A bag contains 4 red and 3 black balls. One b | vedclass

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(D) Total number of balls $= 4 + 3 = 7$.Probability of drawing a black ball,$p = \frac{3}{7}$.Probability of drawing a red ball (not black),$q = 1 - p

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(D) Total number of balls $= 4 + 3 = 7$.Probability of drawing a black ball,$p = \frac{3}{7}$.Probability of drawing a red ball (not black),$q = 1 - p = 1 - \frac{3}{7} = \frac{4}{7}$.Since the ball is replaced,the trials are independent,and we use the binomial distribution $P(X = x) = {}^nC_x p^x q^{n-x}$,where $n = 3$.For $X = 0$: $P(X = 0) = {}^3C_0 (\frac{3}{7})^0 (\frac{4}{7})^3 = (\frac{4}{7})^3$.For $X = 1$: $P(X = 1) = {}^3C_1 (\frac{3}{7})^1 (\frac{4}{7})^2 = 3 	imes \frac{3}{7} 	imes (\frac{4}{7})^2 = \frac{9}{7} (\frac{4}{7})^2$.For $X = 2$: $P(X = 2) = {}^3C_2 (\frac{3}{7})^2 (\frac{4}{7})^1 = 3 	imes (\frac{3}{7})^2 	imes \frac{4}{7} = \frac{12}{7} (\frac{3}{7})^2$.For $X = 3$: $P(X = 3) = {}^3C_3 (\frac{3}{7})^3 (\frac{4}{7})^0 = (\frac{3}{7})^3$.Comparing this with the given options,option $D$ matches the calculated distribution.

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