Assume that the probability a child is a boy is 0.51 and that the sexes of children born into a family are independent. What is the probability that a family of five children has:. a) exactly three boys?
b) at least one boy?
c) at least one girl?
d) all children of the same sex?

a) 10*0.51^3*0.49^2=approx 0.32

b)1-0.49^5= approx 0.97

c)1-0.51^5=0.965

d) approx 0.028 =P(all kids are girls)   approx 0.035 =P(all kids are boys)

P(all kids are of the same sexboys or girls)=0.063

Step-by-step explanation:

a) The probability that exactly 3 of 5 kids are boys is

p(3 boys)=C5 3 *p^3*q^2

C5 3 =   5!/(3!*2!)= 4*5/2=10

p=0.51 is the probability a child is boy

q=1-p=1-0.51=0.49 is the probability a child is girl

P(3 boys of 5 kids)= 10*0.51^3*0.49*2

b) The event that at least 1 kid from five is boy is a combination of events

1 boy -4 girls, 2 boys-3 girls, 3 boys-2 girls, 4 boys-1 girl, 5 boys.

The event " all 5 kids are the girls"  is the only event which is not the part of

event " at least 1 kid from five is boy"

So P(at least 1 kid from five is boy)=1 - P(all 5 kids are the girls)

P(all 5 kids are the girls)=0.49^5=approx 0.028

P(at least 1 kid from five is boy)=1-0.028=approx 0.97

c) Similarly like b)  P(at least one of five kids is girl)= 1-P(all 5 kids are the boys)

P(all 5 kids are the boys)=0.51^5=approx 0.035

P(at least one of five kids is girl)= 1-0.035=approx=0.965

d) Probability all kids are of the same sex= probability that all kids are the boys+ probability all kids are the girls.

Using b) and c)  P(all kids of the same sex)= 0.035+0.028=0.063