An urn A contains 4 white and 1 black ball; u | vedclass

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(C) Let $W_A, B_A$ be the events of drawing a white or black ball from $A$. $P(W_A) = \frac{4}{5}, P(B_A) = \frac{1}{5}$.After transferring from $A$ t

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$\frac{101}{180}$

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(C) Let $W_A, B_A$ be the events of drawing a white or black ball from $A$. $P(W_A) = \frac{4}{5}, P(B_A) = \frac{1}{5}$.After transferring from $A$ to $B$,urn $B$ has $6$ balls.Case $1$: If $W_A$ is transferred,$B$ has $4$ white,$2$ black. $P(W_{B|W_A}) = \frac{4}{6}, P(B_{B|W_A}) = \frac{2}{6}$.Case $2$: If $B_A$ is transferred,$B$ has $3$ white,$3$ black. $P(W_{B|B_A}) = \frac{3}{6}, P(B_{B|B_A}) = \frac{3}{6}$.Urn $C$ initially has $2$ white,$3$ black. After transfer from $B$,it has $6$ balls.If $W_B$ is transferred,$C$ has $3$ white,$3$ black. $P(B_C|W_B) = \frac{3}{6}$.If $B_B$ is transferred,$C$ has $2$ white,$4$ black. $P(B_C|B_B) = \frac{4}{6}$.Total probability $P(B_C) = P(B_C|W_B)P(W_B) + P(B_C|B_B)P(B_B)$.$P(W_B) = P(W_B|W_A)P(W_A) + P(W_B|B_A)P(B_A) = (\frac{4}{6} 	imes \frac{4}{5}) + (\frac{3}{6} 	imes \frac{1}{5}) = \frac{16+3}{30} = \frac{19}{30}$.$P(B_B) = 1 - \frac{19}{30} = \frac{11}{30}$.$P(B_C) = (\frac{3}{6} 	imes \frac{19}{30}) + (\frac{4}{6} 	imes \frac{11}{30}) = \frac{57 + 44}{180} = \frac{101}{180}$.

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