Probability

Question

In a family of 7 children, let E be the event that the family has at least 3 girls and F be the event that the second born child is a girl. Find the following probabilities:

A) P(E)=

B) P(F)=

C) P(EnF)=

D) P(E/F)=

Answer 1

n =  7 children in the family

k =  # of girls in the family

 

A)  Binomial distribution

 

Note:  binomial coefficient = (n choose k) = n!/k!(n-k)!

 

P(E) = P(k≥3) = (7 choose 3)(0.5)³(0.5)⁴ + (7 choose 4)(0.5)⁴(0.5)³ + (7 choose 5)(0.5)⁵(0.5)² + (7 choose 6)(0.5)⁶(0.5)¹ + (7 choose 7)(0.5)⁷(0.5)⁰  = 0.2734 + 0.2734 + 0.1641 + 0.05469 + 0.007813 ≅ 0.7734

 

B)  I use 2ⁿ where n is the number of children in the family.  So 2⁷ = 128 possible combinations of boys and girls.  Then create a tree diagram.  You will notice that 64 possible combinations will have a second child be a girl and the other 64 will have a second child be a boy.  So, P(F) = P(2nd child is a girl) = 64/128 = 0.5

 

C) P(E∩F) = P(E)*P(F) = (0.7734)(0.5) = 0.3867

 

D) P(E|F) = P(E∩F)/P(F) = 0.3867/0.5 = 0.7734