Let U_1 and U_2 be two urns such that U_1 con | vedclass

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(D) Let $H$ be the event that the coin shows heads and $T$ be the event that the coin shows tails. $P(H) = P(T) = 1/2$.Let $W$ be the event that the b

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$12$

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(D) Let $H$ be the event that the coin shows heads and $T$ be the event that the coin shows tails. $P(H) = P(T) = 1/2$.Let $W$ be the event that the ball drawn from $U_2$ is white.If $H$ occurs,$1$ ball is transferred from $U_1$ to $U_2$. $U_1$ has $3W, 2R$. Probability of transferring $W$ is $3/5$,and $R$ is $2/5$.If $W$ is transferred,$U_2$ has $2W$. $P(W|H_W) = 1$. If $R$ is transferred,$U_2$ has $1W, 1R$. $P(W|H_R) = 1/2$.$P(W|H) = (3/5 	imes 1) + (2/5 	imes 1/2) = 3/5 + 1/5 = 4/5$.If $T$ occurs,$2$ balls are transferred from $U_1$ to $U_2$. Possible transfers: $2W$ (prob $^3C_2/^5C_2 = 3/10$),$1W, 1R$ (prob $(^3C_1 	imes ^2C_1)/^5C_2 = 6/10$),$2R$ (prob $^2C_2/^5C_2 = 1/10$).If $2W$ transferred,$U_2$ has $3W$. $P(W|T_{2W}) = 1$.If $1W, 1R$ transferred,$U_2$ has $2W, 1R$. $P(W|T_{1W1R}) = 2/3$.If $2R$ transferred,$U_2$ has $1W, 2R$. $P(W|T_{2R}) = 1/3$.$P(W|T) = (3/10 	imes 1) + (6/10 	imes 2/3) + (1/10 	imes 1/3) = 3/10 + 4/10 + 1/30 = 9/30 + 12/30 + 1/30 = 22/30 = 11/15$.Using Bayes' Theorem: $P(H|W) = \frac{P(W|H)P(H)}{P(W|H)P(H) + P(W|T)P(T)} = \frac{(4/5 	imes 1/2)}{(4/5 	imes 1/2) + (11/15 	imes 1/2)} = \frac{4/10}{4/10 + 11/30} = \frac{12/30}{12/30 + 11/30} = 12/23$.

Box	White	Green	Red	Probability
$A$	$1$	$2$	$3$	$\frac{1}{2}$
$B$	$2$	$3$	$1$	$\frac{1}{3}$
$C$	$3$	$1$	$2$	$\frac{1}{6}$

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