A bag contains 9 identical black balls number | vedclass

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Question

(A) Total number of balls $= 9 + 4 = 13$. Total ways to draw $3$ balls from $13$ is $^{13}C_3 = \frac{13 	imes 12 	imes 11}{3 	imes 2 	imes 1} = 2

Vedclass · Accepted Answer

$\frac{101}{143}$

Vedclass · Answer

(A) Total number of balls $= 9 + 4 = 13$. Total ways to draw $3$ balls from $13$ is $^{13}C_3 = \frac{13 	imes 12 	imes 11}{3 	imes 2 	imes 1} = 286$. The probability of getting at least one white ball is $1 - P(	ext{no white ball})$. If no white ball is drawn, all $3$ balls must be black. Number of ways to draw $3$ black balls $= ^9C_3 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 84$. $P(	ext{no white ball}) = \frac{84}{286} = \frac{42}{143}$. $P(	ext{at least one white ball}) = 1 - \frac{42}{143} = \frac{101}{143}$. Therefore, option $A$ is correct.

$A$ bag contains $9$ identical black balls numbered $1$ to $9$ and $4$ identical white balls numbered $1$ to $4$. If $3$ balls are drawn at a time randomly from that bag, then the probability of getting at least one white ball is

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