Question

Discrete Math:

A couple is planning to have a family. Let us assume that the probability of having a girl is 0.48 and a boy is 0.52, and that the gender of this couple's children are pairwise independent. They want to have at least one girl and at least one boy. At the same time, they know that raising too many kids is difficult. So here's what they plan to do: they'll keep trying to have children until they have at least one girl and at least one boy or until they have four kids. Once one of these two conditions are satisfied, they'll stop. Our goal is to determine the expected number of children this couple will have. Let X (s) be equal to the number of children with outcomes; e.g., X(GGB) = 3. a. What are the possible values of X($)? b. For each such value i, list the outcomes in the event (X = i). For example, the outcome GGB is part of the event (X=3). C. For each such value i, what is P(X = i)? Keep in mind that P(G) = 0.48, P(B) = 0.52 and the gender of the couple's children a independent of each other. d. Finally, what is E[X]? That is, on average, how many kids will such a couple have?

Answer 1

a)

X(s) = GB,BG       GGB,BBG GGGB,BBBG

b)

X= 2 , GB,BG  

X = 3 GGB,BBG

X =4 GGGB,BBBG

c)

P(X = 2) = P(GB) +P(BG) = 2 * 0.48 *0.52 = 0.4992

P(X = 3) = P(   GGB) + P(BBG) = 0.48^2*0.52 + 0.52^2*0.48 = 0.2496

P(X = 4) = 1- (P(X = 2) + P(X=3)

= 1-(0.4992 + 0.2496)

= 0.2512

d)

E(X) = 2 PX( =2) + 3 P(X = 3) + 4*P(X = 4)

= 2* 0.4992 + 3* 0.2496 + 4* 0.2512

= 2.752