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

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(B) Total number of balls $= 6 + 4 + 8 = 18$.Number of balls drawn $= 3$.The total number of ways to draw $3$ balls from $18$ is given by $^{18}C_{3}

Vedclass · Accepted Answer

$\frac{3}{68}$

Vedclass · Answer

(B) Total number of balls $= 6 + 4 + 8 = 18$.Number of balls drawn $= 3$.The total number of ways to draw $3$ balls from $18$ is given by $^{18}C_{3} = \frac{18 	imes 17 	imes 16}{3 	imes 2 	imes 1} = 3 	imes 17 	imes 16 = 816$.The number of ways to draw $1$ red ball from $6$ is $^{6}C_{1} = 6$.The number of ways to draw $2$ white balls from $4$ is $^{4}C_{2} = \frac{4 	imes 3}{2 	imes 1} = 6$.The number of favorable outcomes $= ^{6}C_{1} 	imes ^{4}C_{2} = 6 	imes 6 = 36$.Required probability $= \frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{36}{816}$.Dividing both numerator and denominator by $12$,we get $\frac{36 \div 12}{816 \div 12} = \frac{3}{68}$.

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

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