An urn contains 4 black and 6 white balls. Two balls are drawn without replacement. All black balls in this sample are reserved. All white balls (if any) are returned to the urn and the same number of balls is drawn from the urn. What is the total expected number of black balls obtained in this experiment?

So I was thinking in terms of total expectation where X is the total number of black balls and Y is the number of white balls in the first draw:

$E(X) = E(E(X|Y)) = (4/10)(3/9) E(X|Y=0) + (6/10)(4/9)*2*E(X|Y=1) + (6/10)(5/10)E(X|Y=2) $

Is this the right track? Because indicator variables seem to be possible here too.

  • $\begingroup$ What do you mean by 'the same number of balls'? Again $2$ or a number of balls that equals the number of white balls that are returned? $\endgroup$ – drhab Dec 18 '13 at 23:22
  • $\begingroup$ The problem most likely means 2. $\endgroup$ – user108431 Dec 18 '13 at 23:30
  • $\begingroup$ There is ambiguity, my tentative interpretation is that if for example $2$ black are drawn, then second draw does not take place. The conditional expectation argument seems fine, minor typo should be $(6/10)(5/9)$ in the last part. Indicator rv is another way, $X_i=1$ if $i$-th black is obtained. Then our expectation is $4\Pr(X_i=1)$. But calculation is still needed for the probability, there may be no saving. $\endgroup$ – André Nicolas Dec 18 '13 at 23:52

$E\left(X\right)=E\left(X\mid Y=0\right)P\left\{ Y=0\right\} +E\left(X\mid Y=1\right)P\left\{ Y=1\right\} +E\left(X\mid Y=2\right)P\left\{ Y=2\right\} $

Here $P\left\{ Y=0\right\} =\binom{10}{2}^{-1}\binom{4}{2}\binom{6}{0}=\frac{2}{15}$ , $P\left\{ Y=1\right\} =\binom{10}{2}^{-1}\binom{4}{1}\binom{6}{1}=\frac{8}{15}$ and $P\left\{ Y=2\right\} =\binom{10}{2}^{-1}\binom{4}{0}\binom{6}{2}=\frac{5}{15}$

This agrees with what you are saying in your question (with exception of the typo mentioned by André).

$E\left(X\mid Y=0\right)=2+EU$ where $U$ is the number of obtained blacks by drawing $2$ balls from an urn with $0$ blacks and $6$ whites.

$E\left(X\mid Y=1\right)=1+EV$ where $V$ is the number of obtained blacks by drawing $2$ balls from an urn with $1$ blacks and $6$ whites.

$E\left(X\mid Y=2\right)=0+EW$ where $W$ is the number of obtained blacks by drawing $2$ balls from an urn with $2$ blacks and $6$ whites.

It is clear that $U=0$, hence $EU=0$. As André points out in his comment you could also agree that in this situation the second drawing does not take place. There are no black balls to catch anymore. Next to that $EV=\frac{2}{7}$ and $EW=\frac{1}{2}$. They can be calculated as mean of hypergeometric distributions (or more directly by using indicator rv's). This leads to:



multiply the probability of getting n balls by n and add all the terms up to get expected number of balls.

Probability of drawing 4 balls: $\dfrac{\binom{4}{2}}{\binom{10}{2}}\cdot\dfrac{\binom{2}{2}}{\binom{8}{2}}=\dfrac{1}{210}$

Probability of drawing 3 balls: $\dfrac{4\cdot6}{\binom{10}{2}}\cdot\dfrac{\binom{3}{2}}{\binom{9}{2}}+\dfrac{\binom{4}{2}}{\binom{10}{2}}\cdot\dfrac{2\cdot6}{\binom{8}{2}}=\dfrac{32}{315}$

probability of drawing 2 balls:$\dfrac{\binom{4}{2}}{\binom{10}{2}}\cdot\dfrac{\binom{6}{2}}{\binom{8}{2}}+\dfrac{\binom{6}{2}}{\binom{10}{2}}\cdot\dfrac{\binom{4}{2}}{\binom{10}{2}}+\dfrac{6\cdot4}{\binom{10}{2}}\cdot\dfrac{3\cdot4}{\binom{9}{2}}=\dfrac{37}{126}$

Probability of drawing 1 ball:$\dfrac{4\cdot6}{\binom{10}{2}}\cdot\dfrac{\binom{6}{2}}{\binom{9}{2}}+\dfrac{\binom{6}{2}}{\binom{10}{2}}\cdot\dfrac{4\cdot 6}{\binom{10}{2}}=\dfrac{2}{5}$

Probability of drawing 0 balls: $\dfrac{\binom{6}{2}}{\binom{10}{2}}\cdot\dfrac{\binom{6}{2}}{\binom{10}{2}}=\dfrac{1}{9}$

So the average number of black balls is $4\cdot\dfrac{1}{210}+3\cdot\dfrac{32}{315}+2\cdot\dfrac{37}{126}+\dfrac{2}{5}=\dfrac{59}{45}\approx1.31$

  • $\begingroup$ Nooooooo! this is wrong!! it adds up to 41/45!(Bam!(commits suicide). Well one of those has a mistake, Ill go back to it later, I'm too exhausted to fix now. $\endgroup$ – Jorge Fernández Hidalgo Dec 19 '13 at 0:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.