Nine balls are drawn simultaneously from a ba | vedclass

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(B) Total number of balls = $5 + 7 = 12$. We need to draw $9$ balls from $12$ balls. The total number of ways to select $9$ balls is $n(S) = {}^{12}C_

Vedclass · Accepted Answer

$\frac{7}{22}$

Vedclass · Answer

(B) Total number of balls = $5 + 7 = 12$. We need to draw $9$ balls from $12$ balls. The total number of ways to select $9$ balls is $n(S) = {}^{12}C_9 = {}^{12}C_3 = \frac{12 	imes 11 	imes 10}{3 	imes 2 	imes 1} = 220$. We need to select $3$ white balls from $5$ white balls and $6$ black balls from $7$ black balls. The number of ways to select $3$ white balls is ${}^5C_3 = {}^5C_2 = \frac{5 	imes 4}{2 	imes 1} = 10$. The number of ways to select $6$ black balls is ${}^7C_6 = {}^7C_1 = 7$. The number of favorable outcomes is $n(P) = 10 	imes 7 = 70$. The probability is $P = \frac{n(P)}{n(S)} = \frac{70}{220} = \frac{7}{22}$.

Nine balls are drawn simultaneously from a bag containing $5$ white and $7$ black balls. The probability of drawing $3$ white and $6$ black balls is:

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