Found artwork probabilities A stolen piece of art is found. The probability of it being the real one is 0.70. To find out if it is the real one, two experts are asked to give their verdict. Expert 1 is correct in 80% of the cases. Expert 2 is correct in 90% of the cases. What is the probability of 


expert 1 saying it is real and expert 2 saying it is fake; 
the art being real if expert 1 says it is real and expert 2 says it is fake ?


My attempts


$A$: it is real
$B$:  expert 1 is correct
$C$: expert 2 is correct


So 


$P(A) = 0.70$ 
$P(B) = 0.80$
$P(C) = 0.90$





I would say the asked probability is $P(B \cap C^c) = P(B)P(C^c) = 0.80 \times 0.10 = 0.08.$
I used this because I'm pretty sure $B$ and $C$ are independent.
I'm just gonna be honest and say I don't have a clue for this one. I would maybe try $P(A|B\cap C^c) =\frac{ P(B \cap C^c|A)}{P(A)} = \frac{0.8 \times 0.9}{0.7} = 0.80.$


I also thought about making a Venn diagram, but that didn't work. Can you guys please give some hints? I'd be very grateful!
Let's perform $1000$ trials.  The art is real in $700$ of those trials and fake in the other $300$.

If the art is real, the first expert will say it's real in $560$ of the cases.  Of those, the second expert will agree that it's real in $504$ of the cases and will declare it fake the remaining $56$ times.  In the remaining $140$ "real" cases (where the first expert is wrong), the second expert will declare the art real $126$ times and also will be fooled $14$ times.

In the $300$ cases where the art is fake, the first expert will be fooled $60$ times and will declare it fake the other $240$ times.  Of the $60$ times the first expert is fooled, the second expert also will be fooled $6$ times and will declare the art fake the other $54$ times.  The second expert will be fooled in $24$ of the "fake" cases where the first expert gets it right, and the second expert will agree with the first expert that the art is fake the other $216$ times.

So the first expert will say the art is real and the second expert will say it's fake on $56+54=110$ occasions (out of $1000$), or with probability $0.11$.  If that is their collective verdict, the art will be real $56$ times and fake $54$ times, so the likelihood that it's real in that case is $0.50909...$
Assume Expert 1 and Expert 2 do their analysis independently and give their verdicts.

Let E1 be Expert 1 and E2 be Expert 2.

P(E1 saying real) = P(A).P(E1 saying real|A) + P($A^c$).P(E1 saying real|$A^c$) = (0.7*0.8) + (0.3*0.2) = 0.62

P(E2 saying fake) = P(A).P(E2 saying fake|A) + P($A^c$).P(E2 saying fake|$A^c$) = (0.7*0.1) + (0.3*0.9) = 0.34

P(E1 saying real and E2 saying fake) = P(E1 saying real) * P(E2 saying fake) = 0.62*0.34 = 0.21

P(E1 saying real and the artwork is real) = P(A).P(E1 saying real|A)/P(E1 saying real) = (0.7*0.8)/0.62 = 0.9032

P(E2 saying fake and the artwork is real) = P(A).P(E2 saying fake|A)/P(E2 saying fake) = (0.7*0.1)/0.34 = 0.2059

P(The artwork is real and E1 saying real and E2 saying fake) = P(E1 saying real and the artwork is real) * P(E2 saying fake and the artwork is real) = 0.9032*0.2059 = 0.186