Question

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

• A. $\frac{12}{13}$
• B. $\frac{11}{17}$
• C. $\frac{16}{21}$
• D. $\frac{17}{21}$

Answer 1

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

Total ways of selecting $11$ members from $15$ members is ${}^{15}{C}_{11}.$
Favourable ways of selecting at least $3$ bowlers is
$\left(i\right)$ Select $3$ bowlers and rest $8$ others players out of the remaining $10$ players.
$\left(ii\right)$ Select $4$ bowlers and rest $7$ others players out of the remaining $10$ players.
$\left(iii\right)$ Select $5$ bowlers and rest $6$ others players out of the remaining $10$ players.
$\therefore$ Total favourable ways $={}^5{C}_3\times {}^{10}{C}_8+{}^5{C}_4\times {}^{10}{C}_7+{}^5{C}_5\times {}^{10}{C}_6$
$\Rightarrow$ Required probability $=\frac{{}^5{C}_3\times {}^{10}{C}_8+{}^5{C}_4\times {}^{10}{C}_7+{}^5{C}_5\times {}^{10}{C}_6}{{}^{15}{C}_{11}}$
$=\frac{1260}{1365}=\frac{12}{13}$