# $P\{X>r\}$ for a geometric distribution

In the geometric distribution section of the Head First statistics book, it is said that the $P\{X>r\}$ is $q^r$ because for the success to occur after $r$ trials it means that the first $r$ trials must have ended in failure. And, according to the book, we don't need $p$ in this formula because we do not want to know which trial was successful, only that we needed more than $r$ trials. However, I do not really understand this concept. Shouldn't the probability rather be $q^rp + q^r+p$ and so on until infinity. Could anyone please help explain this?

The book does not cheat you. In order to know that $X>r$, you only need to ensure that $r$ first trials are unsuccessful. These $r$ consecutive failures have probability $q^r$, which is the answer to your question.

• Thanks for the answer, I understand the mathematical proof now but intuitively the only thing that confuses me is that when we consider the success occurring in either r+1 or r+2 or r+3 or r+4 trials (which is what is implied by P{X>r} in my mind) what we usually do is add up the probabilities for each of those, since the probability of it being > r can occur in any of those and not necessarily the r+1th trial which is what q^r would suggest.
– QPTR
Commented Sep 26, 2015 at 16:26
• @QPTR, if you always think about what you "usually do", you'll never learn anything new. So I recommend to try hard getting used to new ideas and approaches. Commented Sep 26, 2015 at 18:33
• well, true, but I do want to learn, which is why I was curious about the intuition behind this new way.
– QPTR
Commented Sep 27, 2015 at 2:02

Let us remember how geometric distribution works. Let $X$ is geometric distr. with parameter $p$.

Then, we calculate:

$$Pr(X\le r)=\sum_{k=1}^rpq^{k-1}.$$

Now we are looking for opposite probability (remember that $1=p+q$), so: $$Pr(X>r)=1-\sum_{k=1}^rpq^{k-1} =1-\sum_{k=1}^r((1-q)q^{k-1}).$$

$$Pr(X>r)=1-\sum_{k=1}^r(q^{k-1}-q^k) =1-(1-q^r)=q^r.$$

Where in the last equality survives only first and last member of the sum. Hence the answer $q^r$.

• Thanks for the answer. I understand the derivation now!
– QPTR
Commented Sep 26, 2015 at 11:38
• I also tried doing it the other way round that is the sum of q^r and q^r+1 and so on and it gives the same answer!
– QPTR
Commented Sep 26, 2015 at 11:40