For a continuous random variable, when can you use the PMF vs. the CDF? I think I am missing a core concept in my (fairly basic) probability class.  I will try to explain my confusion as best I can...
If I am given a continuous random variable
(for instance, say that we have $X$ which follows an exponential distribution with expected value $E(X)=3$)
and we want to to find the probability that the outcome is greater/less than a particular value (continuing with my example, say we are trying to find $P(x>5)$,) my intuition tells me that I would need to integrate the exponential PMF from 5 to infinity.  However, in looking at the solution for this particular problem, I can actually calculate the answer by simply plugging 5 into the exponential PMF.
Can anybody explain to me when I would be able to simply plug the value into the PMF and when I would need to instead integrate the PMF?
I apologize if this is unclear! My best guess here is that I am missing some key difference between the PMF and the CDF.
If anybody could try and elucidate this for me I would be very grateful!  Thanks in advance.
 A: The answer is the integral of the PDF from $5$ to $\infty$, and is not the PDF evaluated at $5$. The reason you might be confused here is that the exponential distribution has a PDF whose value is very similar to its integral.
If $X$ is exponential with a mean of $3$ (a rate of $\frac13$), then:


*

*The PDF of $X$ is $f(x) = \frac13 e^{-x/3}$ (for $x \ge 0$)

*The CDF of $X$ is $F(x) = 1 - e^{-x/3}$ (for $x \ge 0$)


To calculate $\Pr[X > 5]$, we can do one of two things:


*

*Use the PDF. In that case, we integrate $f$ from $5$ to $\infty$, getting $$\int_5^\infty \frac13 e^{-x/3}\,dx = \left. -e^{-x/3}\vphantom{\int}\right|_5^{\infty} = e^{-5/3}.$$

*Use the CDF: evaluating $F$ at $5$ tells us that $\Pr[X \le 5] = F(5) = 1 - e^{-5/3}$, so $\Pr[X > 5] = 1 - \Pr[X \le 5] = e^{-5/3}.$


But if we'd just evaluated the PDF at 5, we'd have gotten $\frac13 e^{-5/3}$, which is not the right answer.
Note that the second method is just the first method in disguise: in cases where the PDF exists, the CDF is just the integral of the PDF, so here we have $$F(x) = \int_0^x f(t)\,dt = \int_0^x \frac13 e^{-t/3}\,dt = \left. -e^{-t/3}\vphantom{\int}\right|_0^{x} = 1 - e^{-x/3}.$$
