A certain virus infects one in every 150 people. A test used to detect the virus in a person is positive 90% of the time when the person has the virus and 5% of the time when the person does not have the virus. (This 5% 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.

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valenbraca valenbraca · Answer 1 · 2019-10-26T23:37:08+02:00

Answer:

a) P(A|B)=0.108

b) [tex]P(A^c|B^c)[/tex]=0.999

Step-by-step explanation:

Given the events:

A: the person is infected

B: the person tests positive

[tex]A^c[/tex]: the person is not infected

[tex]B^c[/tex]: the person tests negative

a) If we check the attached picture, we can see that:

[tex]P(A)=\frac{1}{150}[/tex]

[tex]P(A^c)=\frac{149}{150}[/tex]

P(B|A)=[tex]\frac{90}{100}[/tex]

[tex]P(B|A^c)=\frac{5}{100}[/tex]

Bayes' Theorem:

P(B)= P(B∩A) + P(B∩[tex]A^c[/tex])

P(B)= P(B|A)×P(A) + P(B|[tex]A^c[/tex])×P([tex]A^c[/tex])

P(B)=[tex]\frac{90}{100} \frac{1}{150} +\frac{5}{100} \frac{149}{150}=\frac{167}{3000}[/tex]

We have to find the probability that the person is infected given that a person tests positive.

P(A|B)=[tex]\frac{P(B|A)P(A)}{P(B)}[/tex]=[tex]\frac{\frac{90}{100}.\frac{1}{150} }{\frac{167}{3000} }=\frac{18}{167}=0.108[/tex]

b) We have to find the probability that the person is not infected given that a person tests negative:

[tex]P(A^c|B^c)=\frac{P(B^c|A^c)P(A^c)}{P(B^c)}=\frac{P(B^c|A^c)P(A^c)}{1-P(B)}[/tex]

=[tex]\frac{\frac{95}{100} . \frac{149}{150} }{1-\frac{167}{3000}}=\frac{2831}{2833}=0.999[/tex]