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A certain virus infects one in every 400 people. A test used to detect the virus in a person is positive 90% of the time if the person has the virus and 10% of the time if the person does not have the virus. Let A be the event "the person is infected" and B be the event "the person tests positive." (a) Find the probability that a person has the virus given that they have tested positive. (b) Find the probability that a person does not have the virus given that they have tested negative.

Respuesta :

Answer:Given:

P(A)=1/400

P(B|A)=9/10

P(B|~A)=1/10

By the law of complements,

P(~A)=1-P(A)=399/400

By the law of total probability,

P(B)=P(B|A)*P(A)+P(B|A)*P(~A)

=(9/10)*(1/400)+(1/10)*(399/400)

=51/500

Note: get used to working in fraction when doing probability.

(a) Find P(A|B):

By Baye's Theorem,

P(A|B)

=P(B|A)*P(A)/P(B)

=(9/10)*(1/400)/(51/500)

=3/136

(b) Find P(~A|~B)

We know that

P(~A)=1-P(A)=399/400

P(~B)=1-P(B)=133/136

P(A∩B)

=P(B|A)*P(A) [def. of cond. prob.]

=9/10*(1/400)

=9/4000

P(A∪B)

=P(A)+P(B)-P(A∩B)

=1/400+51/500-9/4000

=409/4000

P(~A|~B)

=P(~A∩~B)/P(~B)

=P(~A∪B)/P(~B)

=(1-P(A∪B)/(1-P(B)) [ law of complements ]

=(3591/4000) ÷ (449/500)

=3591/3592

The results can be easily verified using a contingency table for a random sample of 4000 persons (assuming outcomes correspond exactly to probability):

===....B...~B...TOT

..A . 9 . . 1 . . 10

.~A .399 .3591 . 3990

Tot .408 .3592 . 4000

So P(A|B)=9/408=3/136

P(~A|~B)=3591/3592

As before.

Step-by-step explanation: its were the answer is

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