Question

Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that i. the youngest is a girl, ii. at least one is a girl? Select the correct answer from above options

Answer 1

We may write the sample space as `S={G_(1)G_(2), G_(1)B_(2), B_(1)G_(2), B_(1)B_(2)}`, where the youngest child appears later. (i) Let A = event that both the children are girls, and B = event that the youngest child is a girl. Then, `A={G_(1)G_(2)}, B={G_(1)G_(2), B_(1)G_(2)}` and `A nn B = {G_(1)G_(2)}`. `:. P(A) = (n(A))/(n(S))=1/4, P(B)=(n(B))/(n(S))=2/4=1/2` and `P(A nn B)=(n(A nn B))/(n(S))=1/4` Suppose B has already occurred and then A occurs. So, we have to find `P(A//B)` Now, `P(A//B)=(P(Ann B))/(P(B))=((1//4))/((1//2))=(1/4xx2/1)=1/2`. Hence, the required probability is `1/2`. (ii) Let `A=` event that both the children are girls, and `E=` event that at least one of the children is a girl. Then, `A={G_(1)G_(2)}, E={G_(1)B_(2), B_(1)G_(2), G_(1)G_(2)}` and `A nn E={G_(1)G_(2)}`. `:. P(A)=(n(A))/(n(S))=1/4, P(E)=(n(E))/(n(S))=3/4` and `P(A nn E)=(n(A nn E))/(n(S))=1/4`. Suppose E has already occurred and then A occurs. So, we have to find `P(A//E)`. Now, `P(A//E)=(P(A nn E))/(P(E))=((1//4))/((3//4))=(1/4xx4/3)=1/3`.