A box P contains one white ball,three red bal | vedclass

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Question

(C) Total balls in box $P = 1 + 3 + 2 = 6$. Total balls in box $Q = 2 + 3 + 4 = 9$. Let $W_P, R_P, B_P$ be the events of drawing a white,red,and black

Vedclass · Accepted Answer

$\frac{35}{54}$

Vedclass · Answer

(C) Total balls in box $P = 1 + 3 + 2 = 6$. Total balls in box $Q = 2 + 3 + 4 = 9$. Let $W_P, R_P, B_P$ be the events of drawing a white,red,and black ball from box $P$ respectively,and $W_Q, R_Q, B_Q$ be the corresponding events for box $Q$. The probabilities are: $P(W_P) = \frac{1}{6}, P(R_P) = \frac{3}{6}, P(B_P) = \frac{2}{6}$ $P(W_Q) = \frac{2}{9}, P(R_Q) = \frac{3}{9}, P(B_Q) = \frac{4}{9}$ The probability that the balls are of the same colour is: $P(	ext{same}) = P(W_P)P(W_Q) + P(R_P)P(R_Q) + P(B_P)P(B_Q)$ $P(	ext{same}) = (\frac{1}{6} 	imes \frac{2}{9}) + (\frac{3}{6} 	imes \frac{3}{9}) + (\frac{2}{6} 	imes \frac{4}{9}) = \frac{2 + 9 + 8}{54} = \frac{19}{54}$ The probability that the balls are of different colours is: $P(	ext{different}) = 1 - P(	ext{same}) = 1 - \frac{19}{54} = \frac{35}{54}$

$A$ box $P$ contains one white ball,three red balls and two black balls. Another box $Q$ contains two white balls,three red balls and four black balls. If one ball is drawn at random from each one of the two boxes,then the probability that the balls drawn are of different colour is

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