In a box,there are 2 red,3 black,and 4 white | vedclass

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(C) Total number of ways to draw $3$ balls from $9$ balls is ${}^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$. For the balls to be of

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$\frac{5}{84}$

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(C) Total number of ways to draw $3$ balls from $9$ balls is ${}^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$. For the balls to be of the same color,they must be either all black or all white (since there are only $2$ red balls,we cannot draw $3$ red balls). Number of ways to draw $3$ black balls = ${}^3C_3 = 1$. Number of ways to draw $3$ white balls = ${}^4C_3 = 4$. Total favorable outcomes = $1 + 4 = 5$. Required probability = $\frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{5}{84}$.

In a box,there are $2$ red,$3$ black,and $4$ white balls. If $3$ balls are drawn together,what is the probability that they are of the same color?

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