# Conditional Probability

• Dec 20th 2008, 04:46 PM
Aryth
Conditional Probability
Suppose there is a medical diagnostic test for a disease. The sensitivity of the test is .95. This means that if a person has the disease, the probability that the test gives a positive response is .95. The specificity of the test is .90. This means that if a person does not have the disease, the probability that the test gives a negative response is .90, or that the false positive rate of the test is .10. In the population, 1% of the people have the disease. What is the probability that a person tested has the disease, given the results of the test is positive? Let D e the event "the person has the disease" and let T be the event "the test gives a positive result."

All I really need to know is how to find $P(D)$. The rest I can handle on my own.

Thanks.

I need to find $P(D|T) = \frac{P(T|D)P(D)}{P(T|D)P(D) + P(T|D')P(D')}$

$P(D|T) = \frac{(0.95)(0.01)}{(0.95)(0.01) + (0.10)(0.99)}$

$P(D|T) = \frac{0.0095}{0.0095 + 0.099}$

$P(D|T) = \frac{0.0095}{0.1085}$

$P(D|T) = 0.0875$

Is that right?
• Dec 20th 2008, 07:03 PM
tester85
I attempted the question and used bayes theorem and got the same answer as what you got.
• Dec 20th 2008, 07:16 PM
mr fantastic
Quote:

Originally Posted by Aryth
Suppose there is a medical diagnostic test for a disease. The sensitivity of the test is .95. This means that if a person has the disease, the probability that the test gives a positive response is .95. The specificity of the test is .90. This means that if a person does not have the disease, the probability that the test gives a negative response is .90, or that the false positive rate of the test is .10. In the population, 1% of the people have the disease. What is the probability that a person tested has the disease, given the results of the test is positive? Let D e the event "the person has the disease" and let T be the event "the test gives a positive result."

All I really need to know is how to find $P(D)$. The rest I can handle on my own.

Thanks.

I need to find $P(D|T) = \frac{P(T|D)P(D)}{P(T|D)P(D) + P(T|D')P(D')}$

$P(D|T) = \frac{(0.95)(0.01)}{(0.95)(0.01) + (0.10)(0.99)}$

$P(D|T) = \frac{0.0095}{0.0095 + 0.099}$

$P(D|T) = \frac{0.0095}{0.1085}$

$P(D|T) = 0.0875$

Is that right?

Yes. Well done.
• Dec 20th 2008, 07:21 PM
Aryth
Quote:

Originally Posted by mr fantastic
Yes. Well done.

The reason why I posted this was because I multiplied 0.10 and 0.99 wrong and couldn't arrive at the correct answer. But I do have one more question, the answer in my book provides this:

$P(D|T) = \frac{P(D \cap T)}{P(T)} = 0.0875$

But, what is P(T) without knowing whether or not the person has the disease? Or are you supposed to assume that they have the disease? I tried both scenarios and couldn't arrive at that solution.
• Dec 20th 2008, 07:27 PM
mr fantastic
Quote:

Originally Posted by Aryth
The reason why I posted this was because I multiplied 0.10 and 0.99 wrong and couldn't arrive at the correct answer. But I do have one more question, the answer in my book provides this:

$P(D|T) = \frac{P(D \cap T)}{P(T)} = 0.0875$

But, what is P(T) without knowing whether or not the person has the disease? Or are you supposed to assume that they have the disease? I tried both scenarios and couldn't arrive at that solution.

$P(D|T) = \frac{P(T|D)P(D)}{P(T|D)P(D) + P(T|D')P(D')}$ and $P(D|T) = \frac{P(D \cap T)}{P(T)}$ are identical statements.

Pr(T) is just the total probability of testing positive and is given by $P(T|D)P(D) + P(T|D')P(D')$. Think about it ...
• Dec 20th 2008, 07:31 PM
Aryth
I understand that the two are identical, and I understand the law of total probability.

But the way that the answer is formed in the book makes the solution look immediately obvious from the conditional probability statement alone. Is that the case?
• Dec 20th 2008, 11:50 PM
mr fantastic
Quote:

Originally Posted by Aryth
I understand that the two are identical, and I understand the law of total probability.

But the way that the answer is formed in the book makes the solution look immediately obvious from the conditional probability statement alone. Is that the case?

Maybe for some ..... But not for me.
• Dec 21st 2008, 07:37 PM
Aryth
Thanks for the help. Its much appreciated.