Probability of number of boys and girls being born on a single day
I know this is kind of tedious and only practical when you are considering no more than 10 children  (which I think is a lot to parent), but anyway, I like to solve these problems with a binomial expansion.

So we have binomial probability so 2 options - either a boy (n) or a girl (g) at equal probability or 50% or 1/2.

(b + g)^n = a binomial expansion according to the coefficients of Pascal's triangle.
Where n = 4 children, then the row of Pascal's triangle is 4 in which the first row of pascal's triangle is technically row 0. 

Equation: (b + g)^4 = 1(b)^4 + 4(b^3)(g) + 6(b^2)(g^2) + 4(b)(g^3) + 1(g^4)

Problem A)
So the probability that the number of boys and girls will be equal is 

Probability((6(b^2)(g^2)), where (b^2)(g^2) means two boys and two girls, an equal number.
When b = 1/2, and g = 1/2, or well ... b + g = 1

6(b^2)(g^2) = 6[(1/2)^2 * (1 / 2)^2] = 6 * (1/2)^4 = 6 / 16 = 3/8

B) All four will be girls: Prob((1(g^4))

= 1(1/2)^4 = 1/16

C) At least one baby will be a girl:
That corresponds to 4(b^3)(g), 6(b^2)(g^2), 4(b)(g^3), and 1(g^4)

The probability that at least one will be a girl is

= (4 + 6 + 4 + 1) * (1/2)^4 = 15 / 16

D) Which combination of boys and girls is most likely?
That would be 6(b^2)(g^2), so when b = g = 1/2

Prob(most likely combination) = 6/16 = 3/8 or two boys and two girls. 

Note that if you add up all the combinations, the result is 100% of all probabilities:

(b + g)^4 = 1(b)^4 + 4(b^3)(g) + 6(b^2)(g^2) + 4(b)(g^3) + 1(g^4)
When b = g = 1/2, then basically:

(1 / 16) + (4 / 16)+ (6 / 16) + (4 / 16) + (1 / 16) = 16 / 16 = 1 = 100%

You did it right. But I suppose this is another way to evaluate a binomial like this.