A box contains 8 red, 3 white and 9 blue ball | vedclass

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Question

(A) Total number of balls $= 8 + 3 + 9 = 20$. Number of ways to select $3$ balls out of $20$ is given by $n(S) = {}^{20}C_3 = \frac{20 	imes 19 	ime

Vedclass · Accepted Answer

$\frac{7}{95}$

Vedclass · Answer

(A) Total number of balls $= 8 + 3 + 9 = 20$. Number of ways to select $3$ balls out of $20$ is given by $n(S) = {}^{20}C_3 = \frac{20 	imes 19 	imes 18}{3 	imes 2 	imes 1} = 20 	imes 19 	imes 3 = 1140$. Number of ways to select $2$ red balls from $8$ and $1$ white ball from $3$ is $n(E) = {}^{8}C_2 	imes {}^{3}C_1 = \frac{8 	imes 7}{2 	imes 1} 	imes 3 = 28 	imes 3 = 84$. Probability $P(E) = \frac{n(E)}{n(S)} = \frac{84}{1140} = \frac{84 \div 12}{1140 \div 12} = \frac{7}{95}$.

$A$ box contains $8$ red, $3$ white and $9$ blue balls. If $3$ balls are drawn from the box at random one after the other without replacement, then the probability that they are $2$ red balls and $1$ white ball is

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