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

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(D) Total number of balls = $4 + 5 + 6 = 15$.We need to select $3$ balls out of $15$. The total number of ways to select $3$ balls is given by ${}^{15

Vedclass · Accepted Answer

$\frac{24}{91}$

Vedclass · Answer

(D) Total number of balls = $4 + 5 + 6 = 15$.We need to select $3$ balls out of $15$. The total number of ways to select $3$ balls 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 select $1$ white,$1$ red,and $1$ green ball is ${}^4C_1 	imes {}^5C_1 	imes {}^6C_1 = 4 	imes 5 	imes 6 = 120$.The required probability is $\frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{120}{455}$.Dividing both numerator and denominator by $5$,we get $\frac{24}{91}$.

$A$ bag contains $4$ white,$5$ red,and $6$ green balls. Three balls are picked up randomly. The probability that a white,a red,and a green ball is drawn is

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