# Hypergeometric or?

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.

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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

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.