Solution:
The possibilities are either girl or boy
p(b) = probability of boy =[tex]\frac{1}{2}[/tex]
p(g) = probability of girl=[tex]\frac{1}{2}[/tex]
A) all boys
If all are boys means the 3 children will be boys
Then
P(all boys) = [tex]p(b) \times p(b) \times p(b)[/tex]
P(all boys) =[tex]\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}[/tex]
P(all boys) =[tex]\frac{1}{8}[/tex]
B) All boys or all girls
If all are boys means the 3 children will be boys or girls
From eq (1)
P(all boys) =[tex]\frac{1}{8}[/tex]
Similarly ,
P(all girls) =[tex]\frac{1}{8}[/tex]
C) Exactly two boys or two girls
P(Exactly two boys) out of 3 children there will be 2 boys and 1 girl
P(Exactly two boys) = [tex]p(b) \times p(b) \times p(g)[/tex]
P( Exactly two boys) =[tex]\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}[/tex]
P( Exactly two boys) = [tex]\frac{1}{8}[/tex]
Similarly P(Exactly two Girls ) means out of 3 children there will be 2 girls and 1 boy
P( Exactly two girls) = [tex]\frac{1}{8}[/tex]
D) At least one child of each gender
P(At least one child of each gender) =[tex]p(b) \times p(g) \times p(g)[/tex]
This means among the 3 children there should be one children of different gender. So lets assume out of three children one child be boy and the remaining 2 be girls
Thus
P(At least one child of each gender) = [tex]\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}[/tex] = [tex]\frac{1}{8}[/tex]
