Question
In a certain society, all married couples use the followingstrategy to determine the number of children that they willhave:
If the first child is a girl, they have no more children.
If the first child is a boy, they have a second child.
If the secondchild is a girl, they have No more children. If the secondchild is a boy, they have exactly one more child.
*ignore twins, and sex of children is independent and random
So (i) what is the probability distribution for the number ofchildren in the family?
I thought something like this:
For 0<= x <=2 then it is a geometric distribution withp =1/2 and n=1,2
For x>2 then it's 3
--> is that correct?
(ii)what is the probability distribution for the number of girlchildren in the family?
I thought it is uniform distribution [0,1]
(iii)A male child is chosen at random from all the male children inthe population. What is the probability distribution for the numberof sisters of this child?
I got really stuck here...
Explanation / Answer
i) Let X be the number of children in the family. X can takevalues of 1,2 or 3 P(X=1) =1/2 The first child has to be a girl for X=1 P(X=2) = (1/2)(1/2) = 1/4 It hasto be a boy, then a girl P(X=3) = (1/2)(1/2) = 1/4 It has to be aboy, then a boy again. The third child could be a boy or a girl, itdoesn't matter here. Notice that P(X=1) + P(X=2) + P(X=3) = 1. This is away to check with our reasoning is correct. Thus, we have the following distribution: X = 1, with prob = 1/2 X = 2, with prob = 1/4 X = 3, with prob = 1/4 ii) Let Y be the number of girls. Y can take values of 0and 1 P(Y=0) = (1/2)(1/2)(1/2) = 1/8. It has to be aboy, then a boy, and then a boy P(Y=1) = 1/2 + (1/2)(1/2) + (1/2)(1/2)(1/2) = 7/8. (3cases: girl, or boy-girl, or boy-boy-girl) Note that P(Y=0) + P(Y=1) = 1 Thus, we have the following distribution: Y = 0, with prob = 1/8 Y = 1, with prob = 7/8 iii) Let Z be the number of sisters for the male child chosen.The boy can either have 0 or 1 sister. So Z can take 0 or 1 P(Z = 0) = (1/2)(1/2)(1/2)= 1/8 In the family, we have aboy, a boy and a boy again (so we don't have any sister) P(Z = 1) = 1 - P(Z = 0) = 1 - 1/8 = 7/8 Thus, Z = 0, with prob = 1/8 Z = 1, with prob = 7/8 X = 2, with prob = 1/4 X = 3, with prob = 1/4 ii) Let Y be the number of girls. Y can take values of 0and 1 P(Y=0) = (1/2)(1/2)(1/2) = 1/8. It has to be aboy, then a boy, and then a boy P(Y=1) = 1/2 + (1/2)(1/2) + (1/2)(1/2)(1/2) = 7/8. (3cases: girl, or boy-girl, or boy-boy-girl) Note that P(Y=0) + P(Y=1) = 1 Thus, we have the following distribution: Y = 0, with prob = 1/8 Y = 1, with prob = 7/8 iii) Let Z be the number of sisters for the male child chosen.The boy can either have 0 or 1 sister. So Z can take 0 or 1 P(Z = 0) = (1/2)(1/2)(1/2)= 1/8 In the family, we have aboy, a boy and a boy again (so we don't have any sister) P(Z = 1) = 1 - P(Z = 0) = 1 - 1/8 = 7/8 Thus, Z = 0, with prob = 1/8 Z = 1, with prob = 7/8 Z = 0, with prob = 1/8 Z = 1, with prob = 7/8