Bayes theorem in probability and Cancer or disease problem There is a man who thinks he got cancer and goes with Doctor A. Dr. A diagnose cancer only to the 60% of the patients who really have cancer and never diagnose cancer to the ones who does not have the disease. Dr. A doesn't diagnose cancer to the man, but just to be sure, he goes with medic B who diagnose cancer to the 80% of the patients who really have cancer and diagnose cancer to the 10% of the patients that does not have the disease. Dr. B neither diagnose cancer to the man.
a) What is the prior probability of having cancer that the man has if the posterior probability of having cancer after the diagnose of Dr. A and Dr. B is 0.5?
b) What is the prior probability of having cancer that the man has if the posterior probability of having cancer after the diagnose of Dr. A and Dr. B is 0.1?
What I've done until now is:
Up to this moment what I've done is this. Let A, B and X be random variables such that:
$A=\begin{cases}1 \text{ if Dr. A diagnose cancer}\\
                0 \text{ if Dr. A DOES NOT diagnose cancer}\end{cases}$
$B=\begin{cases}1 \text{ if Dr. B diagnose cancer}\\
                0 \text{ if Dr. B DOES NOT diagnose cancer}\end{cases}$
$X=\begin{cases}1 \text{ if the man actually has cancer}\\
                0 \text{ if the man DOES NOT have cancer}\end{cases}$
Now, I'm having that


*

*$\mathbb{P}(A=1|X=1)=0.6$


*$\mathbb{P}(A=0|X=1)=0.4$


*$\mathbb{P}(A=1|X=0)=0$


*$\mathbb{P}(A=0|X=0)=1$

And in the other hand I have that


*

*$\mathbb{P}(B=1|X=1)=0.8$


*$\mathbb{P}(B=0|X=1)=0.2$


*$\mathbb{P}(B=1|X=0)=0.1$


*$\mathbb{P}(B=0|X=0)=0.9$

and for a) the  posterior is:

$\mathbb{P}(X=1|A=0,B=0)=0.5$

and for b) the  posterior is:

$\mathbb{P}(X=1|A=0,B=0)=0.1$

I'm assuming that A and B are independent, hence I'm looking so, what I have is that I'm looking for a prior ($P(X=1)$) such that:
$\mathbb{P}(X=1|A=0,B=0)=\frac{\mathbb{P}(X=1,A=0,B=0)}{\mathbb{P}(A=0,B=0)}=\frac{\mathbb{P}(A=0|X=1,B=0)\mathbb{P}(X=1,B=0)}{\mathbb{P}(A=0,B=0)}=\frac{\mathbb{P}(A=0|X=1)\mathbb{P}(B=0|X=1)\mathbb{P}(X=1)}{\mathbb{P}(A=0,B=0)}$
The thing is that... I'm not really sure what to do with the denominator: $\mathbb{P}(A=0,B=0)$
 A: A case could be made that this answer is defective, because I am supposed to
study the OP's work step by step, and find a way to bring it to completion.
This is just too difficult.
3-way Bayes Theorem problems are a nightmare.  Thankfully, you can simplify
the problem(s) greatly, by
(1) Assuming no information re posterior probability of cancer.
(2) Adjust the answer construing prior probability of cancer as a variable,
and solving for it, given the known posterior probability.
Work shown below.
Both doctors said : No Cancer.
If the patient had had Cancer, then given no other info, chances of two false negatives
occurring would have been $0.4 \times 0.2 = 0.08$.
If the patient had not had Cancer, then given no other info, chances of two true negatives
occuring would have been $1.0 \times 0.9 = 0.9$
Therefore, the chance that the patient has Cancer, at this point is
$$Y = \frac{8}{8 + 90} = \frac{4}{49}.$$
Chance of no cancer is (at this point is)
$$N = \frac{45}{49}.$$
Now, new info.
Prior chance that patient has Cancer $= p$, with $q = (1 - p).$
Then, chance that patient has Cancer, with all the info is
$$\frac{p \times Y}{(p\times Y) + (q \times N)}$$
$$= \frac{p \times \frac{4}{49}}
{\left[p \times \frac{4}{49}\right] + \left[(1-p) \times \frac{45}{49}\right]}$$
$$ = \frac{4p}{45-41p}.$$
$$\frac{4p}{45-41p} = 0.5 \implies \frac{8p}{45-41p} = 1
\implies 49p = 45 \implies p = \frac{45}{49}.$$
$$\frac{4p}{45-41p} = 0.1 \implies \frac{40p}{45-41p} = 1
\implies 81p = 45 \implies p = \frac{45}{81}.$$
A: $$\mathbb{P}(A=0,B=0) \\=\mathbb{P}(A=0,B=0\mid X=0)\mathbb{P}(X=0)+\mathbb{P}(A=0,B=0\mid X=1)\mathbb{P}(X=1)$$
where $\mathbb{P}(X=0) = 1- \mathbb{P}(X=1)$
