A family decides to have children until it has three children of the same gender. Assuming P(B) = P(G) = .5, what is the pmf of X = the number of children in the family?

Answer : Step 1 : Given a family decides to have children until it has three children of the same gender. We assuming P(B) = P(G) = 0.5. Then we have to find the pmf of X = the number of children in the family. Here there are 3 of the same. So the probability they have 1 or 2 is zero in each case. Then unless there is a third sex which I haven't heard about. So they have not 6. Then the probability is 0. We consider 3 cases : P(3 children) , P(4 children) : P(5 children). Here B and G are students. So we have 2 children. 3 So 2 = 8 So the possibilities are : First is 3 children is So 2 = 8 Here total number of children is 8. = {GGG, GGB, GBG, BGG, GBB, BGB, BBG, BBB} = 2 8 1 = 4 = 0.25 Therefore the probability of three same-sex children4= Second is 4 children is 2 So 4 = 16 = {GGGG, GGGB, GGBG, GBGG, BGGG, GGBB, GBGB, BGGB, GBBG, BGBG, BBGG, BBBB, BBBG, BBGB, BGBB, GBBB} Here we count only {GGBG, GBGG, BGGG, BBGB, BGBB, GBBB} Then, ={GGBG, GBGG, BGGG, BBGB, BGBB, GBBB} = 6 16 3 = 8 = 0.25 2 So the probability of four children 8or 0.25 And third on is 5 children. So we are finding 5 children : So we have 2 children. 5 So 2 = 32 Then the possibilities are = {GGGGG,GGGGB,GGGBB,GGGBG,GGBGB,GGBBB,GBBGB . . . GBGBG} 12 = 32 3 = 8 = 0.375 And another 16 are obtained by reversing the order of arrival of the children. There are only 6 (marked *) of these possibilities which result in a family of 3 and 2, so in the full listing there will be 12 possibilities. 3 Therefore the probability of a family of 5 8 So the pmf is