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

Answer:

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

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

Step-by-step 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 %