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A college cricket team with $11$ players consists of $4$ batsman, $3$ all-rounders, $3$ bowlers and $1$ wicket keeper. $3$ players are selected randomly. Find the probability that the selection contains a batsman, a bowler and an all-rounder.

  1. $\dfrac{12}{60} \\$
  2. $\dfrac{13}{25} \\$
  3. $\dfrac{12}{55} \\$
  4. $\dfrac{104}{165}$
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$11$ players$:4$ batsman, $3$ all-rounders, $3$ bowlers and $1$ wicket keeper.

$3$ players are selected randomly $\implies ^{11}C_{3} = \dfrac{11!}{8!3!} = \dfrac{11\cdot 10\cdot 9\cdot 8!}{8!\cdot 3\cdot 2\cdot 1} = 165$

The selection contains a batsman, a bowler and an all-rounder $\implies ^{4}C_{1} \times ^{3}C_{1} \times ^{3}C_{1}  = 36$

Probability of an event happening $ = \dfrac{\text{Number of ways it can happen}}{\text{Total number of outcomes}} = \dfrac{36}{165}=\dfrac{12}{55}.$

So, the correct answer is C.
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