# Probability that Independent Exponentially Distributed RVs are Equal

I am wondering about a following problem, which seems to be intuitively trivial, but I am not sure how to approach it (due to my lack of knowledge, I believe).

Imagine I have two machines working. Each of them have may break in an exponentially distributed moments in time with mean $1/\lambda_1$ and $1/\lambda_2$.

My question is, is, can these machines break together in the same instant of time if we consider that they are independent from each other?

My intuition tells me that the probability of them breaking in the same moment is 0, but I don't know how to prove it mathematically. Could anyone shed some light on this issue?

In this case let $f,g$ be the PDFs for your two random variables. Then the probability that they are equal is equal to: $$\int_0^{\infty} \int_0^{\infty} \mathbb{I}(x=y) \cdot f(x) \cdot g(y) dxdy,$$ where $\mathbb{I}$ is the characteristic/indicator function I mentioned before. Note that this is a double integral of a function which is zero everywhere except on a line, and finite on that line. So this integral will clearly be equal to zero.
The nuance is that while this event has zero probability it does have probability density. For instance for any positive number $x$, the probability that your first variable $X = x$ is zero. But it does have density $f(x)$. So for a very small number $\epsilon$, there is about $\epsilon \cdot f(x)$ chance that $X \in [x-\frac{\epsilon}{2},x+\frac{\epsilon}{2}]$.
You could look at the law of the difference of the instant they break. What probability does it assign to $0$ ? This is the difference of two independent exponential distribution.
If your parameters $\lambda_1,\lambda_2$ are non zero (or infinity), the mass assigned to $0$ by a Laplace law is $0$, the probability you are looking for.