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

Updated On: Oct 4, 2024
  • \(\frac{7}{13}\)
  • \(\frac{6}{13}\)
  • \(\frac{11}{15}\)
  • \(\frac{12}{13}\)
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The Correct Option is D

Solution and Explanation

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\(\frac{1260}{1365}=\frac{12}{13}\)
 
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