# Problem of the week #17 - July 23rd, 2012

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#### Jameson

Staff member
A blood test indicates the presence of a particular disease 95% of the time when the disease is actually present. The same test indicates the presence of the disease 0.5% of the time when the disease is not present. One percent of the population actually has the disease. Calculate the probability that a person has the disease given that the test indicates the presence of the disease.

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#### Jameson

Staff member
Congratulations to the following members for their correct solution:

1) Sudharaka
2) veronica1999

Solution:

This is a conditional probability problem.

Let B represent the probability that the blood test indicates the presence of the disease and D represent that a person has the disease. We want to find $$\displaystyle P(D|B)$$.

By definition of conditional probability $$\displaystyle P(D|B) = \frac{P(D \cap B)}{P(B)}$$. From the information given in the problem we know that $$\displaystyle P(D \cap B) = (.01)(.95)=.0095$$

Now to find the probability of B we must consider two cases: the blood test indicates the presence of the disease and the person has the disease (1) and the blood test indicates the presence of the disease and the person does not have the disease (2).

(1) $$\displaystyle (.95)(.01)=.0095$$ (The test was positive and the person has it)
(2) $$\displaystyle (.005)(.99)=.00495$$ (The was positive, meaning the test was incorrect, and the person does not have it)

$$\displaystyle P(B) = (1)+(2) = .0095+.00495=.01445$$

Our final answer is $$\displaystyle P(D|B) = \frac{P(D \cap B)}{P(B)}=\frac{.0095}{.01445}=.6574$$

So even though this test is 95% accurate and only 1% of the population has the disease we arrive a surprisingly low percent of 65.7. This is a good example of how conditional probability can be used to show how wrong our intuitions are about some things

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