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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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: $P(E_2|G) = \frac{\frac{1}{3} 	imes \frac{2}{3}}{\frac{1}{3} 	imes \frac{1}{3} + \frac{1}{3} 	imes \frac{2}{3} + \frac{1}{3} 	imes \frac{1}{2}} = \frac{\frac{2}{9}}{\frac{1}{9} + \frac{2}{9} + \frac{1}{6}} = \frac{\frac{2}{9}}{\frac{2+4+3}{18}} = \frac{\frac{2}{9}}{\frac{9}{18}} = \frac{2}{9} 	imes 2 = \frac{4}{9}$.

Similar Questions

Bag $A$ contains $3$ white and $5$ black balls, while bag $B$ contains $4$ white and $3$ black balls. $A$ ball is selected at random from bag $A$ and put into bag $B$. If a ball is now selected at random from bag $B$, what is the probability that this ball is white?

$A$ and $B$ are two events of a random experiment such that $P(B)=0.4$, $P(A \cap \bar{B})=0.5$, and $P(A \cup B) + P\left(\frac{B}{A \cup \bar{B}}\right) = 1.15$. Then $P(A) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module