Probability Ques 62
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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Answer:
Correct Answer: 62.(a,d)
Solution:
Formula:
Set theoretical notation of probability and some important results:
$P(E \cup F)-P(E \cap F)=\frac{11}{25} \quad …….(i)$
[i.e. only $E$ or only $F$ ]
Neither of them occurs $=\frac{2}{25}$
$\Rightarrow \quad P(\bar{E} \cap \bar{F})=\frac{2}{25}\quad …….(ii)$
From Eq. (i), $P(E)+P(F)-2 P(E \cap F)=\frac{11}{25}\quad …….(iii)$
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} \quad …….(iv)$
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} $
BETA
