Question

Question 26: [4 points]
1% of a population have a certain disease and the remaining 99% are free from this disease. A test is used to detect this disease. This test is positive in 95% of the people with the disease and is also (falsely) positive in 2% of the people free from the disease.
If a person, selected at random from this population, has tested positive, what is the probability that she/he has the disease?
Let D be the event "have the disease" and FD be the event "free from the disease" Let the event TP be the event that the "test is positive".
A diagram with all the above information is shown.

Answer 1

$P(D/TP) = \frac{P(D \: n \: TP)}{P(TP)}$

$P(TP) = P(TP \: n \: D)+P(TP \: n \: FD) \\ P(TP) = 0.95×0.99 + 0.02×0.01 = 0.9407$

$P(D/TP) = \frac{0.9405}{0.9407} = \frac{9405}{9407}$≈0.999787