Probability 3 Question 23
23. 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
(2011)
(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) $P(E)=\frac{3}{5}, P(F)=\frac{4}{5}$
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Solution:
$P(E \cup F)-P(E \cap F)=\frac{11}{25}$
[i.e. only $E$ or only $F$ ]
Neither of them occurs $=\frac{2}{25}$
$\Rightarrow \quad P(\bar{E} \cap \bar{F})=\frac{2}{25}$
From Eq. (i), $P(E)+P(F)-2 P(E \cap F)=\frac{11}{25}$
From Eq. (ii),
$$ (1-P(E))(1-P(F))=\frac{2}{25} $$
$$ \Rightarrow \quad 1-P(E)-P(F)+P(E) \cdot P(F)=\frac{2}{25} $$
From Eqs. (iii) and (iv),
$$ \begin{aligned} & P(E)+P(F)=\frac{7}{5} \text { and } P(E) \cdot P(F)=\frac{12}{25} \\ & \therefore \quad P(E) \cdot \frac{7}{5}-P(E)=\frac{12}{25} \\ & \Rightarrow \quad(P(E))^{2}-\frac{7}{5} P(E)+\frac{12}{25}=0 \\ & \Rightarrow \quad P(E)-\frac{3}{5} \quad P(E)-\frac{4}{5}=0 \\ & \therefore \quad P(E)=\frac{3}{5} \text { or } \frac{4}{5} \Rightarrow P(F)=\frac{4}{5} \text { or } \frac{3}{5} \end{aligned} $$
