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

Answer:

0.0194

Step-by-step 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 $P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|\bar{A})P(\bar{A})}=\frac{(0.99)(0.001)}{(0.99)(0.001)+(0.05)(0.999)}$=0.0194

Answer 2

Answer:

P (A ║ B)  = 1.98 %

Step-by-step 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 %