Probability of one random variable less than two other

Question

I have trouble figuring out how to solve the following problem and can't find a similar on the forum. Given three (or more) independent r.v., say A, B and C, which might or might not have the same distribution(s), find the probability of A being less than the others, or

$P(A<B,A<C)$.

For only two independent r.v. (say A and B) I know that

$P(A<B)=\int _{S_B}\int _{-\infty} ^b f_A(a)f_B(b)dadb = \int _{S_B}F_A(b)f_B(b)dadb$,

if $F_X(x)$ and $f_X(x)$ are the cdf and pdf of r.v. $X$. I have been trying different approaches, as

$P(A<B,A<C)=P(A<B)P(A<C)$ and

$P(A<B,A<C)=P(A<B)P(B<C) + P(A<C)P(C<B)$

but no one seem to be right. Does anyone have any experience from this kind of problem? Right now I'm thinking about (for the three variables case) using a triple integral over the pdf's (but can't figure out how to integrate that either). If you need more information please ask!

Answer 1

$$\int da f_A(a) Pr(B>a)Pr(C>a) = \int f_A(a)(1-F_B(a))(1-F_C(a)) da$$