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

Answer 1

Answer: (A), (D)

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