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

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(C) Total number of balls $= 2 + 3 + 4 = 9$.The total number of ways to draw $3$ balls from $9$ is given by ${}^{9}C_{3} = \frac{9 	imes 8 	imes 7}{

Vedclass · Accepted Answer

$\frac{5}{84}$

Vedclass · Answer

(C) Total number of balls $= 2 + 3 + 4 = 9$.The total number of ways to draw $3$ balls from $9$ is given by ${}^{9}C_{3} = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$.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 ${}^{3}C_{3} = 1$.The number of ways to draw $3$ white balls from $4$ is ${}^{4}C_{3} = 4$.Therefore,the number of favorable outcomes is ${}^{3}C_{3} + {}^{4}C_{3} = 1 + 4 = 5$.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 three balls are drawn together,what is the probability that the balls drawn are of the same colour?

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