A bag contains 3 red,7 white,and 4 black ball | vedclass

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(C) Total number of balls $= 3 + 7 + 4 = 14$.Number of ways to draw $3$ balls out of $14$ is given by $^{14}C_3 = \frac{14 	imes 13 	imes 12}{3 	im

Vedclass · Accepted Answer

$\frac{10}{91}$

Vedclass · Answer

(C) Total number of balls $= 3 + 7 + 4 = 14$.Number of ways to draw $3$ balls out of $14$ is given by $^{14}C_3 = \frac{14 	imes 13 	imes 12}{3 	imes 2 	imes 1} = 14 	imes 13 	imes 2 = 364$.For all three balls to be of the same color,they must be either all red,all white,or all black.Number of ways to choose $3$ red balls $= ^3C_3 = 1$.Number of ways to choose $3$ white balls $= ^7C_3 = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$.Number of ways to choose $3$ black balls $= ^4C_3 = 4$.Total favorable outcomes $= 1 + 35 + 4 = 40$.Required probability $= \frac{40}{364} = \frac{10}{91}$.

$A$ bag contains $3$ red,$7$ white,and $4$ black balls. If three balls are drawn from the bag,the probability that all of them are of the same color is:

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