A box A contains 2 white,3 red and 2 black ba | vedclass

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Question

(A) Let $E$ be the event that one white and one red ball are drawn. Let $H_A$ and $H_B$ be the events of selecting box $A$ and box $B$ respectively. $

Vedclass · Accepted Answer

$\frac{7}{16}$

Vedclass · Answer

(A) Let $E$ be the event that one white and one red ball are drawn. Let $H_A$ and $H_B$ be the events of selecting box $A$ and box $B$ respectively. $P(H_A) = P(H_B) = \frac{1}{2}$.The probability of drawing one white and one red ball from box $A$ is $P(E|H_A) = \frac{^2C_1 	imes ^3C_1}{^7C_2} = \frac{2 	imes 3}{21} = \frac{6}{21} = \frac{2}{7}$.The probability of drawing one white and one red ball from box $B$ is $P(E|H_B) = \frac{^4C_1 	imes ^2C_1}{^9C_2} = \frac{4 	imes 2}{36} = \frac{8}{36} = \frac{2}{9}$.Using Bayes' Theorem,the probability that the balls were drawn from box $B$ given event $E$ is:$P(H_B|E) = \frac{P(H_B)P(E|H_B)}{P(H_A)P(E|H_A) + P(H_B)P(E|H_B)}$$P(H_B|E) = \frac{\frac{1}{2} 	imes \frac{2}{9}}{\frac{1}{2} 	imes \frac{2}{7} + \frac{1}{2} 	imes \frac{2}{9}} = \frac{\frac{2}{9}}{\frac{2}{7} + \frac{2}{9}} = \frac{\frac{2}{9}}{\frac{18+14}{63}} = \frac{2}{9} 	imes \frac{63}{32} = \frac{1}{1} 	imes \frac{7}{16} = \frac{7}{16}$.

Box	White	Green	Red	Probability
$A$	$1$	$2$	$3$	$\frac{1}{2}$
$B$	$2$	$3$	$1$	$\frac{1}{3}$
$C$	$3$	$1$	$2$	$\frac{1}{6}$

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