Considering [this](https://math.stackexchange.com/questions/261073/finding-probability-pxy) I have tried:
$$ P(A>B+C)=P(C<A-B)=\int\_0^1\int\_0^1 P(C<a-b|A=a, B=b)\,da\,db =$$
$$=\int\_0^1\int\_0^1 P(C<a-b)\,f\_A(a)\,f\_b(b)\,da\,db = \int\_0^1\int\_0^1 F\_C(a-b)\,f\_A(a)\,f\_B(b)\,da\,db=$$
$$=\int\_0^1\int\_0^1 (a-b) \cdot 1 \cdot 1 \,da\,db= 0$$
The actual answer is $\frac{1}{6}$.

I feel the mistake is to eliminate both A and B at the same time. Unfortunately I can't figure out the right solution. Any help will be higly appreciated!