conditional probability help

Hi,

I need help solving this problem

About o.1% of people in a given population carry a disease. For those that carry the disease, test are positive in 98% of cases. For those that do not carry the disease, tests are negative 94% of the time.

Given that a person rests positive, what is the probability that the person carries the disease?

p(disease)=0.01%
P(carry disease|test positive)= 98%
P(do not carry|test negative)=94%

P(carry disease[FONT=MathJax_Main]∩[/FONT] test positive) = P(disease) * P(carry disease|test positive)

= (0.0001)(0.98)

but what do I do with the last piece of information (P(do not carry|test negative)=94%) why is that there?

Thanks

WlND said:

Hi,

I need help solving this problem

About o.1% of people in a given population carry a disease. For those that carry the disease, test are positive in 98% of cases. For those that do not carry the disease, tests are negative 94% of the time.

Given that a person rests positive, what is the probability that the person carries the disease?

p(disease)=0.01%....not the same number given in the question
P(carry disease|test positive)= 98%....these are backwards
P(do not carry|test negative)=94%

P(carry disease[FONT=MathJax_Main]∩[/FONT] test positive) = P(disease) * P(carry disease|test positive)

= (0.0001)(0.98)

but what do I do with the last piece of information (P(do not carry|test negative)=94%) why is that there?

Thanks

Click to expand...

Because you have to account for "false positives"; not everyone who tests positive really has the disease. If 94% of non-disease carriers test negative, then the other 6% test positive!

It might make more sense if the conditions are written properly:

P(disease) = 0.1% = 0.001 [Which should it be, 0.1% or 0.01%?]
P(test positive | disease) = 0.98, --> P(test negative | disease) = 0.02
P(test negative | no disease) = 0.94, --> P(test positive | no disease) = 0.06

There are two ways to get a positive test:
P(test positive) = P(test positive | disease)*P(disease) + P(test positive | no disease)*P(no disease)
......................= (0.98)(0.001) + (0.06)(0.999) = ...

I hope you are learning Bayes's Theorem:

P(disease[FONT=MathJax_Main] ∩[/FONT] test positive) = P(disease) * P(test positive | disease)
...................................= P(test positive) * P(disease | test positive)

P(disease | test positive) = P(test positive | disease) / P(test positive)

A test with 6% false positives is not very helpful.

Imagine a population of 100000 people. "About 0.1% of people in a given population carry a disease" so 100 people have the disease, 99900 do not. "For those that carry the disease, test are positive in 98% of cases" so 98 people with the disease test positive. "For those that do not carry the disease, tests are negative 94% of the time." Assuming there are no "null tests", the tests are positive for 6% of the people who do not carry the disease. 6% of 99900 is 5994 so a total of 5994 people who do not carry the disease test positive.

That is, a total of 5994+ 98= 6092 people test positive of whom 98 actually have the disease. If a person test positive, the probability that person actually has the disease is 98/6092= 0.016 or 1.6%.