I'm currently going through some known probability distributions and I'm stuck at the exponential distribution.
Here is an example task:
You are in a super market standing in a queue. You know that the average waiting time is $4$ minutes. Assume that the waiting time $X$ (in minutes) is exponentially distributed. What's the probability that you need to wait...
(i) exactly $3$ minutes?
(ii) more than $4$ minutes?
(iii) between $2$ and $7$ minutes?
If I understood the exponential distribution correctly, I will need to use its distribution function to calculate these probabilities, which is given by
$$F(t)=P(X<t)=\begin{cases} 1-e^{-\lambda t}&\mbox{if }t \geq 0\\ 0&\mbox{else }\end{cases}$$
(i) I'm not sure and leave it away for now.
(ii) $P(X >4) = e^{-\frac{1}{4}\cdot 4}$
(iii) $P(2\leq X \leq 7)=P(X\leq 7)-P(X \leq 2) = 1-e^{-\frac{1}{4}\cdot 7}-\left(1-e^{-\frac{1}{4}\cdot 2}\right)$
(ii) and (iii) should be fine, I hope. But what about (i)?
For a continuous distribution, the probability value at a specific point is zero. It's a measure theory issue:
For discrete probability spaces, the sample space is countable (finite or infinite), and it will either be $\Omega=\{\omega_1,\omega_2,\cdots,\omega_n\}$ or $\Omega=\{\omega_1,\omega_2,\cdots\}$.
We can afford to take the set of all possible subsets of the sample space as the sigma algebra, $\mathcal F = 2^\Omega$, i.e. all subsets of $\Omega$ are events, and hence, will have assigned probabilities.
We assign probabilities to all the subsets of $\Omega$, or in this case, events $\bf A$, via the assigned probabilities to singletons, $P(\{\omega\})$.
So, $P(\bf A)= \displaystyle \sum_{\omega\in A} P(\{\omega\})$.
Whatever way probabilities are assigned, it should satisfy that $\displaystyle \sum_{\omega\in \Omega} P(\{\omega\})=1$.
However, in the case of continuous distributions (uncountable sample spaces over the real line), it is not possible to assign probabilities to $\mathcal F = 2^\Omega$ since $\Omega$ is uncountable. $2^\Omega$ has higher cardinality than the real numbers. We can't assign probabilities to, for example, $[0,1]$,based on adding up the probabilities of singletons.
The most common way to assign probabilities is using intervals $(a,b)$ is $b-a$. We want
$$\mu((a,b)) = \mu((a,b])=\mu([a,b))=\mu([a,b])=b-a.$$
From there we can see that if $b-a=0$ (i.e. the probability assigned to a singleton, $\Pr\left(\{\omega\}\right)$), as in the case of $\Pr(X=3\text{ min})$ in the OP, the probability will be zero.
I bet that was the intention behind the inclusion of the word "exactly" in the question.