# What's the probability that the first four children born are boys and the last two children born are girls?

I'm having some problems with determining how to calculate a question about the gender proportion in newborns in some random family.

A family consists of 6 children. The probability of a boy being born is $1/2$ which is also the probability of a girl being born. So $p = 1/2$.

I'm well aware that the variables follow a binomial distribution.

I want to determine the probability of the first 4 children born being boys and the last 2 being girls.

My "guesses" would be either:

$$\left(\frac{1}{2}\right)^4 \cdot \left(1- \frac{1}{2}\right)^2 \Longleftrightarrow \left(\frac{1}{2}\right)^6 = \frac{1}{64}$$

or

$$\left(\frac{1}{2}\right)^4 + \left(1- \frac{1}{2}\right)^2 \Longleftrightarrow \frac{1}{16} + \frac{1}{4} = \frac{5}{16}$$

I don't know the correct answer, and I'm not quite sure which argument to use.

• You can right-click on the equations, then select Show Math As TeX Commands to see how I formatted the equations. Also, please see this tutorial on how to typeset mathematics on this site. – N. F. Taussig May 26 '15 at 12:15

It's one possibility out of $2^6$ equally likely possibilities. So $\frac 1 {2^6} = \frac 1 {64}$.

The first calculation you did is correct because the probabilities are independent.

In more detail: $$\Pr[\text{first four children are boys and rest are girls}] = \\\Pr[\text{second to fourth children are boys and rest are girls} \mid \text{first child is a boy}] \cdot \Pr[\text{First child is a boy}] = \\\Pr[\text{second to fourth children are boys and rest are girls}] \cdot \Pr[\text{First child is a boy}]\text{ (by independence)}$$