A bag contains 3 white and 6 red balls. Four | vedclass

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(C) Total balls = $3 + 6 = 9$. We draw $4$ balls. Total ways = $^9C_4 = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$. We w

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$\frac{20}{21}$

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(C) Total balls = $3 + 6 = 9$. We draw $4$ balls. Total ways = $^9C_4 = \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} = 126$. We want the probability of getting at least $2$ red balls. This is $1 - [P(0 	ext{ red}) + P(1 	ext{ red})]$. $P(0 	ext{ red}) = \frac{^6C_0 	imes ^3C_4}{126} = 0$ (since we cannot draw $4$ white balls). $P(1 	ext{ red}) = \frac{^6C_1 	imes ^3C_3}{126} = \frac{6 	imes 1}{126} = \frac{6}{126}$. Probability of at least $2$ red balls = $1 - (0 + \frac{6}{126}) = 1 - \frac{6}{126} = \frac{120}{126} = \frac{20}{21}$.

$A$ bag contains $3$ white and $6$ red balls. Four balls are drawn at a time randomly. The probability of getting at least $2$ red balls is:

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