( and (i) 4.3 Diagnostic tests are available for a variety of med- ical conditio

Question

( and (i) 4.3 Diagnostic tests are available for a variety of med- ical conditions. Although these tests are extremely reliable, they sometimes provide incorrect results Suppose that the probability that the test will give a positive diagnosis (indicating that the disease is present) is 0.04. Given that the test diagnosis is positive, the probability that the person has the disease is 0.95. Given that the test diagnosis is negative, the probability that the person does not have the disease is 0.99 (a) Give an example of a simple event. (b) Give an example of a joint event (c) If a diagnostic test result is selected at random, what is the probability that (1) the test diagnosis is negative? (2) the test diagnosis is negative and the patient does not have the disease? (3) the test diagnosis is positive and the patient has the disease?

Explanation / Answer

(a) "The test diagnosis is positive" is an example of a simple event.

(b) "The test diagnosis is positive and the patient has the disease" is a joint event.

(c) Let us define the following events:

P -> diagnosis is positive

N -> diagnosis is negative

D -> patient has the disease

H -> patient does not have the disease (ie, he is healthy)

(1) P(diagnosis is negative) = P(N) = 1-P(diagnosis is positive) = 1-P(P) = 1-0.04 (given in the problem) => P(N) = 0.96.

(2) P(test diagnosis is negative and patient does not have the disease) = P(HN).

Now, P(patient does not have the disease given that diagnosis is negative) = P(H|N) = 0.99 (given in the problem)

But we know P(H|N) = P(HN)/P(N) => P(HN) =P(H|N)×P(N) = 0.99×0.96 => P(HN) =0.9504.

(3) P(test diagnosis is positive and the patient has the disease) = P(DP)

Now, P(patient has the disease given that diagnosis is positive) = P(D|P) = 0.95 (given in the problem)

But P(D|P) = P(DP)/P(P) => P(DP) = P(D|P)×P(P) =0.95×0.04

=> P(DP) = 0.038.

(4) P(D) = P(the disease is present) = P( the disease is present and diagnosis is positive or the disease is present and diagnosis is negative) = P(DP)+P(DN)

Now, from the properties of conditional distribution of exhaustive events, we know: P(DN)+P(HN) = P(N) => P(DN) = P(N)-P(HN) = 0.96-0.9504 = 0.0096

=> P(the disease is present) = 0.038+0.0096 (from (3))

=> P(D) = 0.0476.

(5) P(test diagnosis is positive or the patient has the disease) = P(P or D) = P(P)+P(D)-P(DP) = 0.04+0.0476-0.038

=> P(P or D) = 0.0496

(d) P(test diagnosis is positive given disease is present) = P(P|D) = P(DP)/P(D) = 0.038/0.0476 => P(P|D) = 0.7983

(e) P(disease is present given test diagnosis is positive) = P(D|P) = P(DP)/P(P) = 0.038/0.04 => P(D|P) = 0.95

(f)

CLASSIFICATION TABLE -1

Initially known event:

In (d) it is already known that the disease is present in the given individual; we find the probability that the diagnosis also shows positive result (as shown in row-1 of the table)

In (e) we have the test result that shows positive. We have to find the probability that the individual actually has the disease.

Tge initial information available to us in the beginning is different for the 2 parts (d) and (e). Hence the 2 probabilities are not the same.

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