A harmful particle test for houses is correct 2/3 of the time if the particles are present, and correct 4/5 of the time if the particles are absent. The proportion of houses with harmful particles is 1/6. If a random house is tested and the test comes out negative, what is the probability the particles are absent?
A -> Particle is absent B -> Test is negative
P(A|B) = ?
I get P(B'|A') = 2/3 , --> P(B|A) = 1 - 2/3 = 1/3, P(B'|A) = 4/5 , P(A') = 1/6
The answer is 12/13. I don't understand where my assumptions are wrong.
In order to semplify the calculation I set $\frac{1}{6}=\frac{3}{18}$
$\mathbb{P}[ A \cap B]=\frac{4}{5}\frac{15}{18}=\frac{12}{18}$
$\mathbb{P}[B]=\frac{1}{3}\frac{3}{18}+\frac{4}{5}\frac{15}{18}=\frac{13}{18}$
doing the ratio you get the solution: $\mathbb{P}[A|B]=\frac{12}{13}$
Further Explanation:
$\mathbb{P}[ A \cap B]$ is the probability to get "Negative Test AND No Harmful Particle" (it is a correct test!)
$\mathbb{P}[B]$ is the probability to get a Negative Test: we have 2 cases to be summed