# Find expected red drawn balls when draw $n$ balls?

I have $k_1$ red balls and $k_2$ blue balls in a box. Randomly select $n$ balls in the box without replacement. How to find expected number of red (or blue) balls.

This is my solution

The probability to get r red balls from $k_1+k_2$ balls in the box (event X): $$P_1(X)=\frac{\binom{k_1}{r}\binom{k_2}{n-r}}{\binom{k_1+k_2}{r}}$$

The $r$ can be from $0$ to $k_1$ balls, so the prob. of the event is $$P_2(X)=\sum_{r=0}^{k_1}\frac{\binom{k_1}{r}\binom{k_2}{n-r}}{\binom{k_1+k_2}{n}}$$

Finally, we got the expected value as $$E(X)= \sum_{r=0}^{k_1} r\times \frac{\binom{k_1}{r}\binom{k_2}{n-r}}{\binom{k_1+k_2}{n}}$$

Is it right?

• What do you mean with "in $k$ trails"? It seems to me that $k_1,k_2,n$ are determining and there is no place for an extra parameter $k$. – drhab Aug 23 '16 at 8:25
• $k$ is number of trails and it given. For example, I draw 100 times and count the expected value – Jame Aug 23 '16 at 8:27
• But what is the function of $n$? Aren't you talking about randomly selecting $n=100$ balls here? – drhab Aug 23 '16 at 8:29
• $n$ is number of balls in each drawn times – Jame Aug 23 '16 at 8:30
• Still unclear to me. Can you describe what happens if e.g. $n=2$ and $k=3$? – drhab Aug 23 '16 at 8:33

For $$i=1,\dots,n$$ let $$R_i$$ take value $$1$$ if trial $$i$$ results in a red ball, and let it take value $$0$$ otherwise.

Then $$R:=R_1+\cdots+ R_n$$ stands for the number of red balls.

With linearity of expectation and symmetry we find:$$\mathbb ER=n\mathbb ER_1=\frac{nk_1}{k_1+k_2}$$

edit for explanation:

As described in your question we will randomly select $$n$$ of the $$k_1+k_2$$ balls, but in order to create clarity before doing so we first reserve $$n$$ spots that are labeled by the numbers $$1,2,\dots,n$$. The first ball selected is placed on spot $$1$$, the second on spot $$2$$, et cetera.

Then event $$\{R_i=1\}$$ can be recognized as the event that a red ball will be placed on spot $$i$$.

Essential is here that at the stage of the beginning of the selection all balls have equal chance to become the one that will be placed on spot $$i$$.

There are $$k_1+k_2$$ balls in total, so that chance must be $$\frac{1}{k_1+k_2}$$.

Further $$k_1$$ of the balls are red, so the probability that a red ball will be placed on spot $$i$$ equals: $$\frac{k_1}{k_1+k_2}$$

If that is not clear enough yet then number the red balls with $$1,2,\dots,k_1$$ and let $$A_j$$ denote the event that ball $$j$$ will be placed on spot $$i$$.

Then the event that a red ball is placed on spot $$i$$ is $$A_1\cup\cdots\cup A_{k_1}$$ and its probability is: $$P(A_1\cup\cdots\cup A_{k_1})=\sum_{j=1}^{k_1}P(A_j)=\sum_{j=1}^{k_1}\frac1{k_1+k_2}=\frac{k_1}{k_1+k_2}$$

• Since the R_i are not i.i.d, how does the linearity of expectation apply here? The expected value of, e. g. R_17 is not obvious to me at least. – Zach Boyd Mar 25 at 20:00
• 1) linearity of expectation can always be applied. You don't need i.i.d. for that. 2) Here the $R_i$ are not independent but have equal distribution. I assure you that e.g. balls 1 and 17 have equal chances to be red. This makes it possible to use symmetry in the sense that $R_1$ and $R_{17}$ have the same expectation. – drhab Mar 25 at 21:01
• Agreed on the linearity. Is there an easy way to see why the distributions are equal? – Zach Boyd Mar 25 at 21:13
• Compare it with this: blindfolded you grab $n$ balls and then you place them on spots that are labeled with the numbers $1,2,\dots,n$. Can you think of any reason why e.g. the probability that spot $4$ receives a red ball will be different than the probability that spot $12$ receives a red ball?...I am telling you there is none! This situation does not essentially differ from picking out balls one by one saying: "you are the first, you are the second.." et cetera. – drhab Mar 26 at 9:08
• That is pretty slick. Do you know if there is a way to translate that into a rigorous argument? The assertion essentially seems to be that there is a symmetry in the problem. – Zach Boyd Mar 26 at 13:17

The probability to draw $\ell$ red balls out of $r$ total draws is $$\frac{\binom{k_1}{\ell}\binom{k_2}{r-\ell}}{\binom{k_1+k_2}{r}}.$$ This follows a hypergeometric distibution, and thus has expectation $$r\frac{k_1}{k_1+k_2}.$$ Note that, similarly, the expected value of drawn blue balls equals $$r\frac{k_2}{k_1+k_2}$$ and thus that the sum equals $r$.

• I have updated my equation. Your equation is look like me. Could you check it again. I think it must be summation – Jame Aug 23 '16 at 8:19
• Your formula for the expectation in the form of a sum is indeed correct. A closed form for this sum is the fraction I gave in the answer. – Marc Aug 23 '16 at 8:32