A bag contains 2 white and 4 black balls. A b | vedclass

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(C) Let $p$ be the probability of drawing a white ball and $q$ be the probability of drawing a black ball.Total balls $= 2 + 4 = 6$.$p = \frac{2}{6} =

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$\frac{11}{243}$

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(C) Let $p$ be the probability of drawing a white ball and $q$ be the probability of drawing a black ball.Total balls $= 2 + 4 = 6$.$p = \frac{2}{6} = \frac{1}{3}$.$q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$.Since the balls are drawn $5$ times with replacement,this follows a binomial distribution $B(n, p)$ where $n = 5$.The probability of drawing $k$ white balls is given by $P(X = k) = {^nC_k} p^k q^{n-k}$.We need the probability that at least $4$ balls are white,which is $P(X \ge 4) = P(X = 4) + P(X = 5)$.$P(X = 4) = {^5C_4} \left(\frac{1}{3}ight)^4 \left(\frac{2}{3}ight)^1 = 5 	imes \frac{1}{81} 	imes \frac{2}{3} = \frac{10}{243}$.$P(X = 5) = {^5C_5} \left(\frac{1}{3}ight)^5 \left(\frac{2}{3}ight)^0 = 1 	imes \frac{1}{243} 	imes 1 = \frac{1}{243}$.Therefore,$P(X \ge 4) = \frac{10}{243} + \frac{1}{243} = \frac{11}{243}$.

$A$ bag contains $2$ white and $4$ black balls. $A$ ball is drawn $5$ times with replacement. The probability that at least $4$ of the balls drawn are white is

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