Probability Ques 65
Let $E$ and $F$ be two independent events. If the probability that both $E$ and $F$ happen is $1 / 12$ and the probability that neither $E$ nor $F$ happen is $1 / 2$. Then,
(a) $P(E)=1 / 3, P(F)=1 / 4$
(b) $P(E)=1 / 2, P(F)=1 / 6$
(c) $P(E)=1 / 6, P(F)=1 / 2$
(d) $P(E)=1 / 4, P(F)=1 / 3$
$(1993,2 M)$
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Answer:
Correct Answer: 65.(a,d)
Solution:
Formula:
Set theoretical notation of probability and some important results:
- Both $E$ and $F$ happen $\Rightarrow P(E \cap F)=\frac{1}{12}$
and neither $E$ nor $F$ happens $\Rightarrow P(\bar{E} \cap \bar{F})=\frac{1}{2}$
But for independent events, we have
$ P(E \cap F)=P(E) P(F)=\frac{1}{12} \quad …….(i)$
and
$ \begin{aligned} P(\bar{E} \cap \bar{F}) & =P(\bar{E}) P(\bar{F}) \\ & =\{1-P(E)\}\{(1-P(F)\} \\ & =1-P(E)-P(F)+P(E) P(F) \end{aligned} $
$ \begin{aligned} & \Rightarrow \quad \frac{1}{2}=1-\{P(E)+P(F)\}+\frac{1}{12} \\ & \Rightarrow \quad P(E)+P(F)=1-\frac{1}{2}+\frac{1}{12}=\frac{7}{12} \quad …….(ii) \end{aligned} $
On solving Eqs. (i) and (ii), we get
$\text { either } P(E)=\frac{1}{3} \text { and } P(F)=\frac{1}{4} $
$\text { or } P(E)=\frac{1}{4} \text { and } P(F)=\frac{1}{3}$
BETA
