Basic probability question about two debaters and their reliability. Hello and thanks in advance. The following thing is bugging me. Why is it like this?
We have two debaters Matt and Juliet. It has been shown that what Matt says is right with probability 0.8. It has also been shown that what Juliet says is right with probability 0.9.
Matt and Juliet debate about is C true or false. Matt says that C is true. Juliet says that C is false. What is the probability for C to be true?
P(C is true)
= P(Matt is right and Juliet is wrong)
= P(Matt is right) * (1 - P(Juliet is right))
= 0.8 * (1 - 0.9)
= 0.8 * 0.1
= 0.08

P(C is false)
= P(Matt is wrong and Juliet is right)
= (1 - P(Matt is right)) * P(Juliet is right)
= (1 - 0.8) * 0.9
= 0.2 * 0.9
= 0.18

But then!
Why P(C is true) = 1 - P(C is false) doesn't apply here? Something related to independence?
 A: The result in my comment is the result one is supposed to find but to justify it, we will see that some additional hypotheses are needed...
Call $M$ what Matt says, $J$ what Juliet says, $C$ the event that C is true and $\bar C$ the event that C is false. One assumes that Matt says the truth with probability $m=0.8$, independently on the subject (this is hypothesis (1)), that Juliet says the truth with probability $j=0.9$, independently on the subject (this is hypothesis (2)), and that Matt and Juliet decide what to say independently of the other (this is hypothesis (3)).
One considers the event $A=[M=C,J=\bar C]$ and one asks for $P[C\mid A]=P[A\mid C]\cdot c/a$ with $a=P[A]$ and $c=P[C]$. 


*

*By hypothesis (3), $P[A\mid C]=P[M=C\mid C]\cdot P[J=\bar C\mid C]$. 

*By hypothesis (1), $P[M=C\mid C]=P[M\,\text{says the truth}]=m$. 

*By hypothesis (2), $P[J=\bar C\mid C]=P[J\,\text{does not say the truth}]=1-j$.


Thus, $P[C\mid A]=m(1-j)c/a$. A similar decomposition based on the same hypotheses (1), (2) and (3), yields $P[\bar C\mid A]=(1-m)j(1-c)/a$. Since $P[C\mid A]+P[\bar C\mid A]=1$, summing these yields the value of $a=m(1-j)c+(1-m)j(1-c)$, hence
$$
P[C\mid A]=\frac{m(1-j)c}{m(1-j)c+(1-m)j(1-c)}.
$$
There is still some missing data, namely $c=P[C]$, called the prior probability of the event $C$. Assuming uniform prior (this is hypothesis (4)), one gets $c=1/2$ hence finally, provided assumptions (1), (2), (3) and (4) hold,
$$
P[C\mid A]=\frac{m(1-j)}{m(1-j)+(1-m)j}\left(=\frac4{13}\right).
$$
