# Bayes' Theorem with Multiple Tests

## I'm having difficulties with this problem:

Suppose you have an entire city afflicted with four distinct and exclusive diseases and a laboratory is assigned to test which disease each citizen has.

The reliability of these tests are as follows:

Disease A = 72.7%

Disease B = 81.1%

Disease C = 75.2%

Disease D = 80.1%

The percentage of the population of people afflicted is as follows:

P(B1) = 18.1% (Disease A)

P(B2) = 31.9% (Disease B)

P(B3) = 18.9% (Disease C)

P(B4) = 31.1% (Disease D)

If a random person were to selected from the entire population and then tested positive for disease A, what is the probability that they actually have disease A?

I think that the problem is asking for P(B1|A1), so I used this formula:

P(B1|A) = P(A|B1)P(B1) / ( P(A|B1)P(B1) + P(A|B2)P(B2) + P(A|B3)P(B3) + P(A|B4)P(B4) )

These are the values that I am sure of:

P(A|B1) = .727, because that is the chance of a true positive result of disease A being detected

P(B1) to P(B4) = the population listed above, corresponding to A, B, C and D.

The problem is now, I don't know what values to put inside P(A|B2) to P(A|B4)

Do I put in just the rate of false positive (.273)? Or do I use the corresponding tests for disease B, C and D (.811, .752, .801, respectively)? Or am I missing something here?

• You don’t need them. $\Pr(B2)$ through $\Pr(B4)$ have no bearing on the probability of having disease $A$. – amd Apr 17 '19 at 19:52

1. Scenario 1: A test that gives a positive result for decease A with probability 72.7% whatever the input. So $$P(A|B_1^c) = 0.727$$ and \begin{align*} P(B_1|A) = \frac{P(A|B_1)P(B_1)}{P(A|B_1)P(B_1) + P(A|B_1^c)P(B_1^c)}=0.181 \end{align*}
2. Scenario 2: A test that gives a positive result for decease A with probability 72.7% if you are inflicted by decease A but always gives a negative result if you do not have decease A. So $$P(A|B_1^c) = 0$$ and \begin{align*} P(B_1|A) = \frac{P(A|B_1)P(B_1)}{P(A|B_1)P(B_1) + P(A|B_1^c)P(B_1^c)}=1 \end{align*}