Question

It is known that 2.6% of the population has a certain disease. A new test is developed to screen for the disease. A study has shown that the test returns a positive result for 18% of all individuals, and returns a positive result for 92% of individuals who do have the disease. If a person tests positively for the disease under this test, what is the probability that they actually have the disease? 0.1276 0.1329 0.1435 0.1223 O 0.1382

Answer 1

Answer

Back-up Theory

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then Conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)…….....................................................….(1)

P(A/B) = P(B/A) x {P(A)/P(B)}……………………………..……………............................……..………….(2)

Now to work out the solution,

Let A represent the event that an individual has the disease and B represent the event

that the test returns positive.

Trivially, A^C represents the event that an individual does not have the disease and

B^C represents the event that the test returns negative.

2.6% of the population has the disease = > P(A) = 0.026 …………………………………..........…..(3)

Test returns positive in 18% of cases = > P(B) = 0.18….. ……………………………….........……..(4)

Test returns positive in 92% of individuals who do have the disease

= > P(B/A) = 0.92….. ……………………………………………………………...............……….…..(5)

We want to find the probability that a person has the disease, if the person tests positive. i.e.,

P(A/B)

= P(B/A) x P(A)/P(B) [vide (2)]

= (0.92 x 0.026)/0.18 [vide (5), (3) and (4)]

= 0.1329 Second Option Answer

DONE