Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a question regarding (a homework ) assignment. I've done some research but I couldn't get clear I were on the right track:

$8$ people want to decide who is the designated driver. They each draw a marble in turns without putting it back. One is red, the person who draws that is the designated driver.

$X$ is the amount of drawings needed to conclude a designated driver.

I assumed this is a hypergeometric experiment, but it is no where covert in the material available to us. That made me question if there is another way of doing it. Furthermore, I tought that each individual drawing could be seen as a seperate binomial experiment. We are asked to calculate $E(X)$, $\operatorname{Var}(X)$ and $P(X\geq 3)$. $E(X)$ could then be calculated by $E(X)=np_1 + np_2 + np_3 ... np_8$ and $\operatorname{Var}(X)$ the same way (the sum of each variance).

What is the correct way to approach this?

Sorry if I messed up some terms, because english isn't my native language and the course is given in dutch.

share|cite|improve this question
For some basic information about writing math at this site see e.g. here, here, here and here. – Julian Kuelshammer Dec 3 '12 at 11:57
up vote 0 down vote accepted

While you could well use a hypergeometric distribution, the problem can easily be done without first having knowledge of its formulas.

Let the total number of marbles be $n$. Then the probability of picking the red marble on any specific turn is $\frac{1}{n}$. To see this, imagine that the marbles are picked out of a line from left to right: the red marble is then equally likely to be in any of the $n$ positions.

Hence $P(X\ge3)=1-\frac{2}{n}$, assuming that $n\ge2$.

Also, $E(X^k)=\sum_{i=1}^{n}i^kP(X=i)=\frac{1}{n}\sum_{i=1}^{n}i^k$ for any $k$, so using $k=1,2$ you can find $E(X)$ and $E(X^2)$. Now use $Var(X)=E(X^2)-[E(X)]^2$ to find the variance.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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