What is the probability that a set of nine children will contain three or fewer girls? I can't decide if it is 4/9 because there is the possibility of there being 0, 1, 2, or 3, or if it is 25% using combinations. Thanks!
 A: I will assume that the gender of each child is independently and uniformly chosen.  The question can as such be rephrased as "I flip a fair coin nine times in a row.  What is the probability that I see three or fewer heads."
This is a straightforward application of the binomial distribution.
The probability of getting exactly $k$ heads in flipping $n$ fair coins where probability of heads is $p$ is:
$$Pr(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$$
The probability of there being exactly three girls in your scenario with our assumptions is then:
$$\binom{9}{3}0.5^3\cdot 0.5^{6}=\frac{21}{128}$$
Now, remember that your question asks for the probability of three or fewer children being girls, so add to this result the probability of exactly two girls, exactly one girl, and exactly zero girls.

Your attempt that led to the answer of $4/9$ appears to have been because you thought to say "If my sample space just keeps track of the number of girls I've seen, there are four good outcomes, so if I divide the number of good outcomes by the size of the sample space, I'll get my probability."
This is wrong for a few reasons.  First, notice that there are in fact ten possibilities for the number of girls seen, not nine (namely 0, 1, 2, 3, ..., 8, 9).  Next, you should be made intimately aware of the fact that the formula $Pr(A)=\dfrac{|A|}{|S|}$ is only guaranteed to be true in the scenario where the outcomes in your sample space are equally likely to occur.  There are afterall only two outcomes to playing the lottery, you win or you lose.  The probability of actually winning the lottery is certainly not $50\%$ though since it should be clear that the probability of losing the lottery is much higher than the probability of winning.  Similarly for your problem, the probabilities for the outcomes are not the same.  It is far more likely for there to be four girls than zero girls for example.
A: There are $2^9=512$ possibilities for the genders of the children. Now, let us count the ways that there might be $0,1,2,$ or $3$ girls among them:
$$0:\binom{9}{0}=1\text{ way}$$
$$1:\binom{9}{1}=9\text{ ways}$$
$$2:\binom{9}{2}=36\text{ ways}$$
$$3:\binom{9}{3}=84\text{ ways}$$
Thus, the probability that there will be three or fewer girls is
$$\frac{1+9+36+84}{512}=\frac{130}{512}=\frac{65}{256}=0.2539$$
