If we have:
P(C) = 0.01 , P(!C) = 0.99
P(+|C) = 0.9 , P(-|C) = 0.1
P(-|!C) = 0.8, P(+|!C) = 0.2
How can i compute this probablity?
P (C | (T1 and T2)).
T1 and T2 are independent and T1 = +, T2 = +.
This is where i've reached:
P(+ and C) = P(C).P(+|C) = 0.01 x 0.9 = 0.009
P(+ and !C) = P(!C).P(+|!C) = 0.99 x 0.2 = 0.198
P(+) = P(+ and C).P(+ and !C) = 0.207
T1 and T2 are independent so:
P(T1 and T2) = P(+ and +) = P(+) x P(+) = 0.207 x 0.207
P(C|(T2 and T2)) = P(C and (T1 and T2)) / P(T1 and T2)
What's next? I don't know if i'm in the right place.
Represent the prior probability distribution as $(1,99)$ (We'll leave normalizing for the last step.
Represent the likelihood function for "$+$" as $(9,2)$.
Let "$\circ$" be term-by-term multiplication, thus $(a,b)\circ(c,d)=(ac,bd)$, and understand each pair to be equivalent to any of its scalar multiples, e.g. $(1,4)$ is the same as $(0.2,0.8)$.
Then we have $$ (1,99)\circ(9,2)\circ(9,2)=(81,396)=\left(\frac{81}{477},\frac{396}{477}\right)\approx(0.1698,0.8302). $$ So that's the posterior probablity distribution, i.e. the conditional distribution given the two independent positive tests.
Now the more long-winded version:
\begin{align} \Pr(C\mid +\,+) & = \frac{\Pr(C\text{ and }+\,+)}{\Pr(+\,+)} \\[10pt] & = \frac{\Pr(C\text{ and }+\,+)}{\Pr((C\text{ and }+\,+)\text{ or }((\text{not C})\text{ and } +\,+))} \\[10pt] & = \frac{\Pr(C\text{ and }+\,+)}{\Pr(C\text{ and }+\,+) + \Pr((\text{not C})\text{ and } +\,+)} \\[10pt] & = \frac{\Pr(+\,+\mid C)\Pr(C)}{\Pr(+\,+\mid C)\Pr(C)+\Pr(+\,+\mid\text{not} C)\Pr(\text{not }C)} \\[10pt] & = \frac{(0.9)^2(0.01)}{(0.9)^2(0.01)+ (0.2)^2(0.99)} \approx 0.1698 \end{align}