Original question:
The time it takes a bank clerk to serve a customer is distributed exponentially with the average of 5 minutes. What is the probability that it will take more than 10 minutes to serve 2 customers, one after another? Serving times are independent.
So we have $\lambda = \frac15$ and our limits of integration are from 5 till $\infty$.
My questions are:
1.) Why can't I use the probably density function to calculate them individually and then add them up?
2.) Using the solutions I found out I must use convolution. How do I know when to use it? What do I set my variables to? A definition of convolution I saw earlier gave me:
$$\int_{-\infty}^{\infty} f_1(z-x)f_2(x) dx $$
What exactly is (z-x) in this question for example?
Regarding your first question, the reason why you can't find the individual probabilities and add them is because for two independent events, it is not necessarily the case that $$\Pr[A \cap B] = \Pr[A] + \Pr[B].$$ For example, if $A$ is the event that you roll a number greater than $1$ on a fair six-sided die, then $\Pr[A] = 5/6$. If $B$ is the event that you get heads on a flip of a fair coin, then $\Pr[B] = 1/2$. But it is obviously absurd to suggest that the chance you roll the die and flip the coin and observe both $A$ and $B$ must then be $5/6 + 1/2 = 4/3 > 1$.
Rather, the correct formula is $$\Pr[A \cap B] = \Pr[A] + \Pr[B] - \Pr[A \cup B].$$ Moreover, this formula is true regardless of whether $A$ and $B$ are independent events.
Going back to the original question, you need to model the serving times of the customers. For the first customer, their waiting time is an exponentially distributed random variable, say $X_1$: $$X_1 \sim \operatorname{Exponential}(\lambda = 1/5), \quad f_{X_1}(x) = \frac{1}{5}e^{-x/5}, \quad x > 0.$$ For the second customer, their service time is also exponentially distributed in the same way, and is independent of the waiting time of the first: $$X_2 \sim \operatorname{Exponential}(\lambda = 1/5), \quad f_{X_2}(x) = \frac{1}{5} e^{-x/5}, \quad x > 0.$$ Now, you could calculate $\Pr[X_1 > 10]$ or $\Pr[X_2 > 10]$ but what you really want is $$\Pr[X_1 + X_2 > 10],$$ which is neither of the two probabilities above, nor is it $\Pr[X_1 > 5] + \Pr[X_2 > 5]$, since of course the event $X_1 + X_2 > 10$ could occur if, for example, $X_1 = 11$ and $X_2 = 1$. So you need to think about all the ways in which the sum of the service times exceeds $10$ minutes. And this is what the convolution does.
So, this brings us to your second question. We should motivate it in the following manner, but by first considering the complementary event: that is, $$\Pr[X_1 + X_2 > 10] = 1 - \Pr[X_1 + X_2 \le 10],$$ since either the total service time exceeds $10$ minutes, or it does not exceed $10$ minutes, and that these are the only possible outcomes for the total serving time. And since the serving times are positive, it is perhaps more natural to think of the probability associated with the event $$0 < X_1 + X_2 < 10.$$ To this end, we note that if we observed that the first customer was served in $x_1$ minutes, and $x_1 < 10$, then the second customer would need to be served in at most $X_2 < 10 - x_1$ minutes in order for the total time to remain within the $10$-minute limit. So if we condition on the observed service time of the first customer over all such possible times not exceeding $10$ minutes, we can get the unconditional probability of the total time through the law of total probability: we have $$\Pr[X_1 + X_2 \le 10] = \int_{x_1 = 0}^{10} \Pr[X_2 \le 10 - x_1 \mid X_1 = x_1]f_{X_1}(x_1) \, dx_1.$$ This is in some sense merely a weighted average of the conditional probability of the second customer's service time not resulting in the total time exceeding $10$ minutes, weighted by the probability density of the first customer's service time.
Another way to think about the desired probability is to look at the joint density of the two customers' service times $$f_{X_1, X_2}(x_1, x_2) \overset{\text{ind}}{=} f_{X_1}(x_1) f_{X_2}(x_2),$$ which is the product of the marginal densities of the individual service times because the two are independent. Then we simply integrate the joint density over the space of all possible outcomes satisfying the condition $0 \le x_1 + x_2 \le 10$: $$\Pr[X_1 + X_2 \le 10] = \int_{x_1 = 0}^{10} \int_{x_2 = 0}^{10-x_1} f_{X_1,X_2}(x_1,x_2) \, dx_2 \, dx_1.$$ You will find that this is equal to the previous integral we obtained through conditioning on $X_1$.