Find probability of success on ith attempt in Geometric distribution

Question

I have this question

"An experiment is repeated, and the first success occurs on the 5th attempt. What is the success probability for which this is most likely to happen?"

This sounds like a Geometric distribution because of "first success occurs" so I put in the formula which is $p(1-p)^{4}$. Then I heard you're supposed to get the derivative of that and find when it equals $0$? How does that help us find $p$? What does the derivative of $p(1-p)^{4}$ give me?

Answer 1

For different values of $p$, this probability $p(1-p)^4$ will be different. The question is essentially asking you "which value of $p$ makes $p(1-p)^4$ the largest?" Computing a derivative can be helpful in maximizing this quantity.

Answer 2

Graphical comment on @angryavian's Answer (+1): Using R to graph and search.

p = seq(0,1,by=.001);  pdf = p*(1-p)^4
plot(p, pdf, type="l")
  abline(v=0:1, col="green2")
  abline(h=0, col="green")
p[pdf==max(pdf)]
[1] 0.2