# Conditional Probability

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.

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1. 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.

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2. Note:
The numbers in the contingency table can also be obtained by a tree diagram
considering a sample of 4000 people.

400--A-->10--B-->9
-------------~B->1
----~A-->3990--B-->399
--------------~B-->3591

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3. Probability of 1/300 8/80 1/8

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