Three boxes B_1,B_2 and B_3 contain balls wit | vedclass

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(B) Let $B_1, B_2, B_3$ be the events of selecting boxes $B_1, B_2, B_3$ respectively. Since a die is thrown,$P(B_1) = P(B_2) = P(B_3) = \frac{2}{6} =

Vedclass · Accepted Answer

$\frac{5}{12}$

Vedclass · Answer

(B) Let $B_1, B_2, B_3$ be the events of selecting boxes $B_1, B_2, B_3$ respectively. Since a die is thrown,$P(B_1) = P(B_2) = P(B_3) = \frac{2}{6} = \frac{1}{3}$.Let $R$ be the event of drawing a red ball.For box $B_1$,total balls $= 2+1+2 = 5$,so $P(R|B_1) = \frac{2}{5}$.For box $B_2$,total balls $= 3+2+4 = 9$,so $P(R|B_2) = \frac{4}{9}$.For box $B_3$,total balls $= 4+3+2 = 9$,so $P(R|B_3) = \frac{2}{9}$.Using Bayes' Theorem,the probability that the ball is from $B_2$ given it is red is:$P(B_2|R) = \frac{P(R|B_2)P(B_2)}{P(R|B_1)P(B_1) + P(R|B_2)P(B_2) + P(R|B_3)P(B_3)}$$P(B_2|R) = \frac{(\frac{4}{9} 	imes \frac{1}{3})}{(\frac{2}{5} 	imes \frac{1}{3}) + (\frac{4}{9} 	imes \frac{1}{3}) + (\frac{2}{9} 	imes \frac{1}{3})}$$P(B_2|R) = \frac{\frac{4}{9}}{\frac{2}{5} + \frac{4}{9} + \frac{2}{9}} = \frac{\frac{4}{9}}{\frac{18+20+10}{45}} = \frac{\frac{4}{9}}{\frac{48}{45}} = \frac{4}{9} 	imes \frac{45}{48} = \frac{1}{1} 	imes \frac{5}{12} = \frac{5}{12}$.

Box	White,Black,Red
$B_1$	$2, 1, 2$
$B_2$	$3, 2, 4$
$B_3$	$4, 3, 2$

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