{11}{25}$
$\Rightarrow P\left(E^{\prime }\right)P\left(F^{\prime }\right)=\frac{2}{25}$
or $(1-x)(1-y)=\frac{2}{25}$ .......(2)

Solving (1) and (2), we get
$x=\frac{3}{5},y=\frac{4}{5}$ or $x=\frac{4}{5},y=\frac{3}{5}$

$P\left(E\right)=\frac{3}{5},P\left(F\right)=\frac{4}{5}$ or $P\left(E\right)=\frac{4}{5},P\left(F\right)=\frac{3}{5}$