A bag contains 6 red,4 white and 8 blue balls | vedclass

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(C) Total number of balls = $6 + 4 + 8 = 18$.Number of ways to draw $3$ balls from $18$ is given by ${}^{18}C_3 = \frac{18 	imes 17 	imes 16}{3 	im

Vedclass · Accepted Answer

$\frac{3}{68}$

Vedclass · Answer

(C) Total number of balls = $6 + 4 + 8 = 18$.Number of ways to draw $3$ balls from $18$ is given by ${}^{18}C_3 = \frac{18 	imes 17 	imes 16}{3 	imes 2 	imes 1} = 3 	imes 17 	imes 16 = 816$.We need to draw $2$ white balls from $4$ and $1$ red ball from $6$.Number of ways to choose $2$ white balls = ${}^4C_2 = \frac{4 	imes 3}{2 	imes 1} = 6$.Number of ways to choose $1$ red ball = ${}^6C_1 = 6$.Total favorable outcomes = ${}^4C_2 	imes {}^6C_1 = 6 	imes 6 = 36$.Required probability = $\frac{36}{816} = \frac{3}{68}$.

$A$ bag contains $6$ red,$4$ white and $8$ blue balls. If three balls are drawn at random,then the probability that $2$ are white and $1$ is red,is

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