Probability density function for choosing with replacement Suppose I have a jar of equal numbers of 10 different types of marbles (green, blue, etc.), and I randomly select (and then replace) a marble a thousand times. Consequently I can expect that I will pick each type of marble roughly 100 times, but how can I calculate the formula which describes the probability density function?
For example, if it's a normally distributed value, which I think it is, then I would want to know the mean (which I believe would be 100 for this example) and the standard deviation.
 A: I think the key here is that we can look at this as a sequence of n Bernoulli Trials. Then this becomes trivial:
E(Yn) = n*p
var(Yn) = n*p*(1-p)
sd(Yn) = (n*p*(1-p))^0.5

So, for the example above, we get a expected value of 100, with a standard deviation of 9.4868
I found the following resource helpful: http://www.randomservices.org/random/bernoulli/Introduction.html
A: The probability model for the numbers of each kind ($X_1$ green, $X_2$ blue, $X_3$ red,...) is multinomial. 
The marginal probability model (if you're interested only in the number of a particular colour) is binomial. So if $X_i$ is the observed number of draws of color $i$, then $X_i\sim \text{Bin}(1000,\frac{1}{10})$, so $E(X_i) = np_i = 1000/10 = 100$ and $\text{Var}(X_i)=np_i(1-p_i) = 100 \times 0.9 = 90$  (so the standard deviation = $\sqrt{90}$) and $\text{Cov}(X_i,X_j)=-np_ip_j = -10$; the correlation between the colour-counts is thereby $-1/9$.
In either case (whether looking at counts on all colors or focusing on the count of  one color), the distribution is not normal. However if the number of balls drawn is large, and the probabilities are not very small (both are fine in your problem), then the distributions are approximately (degenerate) multivariate normal and approximately univariate normal respectively.
