Urn A contains 6 red and 4 black balls and ur | vedclass

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(A) Let $R_1$ be the event that a red ball is drawn from $A$ and placed in $B$,and $B_1$ be the event that a black ball is drawn from $A$ and placed i

Vedclass · Accepted Answer

$\frac{32}{55}$

Vedclass · Answer

(A) Let $R_1$ be the event that a red ball is drawn from $A$ and placed in $B$,and $B_1$ be the event that a black ball is drawn from $A$ and placed in $B$. $P(R_1) = \frac{6}{10} = \frac{3}{5}$,$P(B_1) = \frac{4}{10} = \frac{2}{5}$. After transferring a ball,urn $B$ has $11$ balls. If $R_1$ occurred,$B$ has $5$ red and $6$ black balls. $P(R_2|R_1) = \frac{5}{11}$,$P(B_2|R_1) = \frac{6}{11}$. If $B_1$ occurred,$B$ has $4$ red and $7$ black balls. $P(R_2|B_1) = \frac{4}{11}$,$P(B_2|B_1) = \frac{7}{11}$. After transferring back to $A$,urn $A$ has $10$ balls. If $R_1$ and $R_2$ occurred,$A$ has $6$ red balls. $P(R|R_1, R_2) = \frac{6}{10}$. If $R_1$ and $B_2$ occurred,$A$ has $5$ red balls. $P(R|R_1, B_2) = \frac{5}{10}$. If $B_1$ and $R_2$ occurred,$A$ has $7$ red balls. $P(R|B_1, R_2) = \frac{7}{10}$. If $B_1$ and $B_2$ occurred,$A$ has $6$ red balls. $P(R|B_1, B_2) = \frac{6}{10}$. The total probability is $P(R) = P(R_1)P(R_2|R_1)P(R|R_1, R_2) + P(R_1)P(B_2|R_1)P(R|R_1, B_2) + P(B_1)P(R_2|B_1)P(R|B_1, R_2) + P(B_1)P(B_2|B_1)P(R|B_1, B_2)$. $P(R) = (\frac{6}{10} 	imes \frac{5}{11} 	imes \frac{6}{10}) + (\frac{6}{10} 	imes \frac{6}{11} 	imes \frac{5}{10}) + (\frac{4}{10} 	imes \frac{4}{11} 	imes \frac{7}{10}) + (\frac{4}{10} 	imes \frac{7}{11} 	imes \frac{6}{10})$. $P(R) = \frac{180 + 180 + 112 + 168}{1100} = \frac{640}{1100} = \frac{64}{110} = \frac{32}{55}$.

	Red	Blue	Green
Bag $I$	$3$	$2$	$5$
Bag $II$	$4$	$3$	$3$
Bag $III$	$5$	$1$	$4$

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