I've been having a hard time trying to solve a homework question so any hint will be appreciated. Suppose that we have three continuous random variables A,B,C and knowing the fallowing statement:
$$P(A=B)=P(B=C)=P(A=C)=0$$
prove that the fallowing statement is impossible:
$$P(A>B)=P(B>C)=P(C>A)= 0.7$$
the hint is saying that I must use Kolmogorov axioms but I can't find a way to use that.
I tried finding $P(A<B)=P(B<C)=P(C<A)=0.3$ and another idea would be finding any of the operands of the second equation by others and showing that it's not equal to $0.7$ but I can't find a way.
Let $p_1=P(A>B>C)$, $p_2=P(A>C>B)$, $p_3=P(B>A>C)$, $p_4=P(B>C>A)$, $p_5=P(C>A>B)$, $p_6=P(C>B>A)$.
Then $\sum{p_i}=1$, $P(A>B)=p_1+p_2+p_5$, $P(B>C)=p_1+p_3+p_4$, $P(C>A)=p_4+p_5+p_6$.
If $P(A>B)=P(B>C)=P(C>A)=0.7$, then $2p_1+p_2+p_3+2p_4+2p_5+p_6=2.1\Rightarrow p_1+p_4+p_5=1.1$. It's impossible.