Question

A certain virus infects one in every 600 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.

Answer 1

To understand the solution given below, you will need to know a few definitions or identities:

1. P(A)=probability of event A will happen.

2. definition of conditional probability:
P(A|B)=P(A∧B)/P(B)
i.e. Probability of event A happening GIVEN that B has already happened.

3. probability of a complement
P(~A)=1-P(A)

4. identity:
P(A∨B)=P(A)+P(B)-P(A∧B)

5. De Morgan's law:
~A ∧ ~B \equiv; ~(A ∨ B)
so
P(~A ∧ ~B)=1-P(A ∨ B)



We are given
P(A)=1/600;
=> P(~A)=1-1/600=599/600;

P(B|A)=0.9;
=> P(B∧A)/P(A)=0.9
=> P(B∧A)=0.9/600=11500=3/2000
P(B|~A)=0.1;

=> P(B∧~A)/P(~A)=0.1
=> P(B∧~A)=0.1*P(~A)=599/6000

This also means that
P(B∧A)+P(B∧~A)=3/2000+599/6000
=P(B∧(A∨~A)=608/6000
=P(B)
=608/6000
=38/375
=> P(~B)=1-P(B)=1-38/375=337/375
(a) find P(A|B)
P(A|B)
=P(A∧B)/P(B)
=P(B∧A)/P(B)
Substitute P(B∧A) and P(B)
to solve for P(A|B)

Hint: answer to part (a) is much greater than P(A).

(b) find P(~A|~B)
P(~A|~B)
=P(~A∧~B)/P(~B)
=(1-P(A∨B))/P(~B)
=(1-(P(A)+P(B)-P(A∧B))/P(~B)

All the quantities P(A), P(B), P(A∧B) and P(~B) have been calculated previously, so just substitute to find the answer.

Hint: answer for part B is very close to 1.

Finally, if you need detailed explanations or discussions, please post.