Probability 3 Question 26
26. 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) $P(E)=1 / 3, P(F)=1 / 4$
$(1993,2 M)$
(b) $P(E)=1 / 2, P(F)=1 / 6$
(c) $P(E)=1 / 6, P(F)=1 / 2$
(d) $P(E)=1 / 4, P(F)=1 / 3$
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Solution:
- 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
$$ P(E \cap F)=P(E) P(F)=\frac{1}{12} $$
and
$$ \begin{aligned} P(\bar{E} \cap \bar{F}) & =P(\bar{E}) P(\bar{F}) \\ & ={1-P(E)}{(1-P(F)} \\ & =1-P(E)-P(F)+P(E) P(F) \end{aligned} $$
$$ \begin{aligned} & \Rightarrow \quad \frac{1}{2}=1-{P(E)+P(F)}+\frac{1}{12} \\ & \Rightarrow \quad P(E)+P(F)=1-\frac{1}{2}+\frac{1}{12}=\frac{7}{12} \end{aligned} $$
On solving Eqs. (i) and (ii), we get
$$ \begin{array}{lll} \text { either } & P(E)=\frac{1}{3} \text { and } & P(F)=\frac{1}{4} \\ \text { or } & P(E)=\frac{1}{4} \text { and } & P(F)=\frac{1}{3} \end{array} $$
