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.

  • $\begingroup$ 1. You don't specify the relative numbers of each color -- are there twice as many green balls as blue balls? Ten times? Half? ... Your conclusion that you expect 100 of each is not justified by the information you gave (which doesn't specify enough to tell). 2. The phrase "normal function" wouldn't fit here (it's not normal). Do you mean "How do I find the probability function for the number of green balls?" Or do you mean "How do I find the standard deviation of the number of green balls?" Or something else? $\endgroup$ – Glen_b -Reinstate Monica Mar 2 '17 at 3:49
  • $\begingroup$ Please edit your question to contain the necessary information to have an answerable question. $\endgroup$ – Glen_b -Reinstate Monica Mar 2 '17 at 3:50
  • $\begingroup$ @Glen_b if I state that I expect that I can pick each out 100 times out of 1000, then I state the probability of each ball getting selected (10%). I actually believe the answer is normally distributed, and I have verified that it is normally distributed with a quick simulation. I posted an answer below if you're curious to understand the logic. $\endgroup$ – user35581 Mar 3 '17 at 10:14
  • $\begingroup$ I have attempted to improve your question somewhat, making some of the requested changes. However, the question still needs to be fixed.in relation to my original point 2. $\endgroup$ – Glen_b -Reinstate Monica Mar 3 '17 at 15:13
  • $\begingroup$ I have made some edits to address your point #2 - thanks for the feedback. $\endgroup$ – user35581 Mar 3 '17 at 15:30

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.math.uah.edu/stat/bernoulli/Introduction.html


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.


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