• Dec 10th 2007, 12:00 PM
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.
• Dec 10th 2007, 12:23 PM
colby2152
Quote:

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) = ???

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
• Dec 10th 2007, 12:42 PM
galactus
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.