14.Probability
hard

Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happens is $\frac{1}{{12}}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2},$ then

A

$P\,(E) = \frac{1}{3},\,\,P\,(F) = \frac{1}{4}$

B

$P\,(E) = \frac{1}{2},\,\,P\,(F) = \frac{1}{6}$

C

$P\,(E) = \frac{1}{6},\,\,P\,(F) = \frac{1}{2}$

D

None of these

(IIT-1993)

Solution

(a) We are given $P(E \cap F) = \frac{1}{{12}}$ and $P(\bar E \cap \bar F) = \frac{1}{2}$
$ \Rightarrow P(E)\,.\,P(F) = \frac{1}{{12}}$…..$(i)$
and $P(\overline E )\,.\,P(\bar F) = \frac{1}{2}$…..$(ii)$
$ \Rightarrow \{ (1 – P(E)\} $ $\{ (1 – P)(F)\} = \frac{1}{2}$
$ \Rightarrow 1 + P(E)P(F) – P(E) – P(F) = \frac{1}{2}$
$ \Rightarrow 1 + \frac{1}{{12}} – [P(E) + P(F)] = \frac{1}{2}$
$ \Rightarrow P(E) + P(F) = \frac{7}{{12}}$…..$(iii)$
On solving $(i)$ and $(iii),$ we get
$P(E) = \frac{1}{3},\,\frac{1}{4}$ and $P(F) = \frac{1}{4},\,\frac{1}{3}$.

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.