Question

A family has 3 children. Find the probabilities. A) all boys B) All boys or all girls C) Exactly two boys or two girls D) At least one child of each gender What are the probabilities

Answer 1

The possibilities are either girl or boy

p(b) = probability of boy =
`(1)/(2)`

p(g) = probability of girl=
`(1)/(2)`

A) all boys

If all are boys means the 3 children will be boys

Then

P(all boys) =
`p(b) * p(b) * p(b)`

P(all boys) =
`(1)/(2) * (1)/(2) * (1)/(2)`

P(all boys) =
`(1)/(8)`

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) =
`(1)/(8)`

Similarly ,

P(all girls) =
`(1)/(8)`

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) =
`p(b) * p(b) * p(g)`

P( Exactly two boys) =
`(1)/(2) * (1)/(2) * (1)/(2)`

P( Exactly two boys) =
`(1)/(8)`

Similarly P(Exactly two Girls ) means out of 3 children there will be 2 girls and 1 boy

P( Exactly two girls) =
`(1)/(8)`

D) At least one child of each gender

P(At least one child of each gender) =
`p(b) * p(g) * p(g)`

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) =
`(1)/(2) * (1)/(2) * (1)/(2)` =
`(1)/(8)`