There are three families F_1, F_2, F_3. F_1 h | vedclass

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(A) Let $E_1, E_2, E_3$ be the events of choosing families $F_1, F_2, F_3$ respectively. Since the family is chosen randomly,$P(E_1) = P(E_2) = P(E_3)

Vedclass · Accepted Answer

$\frac{4}{9}$

Vedclass · Answer

(A) Let $E_1, E_2, E_3$ be the events of choosing families $F_1, F_2, F_3$ respectively. Since the family is chosen randomly,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.Let $G$ be the event that the selected child is a girl.The probabilities of selecting a girl from each family are:$P(G|E_1) = \frac{1}{3}$$P(G|E_2) = \frac{2}{3}$$P(G|E_3) = \frac{1}{2}$Using Bayes' Theorem,the probability that the girl is from $F_2$ is $P(E_2|G) = \frac{P(E_2)P(G|E_2)}{P(E_1)P(G|E_1) + P(E_2)P(G|E_2) + P(E_3)P(G|E_3)}$.Substituting the values: $P(E_2|G) = \frac{\frac{1}{3} 	imes \frac{2}{3}}{\frac{1}{3} 	imes \frac{1}{3} + \frac{1}{3} 	imes \frac{2}{3} + \frac{1}{3} 	imes \frac{1}{2}} = \frac{\frac{2}{9}}{\frac{1}{9} + \frac{2}{9} + \frac{1}{6}} = \frac{\frac{2}{9}}{\frac{2+4+3}{18}} = \frac{\frac{2}{9}}{\frac{9}{18}} = \frac{2}{9} 	imes 2 = \frac{4}{9}$.

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