A) 7/13
B) 6/13
C) 11/15
D) 12/13
Correct Answer: D
Solution :
[d] 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 mutually 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, required probability \[=P(I)+P(II)+P(III)\] \[=\frac{^{5}{{C}_{3}}{{\times }^{10}}{{C}_{8}}}{^{15}{{C}_{11}}}+\frac{^{5}{{C}_{4}}{{\times }^{10}}{{C}_{7}}}{^{15}{{C}_{11}}}+\frac{^{5}{{C}_{5}}{{\times }^{10}}{{C}_{6}}}{^{15}{{C}_{11}}}=\frac{12}{13}.\]
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