Question

A cricket team of 11 players is to be selected at random out of 15 players of whom 4 are bowlers. The probability that at least 3 bowlers will be selected is

• A. $\frac{{(}^4{C}_3{+}^4{C}_4)}{{}^{11}{C}_5}$
• B. $\frac{{}^4{C}_4}{{}^{15}{C}_{11}}$
• C. $\frac{{(}^4{C}_3{\times }^{11}{C}_8{+}^4{C}_4{\times }^{11}{C}_7)}{{}^{15}{C}_{11}}$
• D. $\frac{{}^4{C}_3}{{}^{15}{C}_{11}}$

Answer 1

Answer: (C)

$\frac{{(}^4{C}_3{\times }^{11}{C}_8{+}^4{C}_4{\times }^{11}{C}_7)}{{}^{15}{C}_{11}}$

Let us choose the bowlers first.
Since we want atleast 3, we can have 2 possibilities - 3 or 4.
If there are 3 bowlers, choose the remaining 8 from the 11 players left $⟹{}^4{C}_3{\times }^{11}{C}_8$.
If there are 4 bowlers, choose the remaining 7 from the 11 players left $⟹{}^4{C}_4{\times }^{11}{C}_7$.
The total number of cases is ${}^{15}{C}_{11}$.
Hence, the required probability $=\frac{{}^4{C}_3{\times }^{11}{C}_8{+}^4{C}_4{\times }^{11}{C}_7}{{}^{15}{C}_{11}}$.