Question

A cricket club has 15 members, of whom only 5 can bowl. If the names of the 15 members are put into a box and 11 drawn at random, then the chance of obtaining 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

Answer: (D)

$\frac{12}{13}$

Answer 2

The correct is (D); Number of ways of selecting 11 players out of 15 are $^{11}C_{15}$
i.e.$\frac{15\times14\times13\times12}{4\times3\times2\times}=1365$
Number of bowlers = 5 and other players = 10
Favorable outcomes = (3 bowlers and 8 others) or (4 bowlers and 7 others) or (5 bowlers and 6 others)
(^5C_3$$\times$$^{10}C_8)+(^5C_4$$\times$$^{10}C_7)+(^5C_5$$\times$$^{10}C_6)
$(10 × 45) + (5 × 120) + (1 × 210)$
$450 + 600 + 210 = 1260$
$\therefore$Probability is$\frac{1260}{1365}=\frac{12}{13}$