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

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Question

(C) The total number of balls in the box is $2 + 3 + 4 = 9$.The total number of ways to draw $3$ balls out of $9$ is given by ${}^9C_3 = \frac{9 	ime

Vedclass · Accepted Answer

$\frac{5}{84}$

Vedclass · Answer

(C) The total number of balls in the box is $2 + 3 + 4 = 9$.The total number of ways to draw $3$ balls out of $9$ is given by ${}^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.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,it is impossible to draw $3$ red balls).The number of ways to draw $3$ black balls from $3$ is ${}^3C_3 = 1$.The number of ways to draw $3$ white balls from $4$ is ${}^4C_3 = 4$.Therefore,the total number of favorable ways is $1 + 4 = 5$.The required probability is $\frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{5}{84}$.

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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