# Conditional probability.

• July 5th 2011, 08:32 PM
wopashui
Conditional probability.
Half percent of a population has a particular disease. A test is developed for the disease. The test gives a false positive 3% of the time and a false negative 2% of the time.

(a) What is the probability that joe (a random person) tests positive?

(b) Joe just got the bad news that the test came back positive; what is the probability that Joe has the disease?

I'm not sure what false positive and false negative mean, does test positive imply he has the disease?
So, the prob that joe tests positive will be 0.5%+3%?

and part b will be 0.5%+2%?
• July 5th 2011, 08:54 PM
mr fantastic
Re: Conditional probability.
Quote:

Originally Posted by wopashui
Half percent of a population has a particular disease. A test is developed for the disease. The test gives a false positive 3% of the time and a false negative 2% of the time.

(a) What is the probability that joe (a random person) tests positive?

(b) Joe just got the bad news that the test came back positive; what is the probability that Joe has the disease?

I'm not sure what false positive and false negative mean, does test positive imply he has the disease?
So, the prob that joe tests positive will be 0.5%+3%?

and part b will be 0.5%+2%?

A false positive occurs when the test indicates that Joe has the disease when in fact he hasn't.

A false negative occurs when the test indicates that Joe doers not have the disease when in fact he does.

Quote:

The test gives a false positive 3% of the time and a false negative 2% of the time.
Pr(+ve | no disease) = 0.03.

Pr(-ve | disease) = 0.02.
• July 6th 2011, 07:20 PM
wopashui
Re: Conditional probability.
Quote:

Originally Posted by mr fantastic
A false positive occurs when the test indicates that Joe has the disease when in fact he hasn't.

A false negative occurs when the test indicates that Joe doers not have the disease when in fact he does.

Pr(+ve | no disease) = 0.03.

Pr(-ve | disease) = 0.02.

so for part a let A be the event of having the disease, B of not having the disease
I got A $\cap$ positive + B $\cap$positive= 0.5%x98%+99.5%x3%

and part b, i use bayees rule {98%x0.5%}/{99.5%x3%+98%x0.5%}

am i on the right track?
• July 7th 2011, 04:13 AM
mr fantastic
Re: Conditional probability.
Quote:

Originally Posted by wopashui
so for part a let A be the event of having the disease, B of not having the disease
I got A $\cap$ positive + B $\cap$positive= 0.5%x98%+99.5%x3%

and part b, i use bayees rule {98%x0.5%}/{99.5%x3%+98%x0.5%}

am i on the right track?

Well, I'd use 0.005, 0.98, 0.995, 0.03 etc. NOT percentages.