Bayes' Theorem with Multiple Tests

Question

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?

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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?

Answer 1

I suppose that by reliability of the test you mean, that if you are inflicted with decease A the test is positive with probability of 72.7%.

This tells us nothing about the probability of the test being positive, when you are not inflicted by decease A. So you do miss some information.

Look at the following two examples.

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*}