Bag A contains 9 white and 8 black balls, whi | vedclass

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(B) Let $W_B$ be the event of picking a white ball from bag $B$, and $B_B$ be the event of picking a black ball from bag $B$.$P(W_B) = \frac{6}{6+4} =

Vedclass · Accepted Answer

$23$

Vedclass · Answer

(B) Let $W_B$ be the event of picking a white ball from bag $B$, and $B_B$ be the event of picking a black ball from bag $B$.$P(W_B) = \frac{6}{6+4} = \frac{6}{10} = \frac{3}{5}$$P(B_B) = \frac{4}{6+4} = \frac{4}{10} = \frac{2}{5}$If a white ball is transferred to bag $A$, bag $A$ now contains $10$ white and $8$ black balls (total $18$). The probability of drawing a white ball from bag $A$ is $P(W_A | W_B) = \frac{10}{18}$.If a black ball is transferred to bag $A$, bag $A$ now contains $9$ white and $9$ black balls (total $18$). The probability of drawing a white ball from bag $A$ is $P(W_A | B_B) = \frac{9}{18}$.Using the law of total probability:$P(W_A) = P(W_B) 	imes P(W_A | W_B) + P(B_B) 	imes P(W_A | B_B)$$P(W_A) = \frac{3}{5} 	imes \frac{10}{18} + \frac{2}{5} 	imes \frac{9}{18}$$P(W_A) = \frac{30}{90} + \frac{18}{90} = \frac{48}{90} = \frac{8}{15}$Thus, $p=8$ and $q=15$. Since $gcd(8,15)=1$, $p+q = 8+15 = 23$.

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

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