An urn A contains 3 white and 5 black balls. | vedclass

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(D) Case $I$: $A$ white ball is transferred from urn $A$ to urn $B$. The probability of selecting a white ball from urn $A$ is $P(W_A) = \frac{3}{3+5}

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$\frac{17}{40}$

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(D) Case $I$: $A$ white ball is transferred from urn $A$ to urn $B$. The probability of selecting a white ball from urn $A$ is $P(W_A) = \frac{3}{3+5} = \frac{3}{8}$. After transferring,urn $B$ contains $7$ white and $8$ black balls (total $15$). The probability of selecting a white ball from urn $B$ is $P(W_B|W_A) = \frac{7}{15}$. The probability of this case is $P(W_A) 	imes P(W_B|W_A) = \frac{3}{8} 	imes \frac{7}{15} = \frac{21}{120} = \frac{7}{40}$. Case $II$: $A$ black ball is transferred from urn $A$ to urn $B$. The probability of selecting a black ball from urn $A$ is $P(B_A) = \frac{5}{3+5} = \frac{5}{8}$. After transferring,urn $B$ contains $6$ white and $9$ black balls (total $15$). The probability of selecting a white ball from urn $B$ is $P(W_B|B_A) = \frac{6}{15}$. The probability of this case is $P(B_A) 	imes P(W_B|B_A) = \frac{5}{8} 	imes \frac{6}{15} = \frac{30}{120} = \frac{10}{40}$. The total probability of drawing a white ball from urn $B$ is $\frac{7}{40} + \frac{10}{40} = \frac{17}{40}$.

An urn $A$ contains $3$ white and $5$ black balls. Another urn $B$ contains $6$ white and $8$ black balls. $A$ ball is picked from $A$ at random and then transferred to $B$. Then,a ball is picked at random from $B$. The probability that it is a white ball is

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