A bag contains 6 white and 4 black balls. A d | vedclass

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(C) Let $X$ be the number obtained on the die. $X \in \{1, 2, 3, 4, 5, 6\}$,each with probability $\frac{1}{6}$.If $X=k$,the number of ways to draw $k

Vedclass · Accepted Answer

$\frac{1}{5}$

Vedclass · Answer

(C) Let $X$ be the number obtained on the die. $X \in \{1, 2, 3, 4, 5, 6\}$,each with probability $\frac{1}{6}$.If $X=k$,the number of ways to draw $k$ white balls from $6$ white balls is $\binom{6}{k}$,and the total number of ways to draw $k$ balls from $10$ balls is $\binom{10}{k}$.The probability of drawing $k$ white balls given $X=k$ is $P(W|X=k) = \frac{\binom{6}{k}}{\binom{10}{k}}$.The total probability is $P(W) = \sum_{k=1}^{6} P(X=k) 	imes P(W|X=k) = \frac{1}{6} \sum_{k=1}^{6} \frac{\binom{6}{k}}{\binom{10}{k}}$.Calculating the terms:$k=1: \frac{6}{10} = \frac{3}{5} = 0.6$$k=2: \frac{15}{45} = \frac{1}{3} \approx 0.333$$k=3: \frac{20}{120} = \frac{1}{6} \approx 0.167$$k=4: \frac{15}{210} = \frac{1}{14} \approx 0.071$$k=5: \frac{6}{252} = \frac{1}{42} \approx 0.024$$k=6: \frac{1}{210} \approx 0.005$Summing these: $\frac{1}{6} \left( \frac{6}{10} + \frac{15}{45} + \frac{20}{120} + \frac{15}{210} + \frac{6}{252} + \frac{1}{210} ight) = \frac{1}{6} \left( \frac{3}{5} + \frac{1}{3} + \frac{1}{6} + \frac{1}{14} + \frac{1}{42} + \frac{1}{210} ight) = \frac{1}{6} \left( \frac{126+70+35+15+5+1}{210} ight) = \frac{1}{6} 	imes \frac{252}{210} = \frac{42}{210} = \frac{1}{5}$.

$A$ bag contains $6$ white and $4$ black balls. $A$ die is rolled once and the number of balls equal to the number obtained on the die are drawn from the bag at random. The probability that all the balls drawn are white is:

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