A disease is present in 2% of the people of a city. A blood test gives a positive result for 95% of the people who have the disease, and also for 10% of the people who do not have it. A person of the city, chosen at random, tests positive. The probability that the person has the disease is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{19}{117}\)
Let \(E_1\) be the event that the person has the disease, \(E_2\) that the person does not, and A that the test is positive.
\(P(E_1) = 0.02,\quad P(A|E_1) = 0.95\)
\(P(E_2) = 0.98,\quad P(A|E_2) = 0.10\)
By the theorem of total probability:
\(P(A) = 0.02\times 0.95 + 0.98\times 0.10\)
\(P(A) = 0.019 + 0.098 = 0.117\)
By Bayes’ theorem:
\( \displaystyle P(E_1|A) = \frac{0.019}{0.117} = \frac{19}{117} \)
The answer is small because the disease is rare: most positive results come from the large healthy group.
Why the other options are incorrect
Option (2) \(\displaystyle\frac{19}{20}\) — Incorrect. This is \(P(A|E_1) = 0.95\), the chance of a positive test for a person who has the disease; the condition is the wrong way round.
Option (3) \(\displaystyle\frac{98}{117}\) — Incorrect. This is \(P(E_2|A)\), the chance that a person who tests positive does not have the disease.
Option (4) \(\displaystyle\frac{117}{1000}\) — Incorrect. This is P(A) = 0.117, the chance that a person tests positive.