In a school there are 3 sections A, B and C. | vedclass

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(D) Let $E_1, E_2, E_3$ be the events of selecting sections $A, B$ and $C$ respectively. Let $G$ be the event of selecting a girl.Given probabilities

Vedclass · Accepted Answer

$\frac{16}{81}$

Vedclass · Answer

(D) Let $E_1, E_2, E_3$ be the events of selecting sections $A, B$ and $C$ respectively. Let $G$ be the event of selecting a girl.Given probabilities of selecting sections are $P(E_1) = 0.2, P(E_2) = 0.3, P(E_3) = 0.5$.The conditional probabilities of selecting a girl from each section are:$P(G|E_1) = \frac{20}{20+30} = \frac{20}{50} = 0.4$$P(G|E_2) = \frac{40}{40+20} = \frac{40}{60} = \frac{2}{3}$$P(G|E_3) = \frac{10}{10+30} = \frac{10}{40} = 0.25$Using Bayes' Theorem,the probability that the girl belongs to section $A$ is $P(E_1|G) = \frac{P(E_1)P(G|E_1)}{P(E_1)P(G|E_1) + P(E_2)P(G|E_2) + P(E_3)P(G|E_3)}$.$P(E_1|G) = \frac{0.2 	imes 0.4}{(0.2 	imes 0.4) + (0.3 	imes \frac{2}{3}) + (0.5 	imes 0.25)}$$P(E_1|G) = \frac{0.08}{0.08 + 0.2 + 0.125} = \frac{0.08}{0.405} = \frac{80}{405} = \frac{16}{81}$.

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

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