Probability that a random number will be the largest from sets

Question

Let's say that I have three ranges of numbers:

$$ A: [0.15, 0.26]\\ B: [0.25, 0.34]\\ C: [0.20, 0.35]$$

If a number was randomly picked from the range of each set, what is the probability that the number from set $A$ will be the largest? From set $B$? etc?

Answer 1

When you pick randomly and independently from three random variables $A$, $B$, and $C$, having (cumulative) distributions $F_A$, $F_B$, and $F_C$ and corresponding distribution functions $f_A$, $f_B$, and $f_C$, then by definition of independence the chance that all three numbers are less than some value $x$ equals

$$\Pr(\max(A,B,C) \le x) = F_A(x)F_B(x)F_C(x).$$

Differentiation with respect to $x$ (via the product rule) gives the PDF of the max as

$$f_{\max(A,B,C)} = f_A(x)F_B(x)F_C(x) + F_A(x)f_B(x)F_C(x) + F_A(x)F_B(x)f_C(x).$$

That sure looks like a decomposition corresponding to $A$, $B$, and $C$. Indeed, the first term (by definition) tells us that the chance that $x \lt A \le x+dx$ and $B \le x$ and $C \le x$ is $f_A(x)F_B(x)F_C(x)dx$. Integrating that over all $x$ would then give the chance that $A$ exceeds both $B$ and $C$.

In the present situation, $\int_\mathbb{R} f_A(x)F_B(x)F_C(x)dx = 17/8910 \approx .00191,$ for instance.

The formula clearly generalizes to any finite number of independent random variables having any distributions you please.