Bag P contains 3 white, 2 red, 5 blue balls a | vedclass

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(C) Let $W_P, R_P, B_P$ be the events of drawing a white, red, or blue ball from bag $P$ respectively.Bag $P$ has $3+2+5 = 10$ balls. So, $P(W_P) = \f

Vedclass · Accepted Answer

$\frac{16}{55}$

Vedclass · Answer

(C) Let $W_P, R_P, B_P$ be the events of drawing a white, red, or blue ball from bag $P$ respectively.Bag $P$ has $3+2+5 = 10$ balls. So, $P(W_P) = \frac{3}{10}, P(R_P) = \frac{2}{10}, P(B_P) = \frac{5}{10}$.After transferring a ball to bag $Q$, bag $Q$ will have $10+1 = 11$ balls.If a white ball is transferred, bag $Q$ has $3$ white, $3$ red, $5$ blue balls. $P(R|W_P) = \frac{3}{11}$.If a red ball is transferred, bag $Q$ has $2$ white, $4$ red, $5$ blue balls. $P(R|R_P) = \frac{4}{11}$.If a blue ball is transferred, bag $Q$ has $2$ white, $3$ red, $6$ blue balls. $P(R|B_P) = \frac{3}{11}$.Using the law of total probability:$P(R) = P(W_P) 	imes P(R|W_P) + P(R_P) 	imes P(R|R_P) + P(B_P) 	imes P(R|B_P)$$P(R) = \frac{3}{10} 	imes \frac{3}{11} + \frac{2}{10} 	imes \frac{4}{11} + \frac{5}{10} 	imes \frac{3}{11}$$P(R) = \frac{9}{110} + \frac{8}{110} + \frac{15}{110} = \frac{32}{110} = \frac{16}{55}$.

Bag $P$ contains $3$ white, $2$ red, $5$ blue balls and bag $Q$ contains $2$ white, $3$ red, $5$ blue balls. $A$ ball is chosen at random from $P$ and is placed in $Q$. If a ball is chosen from bag $Q$ at random, then the probability that it is a red ball is

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