Question

A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 90​% of the time when the person has the virus and ​15% of the time when the person does not have the virus.​ (This 15​% result is called a false positive​.) Let A be the event​ "the person is​ infected" and B be the event​ "the person tests​ positive." ​(a) Using​ Bayes' Theorem, when a person tests​ positive, determine the probability that the person is infected. ​(b) Using​ Bayes' Theorem, when a person tests​ negative, determine the probability that the person is not infected.

Answer 1

a) P[A/B] = 0,019 or P[A/B] = 1,9 %

b) P[A- /B-] = 0,9996 or P[A- /B-] = 99,96 %

Explanation:

Bayes Theorem :

P[A/B] = P(A) * P[B/A] / P(B)

The branches of events are as follows

Condition 1 real infection 1/300 and not infection 299/300

Then

1.- 1/300 299/300

When the test is done (virus present) 0,9 (+) 0,15 (-)

2.- 299/300

When the test is done ( no virus ) 0,15 (+) 0,85 (-)

Then:

P(A) = event person infected P(B) = person test positive

a) P[A/B] = P(A) * P[B/A] / P(B)

where P(A) = 1/300 = 0,0033 P[B/A] = 0,9

Then P(A) * P[B/A] = 0,0033*0,9 = 0,00297

P(B) is ( 1/300 )*0,9 + (299/300)*0,15

P(B) = 0,0033*0,9 + 0,9966*0,15 ⇒ P(B) = 0,1524

Finally

P[A/B] = 0,00297 /0,1524

P[A/B] = 0,019 or P[A/B] = 1,9 %

b) Following sames steps:

P[A- /B-] = (299/300) * 0,85 / (299/300) * 0,85 + (1/300 * 0,1)

P[A- /B-] = 0,8471 /0,8474

P[A- /B-] = 0,9996 or P[A- /B-] = 99,96 %