Garrett loves eating donuts, but enjoys savoring them. The time it takes for him to eat a donut is uniformly distributed between 1 and 5 minutes. If he has been eating for 2 minutes and still isn't done, find the probability that he finishes the donut in the next minute.
My initial attempt was as follows: let $X$ be the number of minutes it takes Garrett to eat this donut. Then we want to find $P(2< X < 3)$. We have $$P(2 < X < 3)=\int_2^3g(x)dx$$ where $g(x)=1/(5-1)$ if $x\in [1,5]$ and $g(x)=0$ otherwise.
Apparently this approach is wrong, and apparently I should be using another $g$, namely $g(x)=1/(5-2)$ iff $x\in[2,5]$. But it is not clear to me why I should be using this $g$ and not the previous one. How do I formally deduce that it is the latter $g$ that I should be using?
One thing that I think I didn't take into account is that the problem is asking for conditional probability. So I suppose I should be using the conditional probability density function, $f_{Y|X}$ from here. In that notation, $X$ = time for which he has been eating the donut, $Y$ = total time for which he will consume the donut -- is that right? What should $f_X, f_{X,Y}$ should be in this case? Which one of them should be my old $g$ ($g(x)=1/4$ for $x\in[1,5]$)? And what are the other functions in this case?
You're using a bazooka when a peashooter will do. Garrett is a priori equally likely to finish his donut in the second minute, the third minute, the fourth minute, and the fifth minute. You know that he did not finish his donut in the second minute. Therefore, the conditional probability that he will finish it in the third minute is $\frac 13$.