Question

Suppose the probability of being infected with a certain virus is 0.001. A test used to detect the virus is positive 99% of the time if the person test has the virus, and positive 5% of the time if the person does not have the virus. Use Bayes Theorem to find the probability that a person is infected with the virus, given that they test positive. Clearly show your work and how you are using the theorem.

Answer 1

P (A ║ B) = 1.98 %

Explanation:

Bayes´ Theorem express

P (A ║ B) = P(A) * P( B ║ A) / P(B)

Now we identify

Event A person infected with a virus. Probability of being infected by a virus is P/A) 0.001

Event B the test was positive. Probability of test positive P(B) = 0,05

Probability of P ( B║ A) is the porbability of test positive given that is infected = 0.99

Then by subtitution in a general equation of the theorem we have

P (A ║ B) = 0.001*0.99/ 0.05

P (A ║ B) = 0.0198 P (A ║ B) = 1.98 %

Answer 2

0.0194

Explanation:

Let's define the following events

A: a person is infected with the virus

B: a person test positive

`\bar{A}`: a person is not infected with the virus

We know that P(A) = 0.001, and then
`P(\bar{A})` = 0.999. A test used to detect the virus is positive 99% of the time if the person test has the virus is equivalent to P(B|A) = 0.99. And positive 5% of the time if the person does not have the virus means
`P(B|\bar{A})`=0.05. We want to find the probability that a person is infected with the virus, given that they test positive, i.e., P(A|B). Bayes Theorem tell us that
`P(A|B)=\fracP(B{P(B|A)P(A)+P(B|\bar{A})P(\bar{A})}=((0.99)(0.001))/((0.99)(0.001)+(0.05)(0.999))`=0.0194