A bag contains 3 white,3 black and 2 red ball | vedclass

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(D) Total number of balls = $3 + 3 + 2 = 8$. Let $R_3$ be the event that the third ball drawn is red. By the principle of symmetry in probability,the

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$\frac{1}{4}$

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(D) Total number of balls = $3 + 3 + 2 = 8$. Let $R_3$ be the event that the third ball drawn is red. By the principle of symmetry in probability,the probability of drawing a red ball at any specific position (first,second,or third) without replacement is equal to the initial proportion of red balls in the bag. $P(R_3) = \frac{	ext{Number of red balls}}{	ext{Total number of balls}} = \frac{2}{8} = \frac{1}{4}$. Alternatively,summing all possible mutually exclusive cases for the first two draws: $P(R_3) = P(W_1 W_2 R_3) + P(W_1 B_2 R_3) + P(B_1 W_2 R_3) + P(B_1 B_2 R_3) + P(R_1 W_2 R_3) + P(R_1 B_2 R_3) + P(W_1 R_2 R_3) + P(B_1 R_2 R_3)$ $= \frac{3}{8} 	imes \frac{2}{7} 	imes \frac{2}{6} + \frac{3}{8} 	imes \frac{3}{7} 	imes \frac{2}{6} + \frac{3}{8} 	imes \frac{3}{7} 	imes \frac{2}{6} + \frac{3}{8} 	imes \frac{2}{7} 	imes \frac{2}{6} + \frac{2}{8} 	imes \frac{3}{7} 	imes \frac{1}{6} + \frac{2}{8} 	imes \frac{3}{7} 	imes \frac{1}{6} + \frac{3}{8} 	imes \frac{2}{7} 	imes \frac{1}{6} + \frac{3}{8} 	imes \frac{2}{7} 	imes \frac{1}{6}$ $= \frac{12 + 18 + 18 + 12 + 6 + 6 + 6 + 6}{336} = \frac{84}{336} = \frac{1}{4}$.

$A$ bag contains $3$ white,$3$ black and $2$ red balls. Three balls are drawn one by one without replacement. The probability that the third ball is red is:

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