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Question

Mathematics Question on Probability

One half percent of the population has a particular disease. A test is developed for the disease. The test gives a false positive 3%3\% of the time and a false negative 2%2\% of the time. What is the probability that Amit (a random person) tests positive?

A

0.4760.476

B

0.0350.035

C

0.5230.523

D

0.2320.232

Answer

0.0350.035

Explanation

Solution

Let DD be the event that Amit has the disease. Let TT be the event that Amit's test comes back positive. We are told that P(D)=0.005P(D) = 0.005, since 1/2%1/2\% of the population has the disease. We also have P(TD)=.98P(T|D) = .98, since 2%2\% of the time a person having the disease is missed ("false negative"). We are told that P(TDc)=.03P(T|D^c) = .03, since there are 3%3\% false positive. Required probability : P(T)=P(TD)P(D)+P(TDc)P(Dc)P(T) = P(T|D) P(D) + P(T|D^c) P(D^c) =(.98)(.005)+(.03)(.995)=0.035= (.98) (.005) + (.03) (.995) = 0.035