Question

The probabilities of four cricketers $A,B,C$ and $D$ scoring more than $50$ runs in a match are $\frac{1}{2},\frac{1}{3},\frac{1}{4}$ and $\frac{1}{10}$. It is known that exactly two of the players scored more than $50$ in a particular match. The probability that these players were $A$ and $B$ is

• A. $\frac{27}{65}$
• B. $\frac{5}{6}$
• C. $\frac{1}{6}$
• D. None of these

Answer 1

The correct option is A $\frac{27}{65}$

Let $E_1$ be the event that exactly two players scored more than $50$ runs then $P\left(E_1\right)=\frac{1}{2}\times \frac{1}{3}\times \frac{3}{4}\times \frac{9}{10}$
$+\frac{1}{2}\times \frac{2}{3}\times \frac{1}{4}\times \frac{9}{10}+\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \frac{1}{10}+\frac{1}{2}\times \frac{1}{3}\times \frac{1}{4}\times \frac{9}{10}$
$+\frac{1}{2}\times \frac{1}{3}\times \frac{3}{4}\times \frac{1}{10}+\frac{1}{2}\times \frac{2}{3}\times \frac{1}{4}\times \frac{1}{10}=\frac{65}{240}$
Let $E_2$ be the event that $A$ and $B$ scored more than $50$ runs, then
$P(E_1\cap E_2)=\frac{1}{2}\times \frac{1}{3}\times \frac{3}{4}\times \frac{9}{10}=\frac{27}{240}$
$\therefore$ Required probability $=P\left(\frac{E_2}{E_1}\right)=\frac{P(E_1\cap E_2)}{P\left(E_1\right)}=\frac{27}{65}$