A) \[\frac{1}{2}and\frac{1}{4}\]
B) \[\frac{1}{3}and\frac{1}{2}\]
C) \[\frac{1}{3}and\frac{1}{4}\]
D) \[\frac{1}{2}and\frac{1}{3}\]
Correct Answer: A
Solution :
[a] Let b and g represent the boy and the girl child, respectively, if a family has two children, the sample space will be \[S=\{(b,b),(b,g),(g,b),(g,g)\}\] Let A be the event that both children are girls. Therefore,\[A=\{(g,g)\}\] (i) Let B be the event that the youngest child is a girl. Therefore, \[B=[(g,g),(g,g)]\] \[\Rightarrow A\cap B=\{(g,g)\}\] \[\therefore P(B)=\frac{2}{4}=\frac{1}{2}\] and \[P(A\cap B)=\frac{1}{4}\] The conditional probability that both are girls, given that the youngest child is a girl, is given by \[P(A/B)\] \[P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{1/4}{1/2}=\frac{1}{2}\] (ii) Let C be the event that at least one child is a girl. Therefore, \[C=\{(b,g),(g,b),(g,g)\}\] \[\Rightarrow A\cap C=\{g,g\}\Rightarrow P(C)=\frac{3}{4}\] and \[P(A\cap C)=\frac{1}{4}\] The conditional probability that both are girls, given that at least one child is a girl, is given by \[P(A|C).\]Therefore, \[P(A|C)=\frac{P(A\cap C)}{P(C)}=\frac{1/4}{3/4}=\frac{1}{3}\].
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