Whats the Probability that

• Oct 16th 2010, 11:26 PM
xian791
Whats the Probability that
Ello all,

I'm doing the practice test for a stats class and I'm stuck on a question.

A special test is given to people complaining of severe headaches. If someone has a brain tumor, the test is positive with a probability of .85(85%) If they have no tumor, the test is positive with a probability of .1(10%) Only 4%(.04) of those tested actually have brain tumors.

Whats is the probability that someone who tested positive has a brain tumor?
The answer is 0.262, but I have no idea how to get to that answer.

At first they ask "what is the probability someone random selected from those who take the test has a brain tumor AND test positive?"
For that I multiplied .85 by .04 and got the correct answer of 0.034

Thanks,
Shawn
• Oct 17th 2010, 12:25 AM
mr fantastic
Quote:

Originally Posted by xian791
Ello all,

I'm doing the practice test for a stats class and I'm stuck on a question.

A special test is given to people complaining of severe headaches. If someone has a brain tumor, the test is positive with a probability of .85(85%) If they have no tumor, the test is positive with a probability of .1(10%) Only 4%(.04) of those tested actually have brain tumors.

Whats is the probability that someone who tested positive has a brain tumor?
The answer is 0.262, but I have no idea how to get to that answer.

At first they ask "what is the probability someone random selected from those who take the test has a brain tumor AND test positive?"
For that I multiplied .85 by .04 and got the correct answer of 0.034

Thanks,
Shawn

A tree diagram makes it very clear (tumor is the first branches, test result is the second branches):

$\Pr(T | +ve) = \displaystyle \frac{(0.04)(0.85)}{(0.04)(0.85) + (0.96)(0.1)} = ....$

(Of course, if it's John Kimble that's being tested, then the probability is zero .... "It's NOT a tumor!")
• Oct 17th 2010, 05:39 AM
HallsofIvy
Here's a slightly different way of dealing with it. First, suppose 10000 people are tested (just because I don't like dealing with fractions). We are told that 4% of the general population have the tumor so 4% of 10000= 400 of our group have the tumor and 10000- 400= 9600 do not. Of those who have the tumor, 85% will test positive- 85% of 400 is 340. Of those who do not have the tumor, 10% will tesp positive: 10% of 9600= 960. So we have a total of 960+ 340= 1300 people who tested positive and 340 of those have the tumor. That is, $\frac{340}{1300}= 0.262$ so about 26.2% of those people who tested positive will actually have the tumor.
• Oct 17th 2010, 09:41 AM
xian791
ahh, thank you both for your input. I would have never thought of using bayes theorem. Question though, why would I use .96 instead of .04?

The method of using actually numbers really was helpful as well.