What is the probability of Being a Druggist after Having 2 tests (First Test Positive and Second Test Negative).
What we have.
T = the person tested positive of drug use
U = the person does drugs
P(U) = The probability of person doing drugs let's say $x$
P(T|U) = The Probability of person's test being positive given he does drugs $V_1$
P($\neg$T|$\neg$U|) = The Probability of person's test being negative given he does not do drugs $V_0$
$T_1$ and $T_2$ are conditionally independent given U
From This we can also derive that
P($\neg$T|U) = The Probability of person's test being negative given he does drugs $(1 - V_1)$
P(T|$\neg$U) = The Probability of person's test being positive given he does not do drugs $(1 - V_0)$
So The approach I used was first, find the probability of person being a druggie after a positive test. And using that as a prior probability for the second test.
Note : I have made Test 1 = $T_1$ and Test 2 = $T_2$
Note : I also used the alternative version of Bayes' Theorem as we do not know P(T)
$P(U|T_1) = \frac{P(T|U)*P(U)}{P(T|U)*P(U) + P(T|\neg U)*P(\neg U)}$
$P(U|T_1) = \frac{V_1*x}{V_1*x + (1-V_0)*(1-x)}$
Now we can use this as a prior probability for the Second Test
$P(U|\neg T_2) = \frac{P(\neg T|U)*P(P_{prior})}{P(\neg T|U)*P(P_{prior}) + P(\neg T|\neg U)*P(\neg P_{prior})}$
$P(U|\neg T_2) = \frac{P(\neg T|U)*P(U|T_1)}{P(\neg T|U)*P(U|T_1) + P(\neg T|\neg U)*P(\neg (U|T_1)))}$
That spiked 2 doubts in my mind.
A) Can I just sub the original value of $(1-V_1)$ and $V_0$ for $P(\neg T|U)$ and $P(\neg T|\neg U)$ respectively, even though this is the second test? Or would that be different given it's a second test?
B) And that, given that the 2 tests are independant. that the $P(T)$ would be same for both the case? Even though for $T_2$ one test has already happened?
And from the alternate version we know
$P(T) = P(T|U)*P(U) + P(T|\neg U)*P(\neg U)$
So Would that mean that
$P(T_1) = P(T_2) = P(T|U)*P(U) + P(T|\neg U)*P(\neg U)$ ?
Hence, can we just use the value and form a new formula
$P(U|\neg T_2) = \frac{P(\neg T|U)*P(P_{prior})}{(1 - P(T))}$
$= \frac{P(\neg T|U)*P(P_{prior})}{(1 - [P(T|U)*P(U) + P(T|\neg U)*P(\neg U)])}$