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In a laboratory, a test was discovered to detect a certain disease. The effectiveness of this test is known as follows: if we denote by E the event that a patient have the disease and for N the event that the test is negative, then it is known that (c|) = 0.92, (N|c) = 0.93,() = 0.01

Obtain the probabilities (|) and (|).

Approach: I've tried to decompose the given data but I havent been able to get any other value. Just that P(Ec) = 0.99

Note:the c next to the characters is "complement"

Thanks.

1 Answers 1

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Bayes's theorem in general says:

$$P(A|B) = \frac{P(B|A) * P(A)}{P(B)}$$

So, to get $P(E|N)$:

$$P(E|N) = \frac{P(N|E) * P(E)}{P(N)}$$

You have $P(N|E)$ and $P(E)$. What is $P(N)$?

$P(N) = P(N\cap E) + P(N \cap E^C) = P(N|E)*P(E) + P(N|E^C)*P(E^C)$

And you have all those values as well, so now you can figure out $P(N)$ and thus $P(E|N)$.

Do something similar for $P(E^C|N)$

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    I dont have P(N|E), or is it 0.07? because it's the complement of this (N|c) = 0.93? – 2017-02-05
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    @ColonelG sorry, you're right, you have to infer those .. And it's $P(N|E)=1-P(N^C|E)$. So it is .08. $P(N^C|E^C) =1-P(N|E^C)=.07$ – 2017-02-05