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(T) 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)

The correct options are
A $P\left(E\right)=\frac{4}{5},P\left(F\right)=\frac{3}{5}$
D $P\left(E\right)=\frac{3}{5},P\left(F\right)=\frac{4}{5}$


$P(E\cup F)-P(E\cap F)=\frac{11}{25}$ . . . (i)
[i.e. only E or only F]

Probability that neither of them occurs $=\frac{2}{25}$
$\Rightarrow P(\overline{E}\cap \overline{F})=\frac{2}{25}\cdots \left(ii\right)$
From Eq. (i), $P\left(E\right)+P\left(F\right)-2P(E\cap F)=\frac{11}{25}$ . . . (iii)
From Eq. (ii), $(1-P(E\left)\right)(1-P(F\left)\right)=\frac{2}{25}$
$\Rightarrow 1-P\left(E\right)-P\left(F\right)+P\left(E\right).P\left(F\right)=\frac{2}{25}$ . . .(iv)
From Eqs. (iii) and (iv),
$P\left(E\right)+P\left(F\right)=\frac{7}{5}andP\left(E\right).P\left(F\right)=\frac{12}{25}\therefore P\left(E\right).[\frac{7}{5}-P(E\left)\right]=\frac{12}{25}\Rightarrow \left(P\right(E){)}^2-\frac{7}{5}P(E)+\frac{12}{25}=0\Rightarrow \left[P\right(E)-\frac{3}{5}]\left[P\right(E)-\frac{4}{5}]=0\therefore P(E)=\frac{3}{5}or\frac{4}{5}\Rightarrow P(F)=\frac{4}{5}or\frac{3}{5}$