Verification of a Bayes Theorem related problem

Question

This was an assignment problem given to me by the professor. I have did it (not sure if its correct). My answer is around $2/10^{12}$. I fear this is wrong. Can someone try this and verify if possible.

I don't know if my understanding is correct. But this is what I did:

Since the 2 tests $T_1$ and $T_2$ are independent, we can write as below (C stands for cancer and C' for not cancer):

$P(C|T_1) = \frac{P(C)P(T_1|C)}{P(C)P(T_1|C)+P(C')P(T_1|C')}$

$P(C|T_2) = \frac{P(C)P(T_2|C)}{P(C)P(T_2|C)+P(C')P(T_2|C')}$

And

$P(C) = 1/10^6$

$P(C') = 1 - 1/10^6$

$P(T_1|C) = 0.01\%$

$P(T_2|C) = 1\%$

$P(T_1|C') = 0.1\%$

$P(T_2|C') = 0.05\%$

The final probability is the product $P(C|T_1).P(C|T_2) = 2/10^5 \times 1/10^7 = 2/10^{12} $

Answer 1

The question's definitions of "false positive rate" and "false negative rate" look wrong and misleading to me.

The false positive rate should be the proportion of people actually without cancer who give a false positive result on the test. Similarly the false negative rate should be the proportion of people actually with cancer who give a false negative on the test.

Imagine there were 10,000,000,000,000 people (many times the world population). Then I suspect the numbers could be

                                total            cancer          no cancer
Population               10,000,000,000,000   10,000,000     9,999,990,000,000 
Positive on first test       10,009,989,000    9,999,000         9,999,990,000 
Positive on both tests           14,899,005    9,899,010             4,999,995 

though you should check these are consistent with the stated false positive and false negative rates.

If so, then looking at the bottom row suggests that the proportion of those testing positive twice who actually have cancer would be about $66.4\%$.