Probablity confusion

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Alexandra

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May 16, 2012
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On my statistics final I was caught offguard by a simple probability exercise that I just did not know how to do. It stated that 3% of the population has the predisposition for a particular disease. Within this 3%, .99 test positive. Meanwhile, out of the 97% that does not have the predisposition, only. 02 tests positive. It t
hen asked that if someone tests positive, what is the probability that this individual has the predisposition.
It has been bothering me for a day now! I must know how to do this. Heeeelp
 
Alexandra said:
On my statistics final I was caught offguard by a simple probability exercise that I just did not know how to do. It stated that 3% of the population has the predisposition for a particular disease. Within this 3%, .99 test positive. Meanwhile, out of the 97% that does not have the predisposition, only. 02 tests positive. It t
hen asked that if someone tests positive, what is the probability that this individual has the predisposition.
It has been bothering me for a day now! I must know how to do this. Heeeelp
Click to expand...

We don't normally do a problem for someone without seeing any work from the person, but I'm feeling generous today...

\(\displaystyle \displaystyle P(Predis|+) = \frac{P(Predis \cap +)}{P(+)}\)

\(\displaystyle \displaystyle P(Predis \cap +)=(.03)(.99)=.0297\)

\(\displaystyle \displaystyle P(+)=(.03)(.99)+(.97)(.02)=.0491\)

\(\displaystyle \displaystyle P(Predis|+) = \frac{.0297}{.0491}=.6049\)
 
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