Question

The Blair family was one of the first to come to the original 13 colonies (now part of the USA). They had 4 children. Assuming that the probability of a child being a girl is 0.5, find the probability that the Blair family had... (a) ...at least 3 girls

Answer 1

31.25% probability that the Blair family had at least 3 girls

Explanation:

For each children, there are only two possible outcomes. Either it was a girl, or it was not. The probability of a child being a girl is independent of any other children. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

They had 4 children.

This means that
`n = 4`

The probability of a child being a girl is 0.5

This means that
`p = 0.5`

Probability of at least 3 children:

`P(X \geq 3) = P(X = 3) + P(X = 4)`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 3) = C_(4,3).(0.5)^(3).(0.5)^(1) = 0.25`

`P(X = 4) = C_(4,4).(0.5)^(4).(0.5)^(0) = 0.0625`

`P(X \geq 3) = P(X = 3) + P(X = 4) = 0.25 + 0.0625 = 0.3125`

31.25% probability that the Blair family had at least 3 girls