Question

Suppose that p = P(male birth) = 0.3, A couple wishes to have exactly two female children in their family. They will have children until this condition is fulfilled.

(a) What is the probability that the family has x male children?

(b) What is the probability that the family has four children?

(c) What is the probability that the family has at most four children?

(d) How many male children would you expect this family to have? How many children would you expect this family to have?

Answer 1

Solution

Suppose that p = P(male birth) = 0.3

(a) What is the probability that the family has x male children

Let X be the number of male children until they get two female children .

Then \( X\sim Nb(2,0.7) \)

we have  \( P(X=x)=C_{x+1}^1\left(0.7\right)^2\left(0.3\right)^x \)

                                  \( =\frac{\left(x+1\right)!}{x!}\left(0.7\right)^2\left(0.3\right)^x \)

                                  \( =\left(x+1\right)\left(0.7\right)^2\left(0.3\right)^x \)

Therefore. \( P(X=x)=\left(x+1\right)\left(0.7\right)^2\left(0.3\right)^x \)

(b) What is the probability that the family has four children

Since  they wish to have 2 female .

If  \( X=2\implies P(X=2)=(2+1)\left(0.7\right)^2\left(0.3\right)^2 \)

Therefore. \( P(X=2)=3\left(0.7\right)^2\left(0.3\right)^2 \)

(c) What is the probability that the family has at most four children

\( \implies P(X\leq 2)=P(0)+P(1)+P(2) \)

                             \( =\left(0.7\right)^2+2\left(0.7\right)^2\left(0.3\right)+3\left(0.7\right)^2\left(0.3\right)^2 \)

                             \( =\left(0.7\right)^2\left(0.6+0.27\right)=\left(0.7\right)^2\left(0.87\right) \)

Therefore.  \( P(X\leq 2)=\left(0.7\right)^2\left(0.87\right) \)

(d) How many male children would you expect this family to have 

For male children  \( E(X)=\frac{2\left(1-0.7\right)}{0.7}=0.85 \)

If they wish to have 2 female children they might get one male children. and the  expectation of the children =3 

Therefore.

a). \( P(X=x)=\left(x+1\right)\left(0.7\right)^2\left(0.3\right)^x \)

b). \( P(X=2)=3\left(0.7\right)^2\left(0.3\right)^2 \)

c). \( P(X\leq 2)=\left(0.7\right)^2\left(0.87\right) \)

d). \( E(X)=0.85 \)