Probability generating function for urn problem without replacement, not using hypergeometric distribution UPDATE: Thanks to those who replied saying I have to calculate the probabilities explicitly. Could someone clarify if this is the form I should end up with:
$G_X$($x$) = P(X=0) + P(X=1)($x$) + P(X=2) ($x^2$) + P(X=3)($x^3$)
Then I find the first and second derivative in order to calculate the expected value and variance?
Thanks!
ORIGINAL POST: We have a probability question which has stumped all of us for a while and we really cannot figure out what to do. The question is:
An urn contains 4 red and 3 green balls. Balls will be drawn from the urn in sequence until the first red ball is drawn (ie. without replacement). Let X denote the number of green balls drawn in this sequence.
(i) Find $G_X$(x), the probability generating function of X.
(ii) Use $G_X$(x) to find E(X), the expected value of X.
(iii) Use $G_X$(x) and E(X) to find $σ^2$(X), the variance of X.
It appears to me from looking in various places online that this would be a hypergeometric distribution, as it is with replacement. However, we have not covered that type of distribution in our course and it seems the lecturer wishes for us to use a different method. We have only covered binomial, geometric and Poisson. I have tried to figure out an alternative way of finding the probability generating function and hence the expected value and variance (just using the derivatives), but, I have not been successful. Would anyone be able to assist?
Thanks! :)
Helen
 A: You don't need to use the formula for a hypergeometric distribution.  Simply observe that the most number of balls you can draw before obtaining the first red ball is $3$, so the support of $X$ is $X \in \{0, 1, 2, 3\}$.  This is small enough to very easily compute explicitly $\Pr[X = k]$ for $k = 0, 1, 2, 3$.
A: This would not be a hypergeometric distribution. You can think of hypergeometric as binomial without replacement, not geometric without replacement (even though the name might suggest otherwise). In other words, hypergeometric doesn't care at which spot the red ball is drawn.
Well, it should be relatively easy to find the probability mass function. Observe that, for example, $$\mathrm{P}(X = 2) = \color{green}{\frac37} \cdot \color{green}{\frac26} \cdot \color{red}{\frac45}$$
You can generalize this in the following manner:
$$
\mathrm{P}(X = x) = p_{X}(x) = \begin{cases}
\displaystyle \color{green}{\frac{\frac{3!}{(3 - x)!}}{\frac{7!}{(7 - x)!}}} \cdot \color{red}{\frac{4}{7 - x}} && x \in \{0, 1, 2, 3\} \\
0 && \text{otherwise}
\end{cases}
$$
Now you can use this to find the probability-generating function by definition.
