Assume that the probability a child is a boy is 06 and that

Assume that the probability a child is a boy is 0.6 and that the sexes of children born into a family are independent. What is the probability that a family of 3 children (a) Exactly two boys and one girl. (b) At least one girl. (c) All children of the same sex.

Solution

P(for a boy) = p =0.6

q = 0.4

Thus no of boys in a family follow binomial as the trials are independent and there are two outcomes.

a) Prob exactly two boys and one girl

= 3C2(0,6)²(0.4) =0.432

b) P(atleast one girl) = P(2 boys or 1 boy or no boy)

= 1-P(3 boys) = 1-0.216 = 0,784

c) P(all children same sex)

= P(no of boys =0) + P(no of boys =3)

= 0.064+0.216

= 0.280