Question

 Consider a family with eight children. Assuming that the sex each child is determined independently of the others and that each child is equally likely to be female or male, 

a. What is the probability that exactly four children are female? Hint: Use the binomial distribution. (10 points) 

b. What is the probability that at most seven children are female? Hint: What is the complement of this event? (10 points)

c. Find the conditional probability that the first two children born are female given that exactly four of the children are female. Hint: Let A be the event that the first two children born are female, let B be the event that exactly four of the eight children are female, and let C be the event that exactly two of the last six children are female. Explain why (A and B)=(A and C) and use this to conclude that P(A and B)=P(A and C). Then explain why A and C are independent and use this to conclude that P(A and C)=P(A) P(C). You may find it helpful to use the following list to organize your thoughts. (30 points)

Child1  Child2 Child3 Child4 Child5 Child6 Child7 Child8

Answer 1

a) part a is correct.

b) P( at most 7 children are female) =1-P(at least 8 children are female)=1-P(8 children are female)=1-[{8!/8!*(8-8)!}*0.5^8*(1-0.5)^8-8]=1-0.5^8=0.9961(approx to 4 decimals)

c) A binominal distribution is made up of sum of bernoulli trials which are independent of each other.

A=First 2 children are born female.

B= 4 out of 8 are female.

P(A and B) =P( First 2 children are born female and 4 out of 8 are female) =P(First 2 children are born female and 2 of last 6 are born female) =P( A and C)=P(A)*P(C) =0.5*0.5*[(6C2)*0.5^2*(1-0.5)^4]=0.05859375.=0.05859

P(8 out of 4 children are female)=8C4*0.5^4*(1-0.5)^8-4=0.27344.=P(B)

P(A|B)=0.05859/0.27344=0.21427(round to 5 decimals).