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

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 $k = 0, 1, 2, 3$.
