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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?

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  • $\begingroup$ You don’t need them. $\Pr(B2)$ through $\Pr(B4)$ have no bearing on the probability of having disease $A$. $\endgroup$ – amd Apr 17 '19 at 19:52
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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*}
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  • 2
    $\begingroup$ Wow, this really helped a lot. I can't understate the perspective this has given me. Thanks a lot! So if we had known the percentage that the test for A would ring positive, if a person had either disease B, C or D, then the problem would be much more clearer and solvable? $\endgroup$ – MTHLily Apr 17 '19 at 9:56
  • $\begingroup$ Yes, you would need both probabilities. The probability of a false negative (test says you don't have decease A but you do) and the probability of a false positive (test says you do have decease A but you don't). If you are satisfied with the answer you can also it ;) $\endgroup$ – EagleOwl Apr 17 '19 at 10:12
  • $\begingroup$ Typically in these problems, “reliability” is synonymous with accuracy: the probability that the test returns a correct result. $\endgroup$ – amd Apr 17 '19 at 19:52

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