Question

A cricket club has $15$ members, of whom only $5$ can bowl. If the names of $15$ members are put into a box and $11$ are drawn at arandom, then the probability of getting an eleven containing at least $3$ bowlers is

• A. $\frac{7}{13}$
• B. $\frac{6}{13}$
• C. $\frac{11}{15}$
• D. $\frac{12}{13}$

Answer 1

The correct option is D $\frac{12}{13}$

The total number of ways of choosing $11$ players out of $15$ is ${}^{15}{C}_{11}$. A team of $11$ players containing at least $3$ bowlers can be chosen in the following exclusive ways:
(I) Three bowlers out of $5$ bowlers and $8$ other players out of the remaining $10$ players.
(II) Four bowlers out of $5$ bowlers and $7$ other players out of the remaining $10$ players.
(III) Five bowlers out of $5$ bowlers and $6$ other players out of the remaining $10$ players.
So, the recquired probability is
$P\left(I\right)+P\left(II\right)+P\left(III\right)=\frac{{}^5{C}_3{\times }^{10}{C}_8}{{}^{15}{C}_{11}}+\frac{{}^5{C}_4{\times }^{10}{C}_7}{{}^{15}{C}_{11}}+\frac{{}^5{C}_5{\times }^{10}{C}_6}{{}^{15}{C}_{11}}=\frac{1260}{1365}=\frac{12}{13}$