This is another binomial distribution problem, we will take that a child is a

boy as the favourable outcome, so in a family with n children the probability

of k boys is b(k;n,0.51).

So the probability of 3 boys in a family with 5 children is b(3;5,0.51)

The probability of at least one boy = 1-prob(no boys)=1-b(0;5,0.51).

Prob all the children the same gender = (prob all girls) + (prob all boys) = b(0,5,0.51)+b(5;5,0.51).

RonL