Question

A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 80% of the time if the person has the virus and 10% of the time if the person does not have the virus. (This 10% result is called a false positive.) Let A be the event "the person is infected" and B be the event "the person tests positive". DRAW A TREE DIAGRAM IN YOUR NOTES AND USE IT TO HELP YOU SOLVE THIS PROBLEM. Find the probability that a person has the virus given that they have tested positive; i.e. find P(AIB). Give your answer as a decimal number and include at least 3 or more non-zero digits. P(AIB)=

Answer 1

In the tree, the first branch will be person has virus or person doesn't have the virus.

P(virus) = 1/300

P(not virus) = 299/300

Now,

Then we branch out from each option. These branches would be positive or negative.

If they have virus:

P(positive) = 0.8

P(negative) = 0.2

If don't have virus:

P(positive) = 0.1

P(negative) = 0.9

Now, solving the question of probability that a person has the virus given that they have tested positive:

We find:

P(A|B)

P(has virus | positive test) = P(positive and has virus) / P(positive test)

P(positive and has virus) = 4/5 * 1/300 = 4/1500

P(positive test) = 1/300 * 4/5 + 1/10 * 299/300

= (4/1500)+(299/3000)

=(8/3000) + (299/3000) = 307/3000

= 0.10233

So,

P(positive and has virus) / P(positive test) = 4/1500 divided by 299/3000 = 0.02675