$11$ 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.