A box contains 12 balls,of which 4 are red,5 | vedclass

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(B) The total number of ways to draw $3$ balls from $12$ is given by $C(12, 3) = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$. We want th

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$\frac{29}{44}$

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(B) The total number of ways to draw $3$ balls from $12$ is given by $C(12, 3) = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$. We want the probability that exactly $2$ balls have the same colour. This means $2$ balls are of one colour and $1$ ball is of a different colour. Case $1$: $2$ red and $1$ non-red (green or white). Number of ways = $C(4, 2) 	imes C(8, 1) = 6 	imes 8 = 48$. Case $2$: $2$ green and $1$ non-green (red or white). Number of ways = $C(5, 2) 	imes C(7, 1) = 10 	imes 7 = 70$. Case $3$: $2$ white and $1$ non-white (red or green). Number of ways = $C(3, 2) 	imes C(9, 1) = 3 	imes 9 = 27$. Total favorable outcomes = $48 + 70 + 27 = 145$. The probability is $\frac{145}{220} = \frac{29}{44}$.

$A$ box contains $12$ balls,of which $4$ are red,$5$ are green,and $3$ are white. If $3$ balls are drawn at random simultaneously from the box,then the probability that exactly $2$ balls have the same colour is

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