There are 3 bags A, B & C. Bag A contains 1 R | vedclass

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(D) Initially: $A = \{1R, 2G\}$,$B = \{2R, 1G\}$,$C = \{1G\}$.Let $X_1$ be the ball transferred from $A$ to $B$,$X_2$ from $B$ to $C$,and $X_3$ from $

Vedclass · Accepted Answer

$\frac{1}{6}$

Vedclass · Answer

(D) Initially: $A = \{1R, 2G\}$,$B = \{2R, 1G\}$,$C = \{1G\}$.Let $X_1$ be the ball transferred from $A$ to $B$,$X_2$ from $B$ to $C$,and $X_3$ from $C$ to $A$.For bag $A$ to have $2R$ and $1G$ at the end,it must have started with $1R$ and $2G$,lost a green ball $(X_1 = G)$,and gained a red ball $(X_3 = R)$.Step $1$: Draw $G$ from $A$ $(P(X_1=G) = \frac{2}{3})$. Now $B$ has ${2R, 2G}$.Step $2$: Draw $R$ from $B$ $(P(X_2=R|X_1=G) = \frac{2}{4} = \frac{1}{2})$. Now $C$ has ${1R, 1G}$.Step $3$: Draw $R$ from $C$ $(P(X_3=R|X_2=R) = \frac{1}{2})$. Now $A$ has ${1R-1G+1R, 2G-1G+1R} = {2R, 1G}$.Total Probability $= P(X_1=G) 	imes P(X_2=R|X_1=G) 	imes P(X_3=R|X_2=R) = \frac{2}{3} 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{6}$.

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