If $\displaystyle X_1, X_2, X_3$ are independent random variables that are uniformly distributed over $\displaystyle (0,1),$ compute the probability that the largest of the three is greater than the sum of the other two.
The probability density for 1 variable is the function:
p(x) = 1 if 0<=x<=1 and 0 otherwise.
The probability density for the variable t = x+y = sum of 2 variables having the above probability density is the function:
p(t) =
t if 0<=t<=1 and
2-t if 1<=t<=2.
Since the 3rd variable (call it w) has a max value of 1, the part of the probability distribution for t going as 2-t is irrelevant - the 3rd variable w has probability 0 that it will be greater than t.
For the other part of the distribution, 0<=t<=1, the probability that w > t is 1-t. That means we can calculate the probability that w > t =
$\displaystyle
\int_0^1 t (1-t) dt = \frac {1}{6}
$