A box contains 24 identical balls,of which 12 | vedclass

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(C) The probability of drawing a white ball in a single draw is $p = \frac{12}{24} = \frac{1}{2}$.For the $4^{th}$ white ball to be drawn on the $7^{t

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$\frac{5}{32}$

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(C) The probability of drawing a white ball in a single draw is $p = \frac{12}{24} = \frac{1}{2}$.For the $4^{th}$ white ball to be drawn on the $7^{th}$ draw,there must be exactly $3$ white balls in the first $6$ draws,and the $7^{th}$ draw must be a white ball.This follows a binomial distribution scenario.The probability is given by: $P = \binom{6}{3} 	imes p^3 	imes (1-p)^3 	imes p$.Substituting $p = \frac{1}{2}$:$P = 20 	imes (\frac{1}{2})^3 	imes (\frac{1}{2})^3 	imes \frac{1}{2} = 20 	imes (\frac{1}{2})^7 = \frac{20}{128} = \frac{5}{32}$.

$A$ box contains $24$ identical balls,of which $12$ are white and $12$ are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the $4^{th}$ time on the $7^{th}$ draw is

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