Question

Only \(0.1 \%\) of the individuals in a certain population have a particular disease (an incidence rate of .001). Of those who have the disease, \(95 \%\) test positive when a certain diagnostic test is applied. Of those who do not have the disease, \(90 \%\) test negative when the test is applied. Suppose that an individual from this population is randomly selected and given the test. a. Construct a tree diagram having two first-generation branches, for has disease and doesn't have disease, and two second-generation branches leading out from each of these, for positive test and negative test. Then enter appropriate probabilities on the four branches. b. Use the general multiplication rule to calculate \(P(\) has disease and positive test). c. Calculate \(P\) (positive test). d. Calculate \(P\) (has disease \(\mid\) positive test). Does the result surprise you? Give an intuitive explanation for the size of this probability.

Short answer

The probabilities are: a) For individuals having the disease and testing positive, it is \(0.00095\); b) The overall probability of testing positive is \(0.10085\); c) The probability of having the disease given a positive test is \(0.00942\). The result may be surprisingly low due to the high incidence of false positive tests arising from low disease prevalence.

Step-by-step solution

Construct a tree diagram

A. Construct a tree diagram with first-generation branches representing having the disease and not having the disease. Under having disease, attach two second-generation branches indicating testing positive and testing negative. Do likewise for the 'not having a disease' branch. The probabilities that go with each branch are: has disease \(0.1\%\), doesn't have disease \(99.9\%\), for those with disease, test positive \(95\%\) and test negative \(5\%\), for those without disease, test positive \(10\%\) and test negative \(90\%\).

Calculation of \(P(\) has disease and positive test).

B. Apply the general multiplication rule, which is \(P(A and B) = P(A)∗P(B|A)\), where A signifies 'has disease' and B denotes 'tests positive'. For those having the disease, the probability of testing positive is \(0.001 * 0.95 = 0.00095\).

Calculation of \(P\) (positive test).

C. The probability of a positive test is the sum of the probabilities of those who have the disease and test positive, plus those who do not have the disease but also test positive. That is \(P(\)Disease Positive) + P(\)No Disease Positive) = \(0.00095 + 0.999 * 0.1 = 0.10085\).

Calculation of \(P\) (has disease | positive test.

D. The probability of having the disease given a positive test is calculated by applying Bayes' theorem: \(P(\)Disease|Positive) = \(\frac{P(Disease Positive)}{P(Positive)}\) = \(\frac{0.00095}{0.10085} = 0.00942\). This probability may seem unexpectedly low. The reason is that even though the test is quite accurate, the disease is very rare. So, most people who test positive are actually healthy people who received a false positive test result.