Fairly easy probability problem

[Hofzinser]

A horrid disease affects 1 in 10,000 people. You are tested for it. The test is accurate 99 times in 100. The result comes through positive. What are the chances that you have the disease?

a) 99/100
b) 99/10,000
c) 1/2
d) 1/100
e) 1/101
f) 1/10,000

Tis isn't too difficult, I know, and I'm fairly sure the answer is (d), based on 10,000/100=100, therefore in a group of 10,000 without a single instance of the disease there will still be 100 false positive results.

But what's confusing me is that in that group there will also be one person who does have the disease - so does that make the answer 1/101? But then there's also a chance that the person with the disease has a false negative result which must also be considered - so does this mean that it is 1/100 after all? Or am I needlessly confusing myself?

I'd be really grateful for an illustration of the precise working needed here.

[undercheck]

This depends on what you exactly mean by 'the test is accurate 99 times of 100'. Does the test give the wrong result exactly 1 % of time both if you have the disease and if you do not? I assume so but this is not clear to me from your post.

Let D={yes, no} denote whether you have the disease,
T={yes, no} denote whether the test result is positive

You want to calculate P(D=yes|T=yes), that is the probability that you have the disease given that the test result is positive.

By Bayes' theorem P(D=yes|T=yes)=P(D=yes)*P(T=yes|D=yes)/P(T=yes) = P(D=yes)*P(T=yes|D=yes) / (P(D=yes)*P(T=yes|D=yes)+P(D=no)*P(T=yes|D=no))

=(1/10000)*(99/100) / ((1/10000)*(99/100) + (9999/10000)*(1/100)) = 99 / (99 + 9999) = 99 / 10098 = 1/102

Or, a perhaps more intuitive approach:
of 1,000,000 people:
- 100 have the disease, of them:
- 1 will have a negative test result
- 99 will have a positive test result
- 999,900 do not have the disease, of them
- 989,901 will have a negative test result
- 9,999 will have a positive test result

So, in total we have 99+9999 positive test results, of which 99 have the disease, the probability of having the disease is 99/(99+9999) = 1/102

[Sphere99991]

The actual wording of the problem is just the chances that you have the disease though, not "given that the test came back positive". So should be still 1/10k. Given that technically the answer should be 1/102, and that is not a choice, I'd say the answer if F.

.01 X 9,999 healthy (wrong) = 99.99) and (.99 X 1 =.99) sick right = 100.98 expected positive tests and .99 positive correct tests. .99/100.98= 1/102

[sebbb]

event B: "you have the disease"
A: "you are positive"

I assume the correctness of the test doesn't depend on if you are sick or not

we want P(B|A)
P(B|A)= P(A and B)/P(A)

Bayes:


P(A)= P(A and B) + P(A and (not B))= P(A|B)*P(B)+P(A|(not B))* P(not B)=(99/100*1/10000+1/100*9999/10000)=(99+9999)/1,000,000=10098/1000000

P(A and B)= P(A|B)*P(B)=99/100*1/10000=99/10^6

final answer: 99/(99+9999)= 99/10098=11/1122 =1/102

[AaronBrown]

Suppose we test 1,000,000 people, 100 (1 in 10,000) of whom have the disease. Of the sick people, 99 will test positive and 1 will test negative. Of the not-sick people, 9,999 will test positive and 989,901 will test negative.

So of the people who test positive, 99/(99 + 9,999) = 1/102 will have the disease.

However, that answers the question, "if you test 1,000,000 people and select one positive result at random, what is the chance that person has the disease?" It's not clear that it is the answer to the question "if you test positive, what is the chance you have the disease?"

To make that logical leap, you have to be a random person. If you sought the test for any reason, say you noticed some symptoms of the disease or knew you were at heightened risk for it, your chance will not be 1/102.

[DiceyPlay]

Apply Bayes Theorem:

Define the following events:
A = you have the disease
B = you do not have the disease
C = you test positive for the disease

P(A|C) = P(C|A)*P(A)/(P(C|A)*P(A)+P(C|B)*P(B)) =

(P(correct positive)*P(you have disease)) / (P(correct positive)*P(you have disease) + P(false positive)*P(you do not have disease))

= (99/100 * 1/10k) / (99/100 * 1/10k + 1/100 * 9999/10k) = (99 / 1M) / (100 / 1M) = 99/100

That would also be the intuitive answer since the test is correct 99/100 times. Usually though a test for a disease results in a correct positive a different proportion of the time than it results in a correct negative. In this case the correct positive and the correct negative occur in the same proportion.

[karlwig]

The answer is not D. It can't be. If you have tested positive, and the test is accurate 99 out of 100 times, of course the chances of you having the disease is 99/100!

The way I understand it, the number of people getting affected has nothing to do with it at all. All you need to know is that you are one of those having a positive result, and that result is almost certainly right - 99 out of 100 times. So the answer is of course A.

[DiceyPlay]

Yes, since the the test is correct for positives and negatives the same proportion of time, the correct answer is the answer one would arrive at by intuition.

Here's a better question that exercises Bayes Theorem:

Suppose that 0.5% of all the students seeking treatment at a school infirmary are eventually diagnosed as having mononucleosis. Of those who do have mono, 90% complain of a sore throat. But 30% of those not having mono also have sore throats. If a student comes to the infirmary and says that he has a sore throat, what is the probability that he has mono?

Problem Source: An Introduction to Mathematical Statistics and Its Applications by Richard J. Larsen and Morris L. Marx Second Edition 1986 - Prentice Hall Question 2.6.25 page 58.

[jjshabado]

[ QUOTE ]
The answer is not D. It can't be. If you have tested positive, and the test is accurate 99 out of 100 times, of course the chances of you having the disease is 99/100!

The way I understand it, the number of people getting affected has nothing to do with it at all. All you need to know is that you are one of those having a positive result, and that result is almost certainly right - 99 out of 100 times. So the answer is of course A.

[/ QUOTE ]

Thank you. I read the OP and was like the answer's A obviously. Then I started reading all of the responses and I couldn't figure out what I was missing.