Question

Suppose a rare disease occurs in about 1 out of 1,000 people who are like you. A test for the disease has sensitivity of 92% and specificity of 87%. Using the technique described in this chapter, compute the probability (as a %) that you actually have the disease, given that your test results are positive. (Round your answer to two decimal places.)

Answer 1

Answer:

The probability that a randomly selected person has the disease given that the test results were positive is 0.01.

Step-by-step explanation:

Denote the events as follows:

D =  a person has the disease.

P = a person's test result is positive.

N = a person's test result is negative.

The information provided is:

$P(D)=0.001\\P(P|D)=0.92\\P(N|D^{c})=0.87$

Consider the tree diagram below.

The probability that a randomly selected person has the disease given that the test results were positive is:

$P(D|P)=\frac{P(P|D)P(D)}{P(P|D)P(D)+P(P|D^{c})P(D^{c})}$

Compute the value of P (D|P) as follows:

$P(D|P)=\frac{P(P|D)P(D)}{P(P|D)P(D)+P(P|D^{c})P(D^{c})}\\=\frac{(0.92\times 0.001)}{(0.92\times 0.001)+(0.13\times 0.999)}\\=0.00703\\\approx0.01$

Thus, the probability that a randomly selected person has the disease given that the test results were positive is 0.01.

Answer 2

The probability that one has the disease, given that your test results are positive is 0.01.

How to calculate the probability?

From the information given, the following can be deduced:

Probability that a person has the disease = 0.001

Probability of a positive test = 0.92

Probability of a negative test = 0.87

Therefore, the probability that one has the disease, given that your test results are positive will be:

= (0.92 × 0.991) / [(0.92 × 0.001) + (0.13 × 0.999)]

= 0.01

In conclusion, the probability is 0.01.

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