Probability of two successes, possible geometric distribution problem

Let Y be the number of trials required to observe r successes. List the outcome space for the variable Y and show that the probabilities for Y are giver by the formula P(Y=k)=(k-1)(p^2)(1-p)^(k-2).

Okay I know the outcome space is Y:={2,3,4,...}, the first two trials could be successes or you could never get two successes. The problem looks geometric but I dont undertand the extra (k-1)*p in front.

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What if $r=3$...? How does this change the range of the random variable $Y$? – user21436 Apr 20 '12 at 3:41

Assuming independence of trials and that each trial has probability $p$ of success and also that $r=2$, as in the title:

$P[Y=k]$ is the probability that the second success occurred on the $k$'th trial. If the second success occurred on the $k$'th trial, then there was exactly one success in the first $k-1$ trials and the $k$'th trial was a success.

Now, there are $k-1$ ways to have exactly one success in the first $k-1$ trials (for example, one way is that the second trial was a success and the other $k-2$ trials were failures), each of which having probability $(1-p)^{k-2} p$. So the probability that exactly one success occurred in the first $k-1$ trials is

$$\tag{1}P( \textstyle{\text{exactly one success}\atop\text{ in the first }k-1\text{ trials} } )=(k-1)(1-p)^{k-2}p.$$

The probability of a success on the $k$'th trial is $$\tag{2}P( \textstyle{\text{ success on the}\atop k\text{'th trial} } ) =p$$

The probability that the second success occurred on the $k$'th trial is the product of $(1)$ and $(2)$: $$P[Y=k]=\bigl(\,(k-1)(1-p)^{k-2} p\,\bigr)\cdot p= (k-1)p^2(1-p)^{k-2}.$$

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