A blood test indicates the presence of a particular disease 95% of the time when the disease is actually present. The same test indicates the presence of the disease 0.5% of the time when the disease is not actually present. One percent of the population actually has the disease.

Calculate the probability that a person actually has the disease given that the test indicates the presence of the disease.

• 0.324
• 0.657
• 0.945
• 0.950
• 0.995

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: B

Let

[math]Y[/math] = Positive test result

[math]D[/math] = disease is pesent (and [math]\sim D[/math] = not [math]D[/math])

Using Baye’s theorem:

[[math]] \operatorname{P}[D | Y] = \frac{\operatorname{P}[Y | D]\operatorname{P}[D]}{\operatorname{P}[Y | D]\operatorname{P}[ D] + \operatorname{P}[Y |\sim D]\operatorname{P}[\sim D]} = \frac{(0.95)(0.01)}{(0.95)(0.01) + (0.005)(0.99)} = 0.657. [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.