Probability Questions

Question

A certain virus infects one in every 2000 people. A test used to detect the virus in a person is positive 96% of the time if the person has the virus and 4% 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, i.e. find P(A | B). b. Find the probability that a person does not have the virus given that they test negative, i.e. find P(not A | not B).

Answer 1

The best way I know to do this is to make a table and assume you have a population of 100000 people.

The table looks like this

test negative test positive

not infected 95952 3998

infected 2 48

How did this table get produced?

In 100000 people 50 are infected.

Since the test is positive 96% of the time when a person is infected, there will be 48 people in the lower right box and that leaves only 2 in the lower left box.

There will be 99950 people not infected. Of those 4% will have a positive test, i.e. 3998 and the other 95952 not infected will have a negative test.

Now your probability should be easy to see.

Part A: 48/(3998+48)

Part B: 95952/(95952+2)