Discrete Conditional Probability
Problem 18 page 152
A doctor assumes that a patient has one of three diseases: \(d_{1}\), \(d_{2}\), \(d_{3}\) Before any test, he assumes an equal probability of each disease He carries out a test that will be positive with probability 0.8 if the patient has \(d_{1}\), 0.6 if he has \(d_{2}\), and 0.4 if he has \(d_{3}\) Given that the outcome of the test was positive, what probabilities should the doctor now assign to the three possible diseases?
Solution:
First, letโs summarize the information that we have.
We will denote the equal probability of having one of the diseases as p
\(P(d_{1}) = p\), \(P(d_{2}) = p\), \(P(d_{3}) = p\), \(P(+|d_{1}) = 0.8\), \(P(+|d_{2}) = 0.6\), \(P(+|d_{3}) = 0.4\)
From the above we can calculate \(P(+)\) as
\[P(+) = P(+|d_{1})P(d_{1}) + P(+|d_{2})P(d_{2}) + P(+|d_{3})P(d_{3}) = p \left ( 0.8 + 0.6 + 0.4\right ) = 1.8p\] We need to calculate
\(P(d_{1}|+)\), \(P(d_{2}|+)\), \(P(d_{3}|+)\)
\(P(d_{1}|+) = \frac{P(d_{1}\cap +)}{P(+)} = \frac{0.8p}{1.8p} = \frac{4}{9}\)
\(P(d_{2}|+) = \frac{P(d_{2}\cap +)}{P(+)} = \frac{0.6p}{1.8p} = \frac{1}{3}\)
\(P(d_{3}|+) = \frac{P(d_{3}\cap +)}{P(+)} = \frac{0.4p}{1.8p} = \frac{2}{9}\)