Question

A couple wishes to have exactly two female child in their family.

Suppose that P(male birth) = 0.4. They will have children until this condition is fulfilled.

- Introduce a random variable X.

b. (5 pts) What is the probability that the family has x male children?

c. (5 pts) What is the probability that the family has four children?

d. (5 pts) What is the probability that the family has at most four children?

e. (5 pts) How many male children would you expect this family to have?

f. (5 pts) What is SD of number of male children?

Answer #1

a)

random variable X is total number of children born in family. this follows geoemetric distribution with parameter r =2 and p=1-0.4 =0.6

b)

probability that the family has x male children =P(1 female child in x+1 children and 2nd female on x+2 th birth) =(x+1C1)*(0.4)x*(0.6)2 =(x+1)*(0.4)x(0.6)2

c)

probability that the family has four children =P(2 male) =(2+1)*(0.4)2*(0.6)2

=0.1728

d)probability that the family has at most four children =P(0 male )+P(1 male)+P(2 male)

=(0+1)*(0.4)0*(0.6)2 +(1+1)*(0.4)1*(0.6)2 +(2+1)*(0.4)2*(0.6)2

=0.8208

e)

expeced male chidren =r/p-r =2/0.6-2 =4/3 =1.3333

f)

SD =sqrt(r(1-p)/p2 )=sqrt(2*(1-0.6)/0.6^2)=1.4907

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