Assume that each born child is equally likely to be a boy or a girl. I

1.	Assume that each born child 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 thati. the youngest is girlii. at least one is a girl.
Answer» Let B= boy G= girl And let us consider, in a sample space, the first child is elder and second child is younger. Total possible outcome = {BB, BG, GB, GG} = 4 Let A = be the event that both the children are girls = 1 Therefore P(A) = \(\cfrac14\) Case 1. Let B = event that youngest is girl = {BG, GG} =2 {Since we have considered second is younger in a sample space} Therefore P(B) = \(\cfrac24\) And (A ∩ B) = both are girls and younger is also girl = (GG) = 1 Therefore , P (A ∩ B) = \(\cfrac14\) We require P(\(\cfrac{A}{B}\)) \(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\) = \(\cfrac{\frac14}{\frac24}=\cfrac12\) Case 2. Let B = event that at least one is girl = {BG,GB GG} =3 {Since we have considered second is younger in a sample space} Therefore P(B) = \(\cfrac34\) And (A ∩ B) = both are girls and atlas one is girl = (GG) = 1 Therefore , P (A ∩ B) = \(\cfrac14\) We require P(\(\cfrac{A}B\)) \(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\) = \(\cfrac{\frac14}{\frac34}\) = \(\cfrac13\)

Assume that each born child 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 thati. the youngest is girlii. at least one is a girl.

Answer»

Let B= boy G= girl

And let us consider, in a sample space, the first child is elder and second child is younger.

Total possible outcome = {BB, BG, GB, GG} = 4

Let A = be the event that both the children are girls = 1

Therefore P(A) = \(\cfrac14\)

Case 1.

Let B = event that youngest is girl = {BG, GG} =2

{Since we have considered second is younger in a sample space}

Therefore P(B) = \(\cfrac24\)

And (A ∩ B) = both are girls and younger is also girl = (GG) = 1

Therefore , P (A ∩ B) = \(\cfrac14\)

We require P(\(\cfrac{A}{B}\))

\(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\)

= \(\cfrac{\frac14}{\frac24}=\cfrac12\)

Case 2.

Let B = event that at least one is girl = {BG,GB GG} =3

{Since we have considered second is younger in a sample space}

Therefore P(B) = \(\cfrac34\)

And (A ∩ B) = both are girls and atlas one is girl = (GG) = 1

Therefore , P (A ∩ B) = \(\cfrac14\)

We require P(\(\cfrac{A}B\))

\(P(\cfrac{A}{B})=\cfrac{P(A\cap B)}{P(B)}\)

= \(\cfrac{\frac14}{\frac34}\) = \(\cfrac13\)