Considering this 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!
The desired chance is the same as the chance that $1-A>B+C$, which is the chance that $A+B+C<1$, which is the volume of a certain tetrahedron, whose vertices are $(0,0,0)$, $(0,0,1)$, $(0,1,0)$, and $(1,0,0)$. This volume is $1/3$ the height ($1$) times the area of the base ($1/2$) so the answer is $1/6$.
The error seems to be the assertion that $P(C<a-b) = F_C(a-b) = (a-b)$. This does not hold for all $(a,b) \in [0,1]\times[0,1]$. For example, consider $a = 0.3, b = 0.5$. In this case $a - b = -0.2$ which is not a probability. Your calculation computes the desired probability and its negative, sums them, and gets zero. To fix this, you need to be careful about the limits of integration.
