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 $B$. Probability of selecting a white ball from $A$ is $P(W_1) = \frac{3}{3+5} = \frac

Vedclass · Accepted Answer

$\frac{17}{40}$

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(D) Case $I$: $A$ white ball is transferred from urn $A$ to urn $B$. Probability of selecting a white ball from $A$ is $P(W_1) = \frac{3}{3+5} = \frac{3}{8}$. After transferring, urn $B$ contains $7$ white and $8$ black balls. Probability of selecting a white ball from $B$ is $P(W_2|W_1) = \frac{7}{7+8} = \frac{7}{15}$. Probability of this case $= P(W_1) 	imes P(W_2|W_1) = \frac{3}{8} 	imes \frac{7}{15} = \frac{21}{120} = \frac{7}{40}$. Case $II$: $A$ black ball is transferred from urn $A$ to urn $B$. Probability of selecting a black ball from $A$ is $P(B_1) = \frac{5}{3+5} = \frac{5}{8}$. After transferring, urn $B$ contains $6$ white and $9$ black balls. Probability of selecting a white ball from $B$ is $P(W_2|B_1) = \frac{6}{6+9} = \frac{6}{15}$. Probability of this case $= P(B_1) 	imes P(W_2|B_1) = \frac{5}{8} 	imes \frac{6}{15} = \frac{30}{120} = \frac{10}{40}$. Total probability $= \frac{7}{40} + \frac{10}{40} = \frac{17}{40}$.

List-$I$	List-$II$
$A$. $P(E_2)$	$(i)$ $\frac{1}{2}$
$B$. $P(\frac{E_1}{E_2})$	$(ii)$ $\frac{5}{6}$
$C$. $P(\frac{\bar{E}_2}{E_1})$	$(iii)$ $\frac{1}{3}$
$D$. $P(\bar{E}_1 \cup \bar{E}_2)$	$(iv)$ $\frac{1}{6}$
	$(v)$ $\frac{2}{3}$

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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