Suppose 30% of the individuals in a community carry a dreaded virus in their blood. Of those with the virus, 5% test negative for the disease and 95% of those who do not have the virus test negative for the disease. Find the probability that a person has the virus even though they test negative for it.

2. Originally Posted by JP101
Suppose 30% of the individuals in a community carry a dreaded virus in their blood. Of those with the virus, 5% test negative for the disease and 95% of those who do not have the virus test negative for the disease. Find the probability that a person has the virus even though they test negative for it.
P(Virus) = .3
P(No Virus) = .7
P(Test Negative | Virus) = .05
P(Test Negative | No Virus) = .95

P(Virus | Test Negative) = ???

Use Bayes' Theorem:

P(Virus | Test Negative) = P(Test Negative | Virus)P(Virus) / [P(Test Negative | Virus)P(Virus) + P(Test Negative | No Virus)P(No Virus)]
= .05(.3)/(.05*.3 + .95*.7)
= .022

3. Sometimes it is easy to build a table with an assumed number. Let's use a population of 1000.

Since 30% are infected, then that's 300 out of our 1000.

Therefore, 700 are not infected. 95% of 700 is 665. Go from there and fill in the table.

Code:
                   Infected                Not infected

Positive          285                         35                     320

Negative          15                         665                   680

300                         700                  1000
Go down the infected column to the negative, then go across the negative row. We get 15/680=0.022

Just as Colby got. Just another way.