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(C) Let $B_I, B_{II}, B_{III}$ be the events of choosing Bag $I$,Bag $II$,and Bag $III$ respectively. Since the bags are chosen at random,$P(B_I) = P(

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Let $B_I, B_{II}, B_{III}$ be the events of choosing Bag $I$,Bag $II$,and Bag $III$ respectively. Since the bags are chosen at random,$P(B_I) = P(B_{II}) = P(B_{III}) = \frac{1}{3}$.Let $R$ be the event of drawing a Red ball and $G$ be the event of drawing a Green ball.Using Bayes' Theorem:$p = P(B_I | R) = \frac{P(B_I)P(R|B_I)}{P(B_I)P(R|B_I) + P(B_{II})P(R|B_{II}) + P(B_{III})P(R|B_{III})}$$p = \frac{\frac{1}{3} 	imes \frac{3}{10}}{\frac{1}{3} 	imes \frac{3}{10} + \frac{1}{3} 	imes \frac{4}{10} + \frac{1}{3} 	imes \frac{5}{10}} = \frac{3}{3+4+5} = \frac{3}{12} = \frac{1}{4}$.Similarly,for the Green ball:$q = P(B_{III} | G) = \frac{P(B_{III})P(G|B_{III})}{P(B_I)P(G|B_I) + P(B_{II})P(G|B_{II}) + P(B_{III})P(G|B_{III})}$$q = \frac{\frac{1}{3} 	imes \frac{4}{10}}{\frac{1}{3} 	imes \frac{5}{10} + \frac{1}{3} 	imes \frac{3}{10} + \frac{1}{3} 	imes \frac{4}{10}} = \frac{4}{5+3+4} = \frac{4}{12} = \frac{1}{3}$.Therefore,$\frac{1}{p} + \frac{1}{q} = \frac{1}{1/4} + \frac{1}{1/3} = 4 + 3 = 7$.

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

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