# Finding the expected number of a certain colored ball drawn from an urn in k draws

Suppose we have an urn containing c yellow balls and d green balls. We draw k balls, without replacement, from the urn. Find the expected number of yellow balls drawn. Hint: Write the number of yellow balls drawn as the sum of c random variables.

I initially thought the solution is the sum of a hypergeometric distribution:

$$E(X) = \sum_{x = 0}^{k}x\frac{\binom{c}{x}\binom{d}{k-x}}{\binom{c+d}{k}}$$

The issue is that this doesn't work for any value of $$k$$ (for example, what if $$k > c$$ or $$k-x > d$$), it also doesn't fit with the hint given in the question. Is there a way to generate a more general solution for $$1?

• Your formula looks right to me. When $x>c$, we have ${c\choose x}=0$ for example. It's only if $c+d>k$ so that the denominator is $0$ that we have a problem. May 17 '19 at 5:16

All you need to do is replace $$k$$ with $$min(k,c)$$. So the expression would become :

$$E(X) = \sum_{x = 0}^{min(k,c)}x\frac{\binom{c}{x}\binom{d}{k-x}}{\binom{c+d}{k}}$$

The comment of saulspatz answers your question, and what I write here is actually not an answer.

I would like to attend you on a simpler way to find $$\mathbb EX$$ (too large for a comment ).

Give the yellow balls numbers $$1,2,\dots,c$$ and let rv $$X_i$$ take value $$1$$ if ball $$i$$ is chosen and value $$0$$ otherwise.

Then $$X=X_1+\cdots +X_c$$.

Now apply linearity of expectation being aware of the fact that the $$X_i$$ have identical distribution with mean $$\frac {k}{c+d}$$.

So you will end up with: $$\mathbb EX=\frac {kc}{c+d}$$

I do not exclude that you already found this yourself on base of the hint.