Question

Let the random variable X denote the number of girls in a five-child family. If the probability of a female birth is 0.5, what are the probabilities of having 0, 1, 2, 3, 4, and 5 girls?

Answer 1

See below:

Explanation:

The probability of having a girl is given as `0.5`. This means that the probability of not having a girl is also `0.5`.

The last thing to look at is the number of ways we can achieve any given result. For instance, there is only 1 way to have 0 girls (and the same goes for 5 girls), but there are 5 different ways we can have 1 girl (and the same goes for 4 girls). This is a combinations question, with:

`C_(n,k)=((n),(k))=(n!)/((k!)(n-k)!)` with `n="population", k="picks"`

Boiling it all down, we end up with:

0 girls `=((5),(0))(1/2)^0(1/2)^5=1xx1xx1/32=1/32`

1 girl `=((5),(1))(1/2)^1(1/2)^4=5xx1/2xx1/16=5/32`

2 girls `=((5),(2))(1/2)^2(1/2)^3=10xx1/4xx1/8=10/32=5/16`

3 girls `=((5),(3))(1/2)^3(1/2)^2=10xx1/8xx1/4=10/32=5/16`

4 girls `=((5),(4))(1/2)^4(1/2)^1=5xx1/16xx1/2=5/32`

5 girls `=((5),(5))(1/2)^5(1/2)^0=1xx1/32xx1=1/32`

Notice that if we add all of these up, we get 1 (or the sum total of all probabilities):

`1/32+5/32+10/32+10/32+5/32+1/32=32/32=1`