A bag contains 4 red,5 white,and 6 black ball | vedclass

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(B) Total number of balls = $4 + 5 + 6 = 15$.Total number of ways to draw $3$ balls from $15$ is given by $^{15}C_3 = \frac{15 	imes 14 	imes 13}{3

Vedclass · Accepted Answer

$24/91$

Vedclass · Answer

(B) Total number of balls = $4 + 5 + 6 = 15$.Total number of ways to draw $3$ balls from $15$ is given by $^{15}C_3 = \frac{15 	imes 14 	imes 13}{3 	imes 2 	imes 1} = 5 	imes 7 	imes 13 = 455$.The number of ways to draw $3$ balls of different colors (one red,one white,and one black) is given by $^4C_1 	imes ^5C_1 	imes ^6C_1 = 4 	imes 5 	imes 6 = 120$.The probability of drawing three balls of different colors is $P = \frac{120}{455}$.Dividing both numerator and denominator by $5$,we get $P = \frac{24}{91}$.

$A$ bag contains $4$ red,$5$ white,and $6$ black balls. If three balls are drawn at random,what is the probability that they are of different colors?

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