Equal number of boys and girls

There are two possible outcomes for the gender of each children: boy or girl. The probability of a boy $b$ is $\frac{1}{2}$ , and the probability of a girl $g$ is $\frac{1}{2}$ .

$(b+g)^4$ $=$ $b^4 + 4b^3g + 6b^2g^2 + 4bg^3 + g^4$

The term $6b^2g^2$ represents $2$ boys and $2$ girls.

$P_{(2 boys,2girls)}$ $=$ $6(\frac{1}{2})^2$ $(\frac{1}{2})^2$ $=$ $\frac{3}{8}$

We can select 2 boys (or girls) from total 4 children in 4C2 (=6) ways. And there are total 16 possible outcomes. Hence the probability would be 6/16 or 3/8.

Each child can be a boy or a girl, so there are $2^4=16$ total different variations of children.

The number of combinations of 4 children which contain 2 boys and 2 girls is $\dbinom{4}{2} \times \dbinom{2}{2} = 6$ .

Thus, the probability of 2 boys and 2 girls is $\dfrac {6}{16} = \dfrac {3}{8}$ .

We can use the formula n! ÷ k!(n-k)!

The possible arrangements of boys and girls can be calculated by calculating the sum of (from when k=0 to n=4 (I can't get that fancy E symbol sorry)) of the formula I gave above. n= 4 as we have 4 children.

When n=4 and k=0 the answer is 1.

When n=4 and k=1 the answer is 4

When n=4 and k=2 the answer is 6

When n=4 and k=3 the answer is 4

When n=4 and k=4 the answer is 1

1+4+6+4+1=16 therefore there are 16 possible outcomes.

We need k to be 2 since we have to arrange 2 boys out of 4 children so if we plug in the values we got 6 ways

Therefore the probability of having 2 boys and 2 girls is 6/16 = 3/8