Probability of one of the 3 independent uniform random variable being greater than rest of two.

Question

Three independent random variables say A,B,C are uniformly distributed in an interval (0,1). Probability of 'A' being the largest?

How to solve this question?

I know how to solve for two random variable A,B by finding prob(A>B)= prob(A-B>0)

considering A-B= C so prob(C>0) can be found. But here in this case for 3 random variable I am not able to solve it.

Answer 1

ok, let's sat you've got A = 1/4

what would the probability be that B is less than 1/4 on the uniform distribution? The answer is 1/4 - do you see/know that? there is 3/4 for it to be bigger and 1/4 for it to be smaller - the chances of a draw disappear to zero because the distribution is continuous

For two independent variables, if they both needed to be smaller then it would be (1/4) x (1/4)

you therefore need to integrate x^2 from 0 to 1

$\int_0^1x^2dx = [({1}/{3})x^3]_0^1 = (1/3) - 0 = 1/3$

Answer 2

Hint:

Is there any reason to believe that e.g. $$A<B<C$$ is a more (or less) probable event than e.g. $$B<C<A$$?