Probability of one of the 3 independent uniform random variable being greater than rest of two.
Solution 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
$\int_0^1x^2dx = [({1}/{3})x^3]_0^1 = (1/3) - 0 = 1/3$
Solution 2
Hint:
Is there any reason to believe that e.g. $$A<B<C$$ is a more (or less) probable event than e.g. $$B<C<A$$?
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Updated on September 27, 2020Comments
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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.