Author 
Message 
Intern
Joined: 12 Oct 2007
Posts: 24

A terrible disease sweeps around the world, luckily only [#permalink]
Show Tags
05 Sep 2008, 15:39
This topic is locked. If you want to discuss this question please repost it in the respective forum.
A terrible disease sweeps around the world, luckily only affecting 1 in 10,000, but for that one, the disease is lethal. Shortly after the disease was discovered, scientists developed a test that is 99% accurate regardless of whether you have the disease. In other words, the test yields the correct positive or negative result 99% of the time. You take the test and a week later, you receive the lab report. The outcome of the test is positive.
What is the probability you have the disease?
Last edited by freestyla86 on 06 Sep 2008, 06:12, edited 1 time in total.



VP
Joined: 05 Jul 2008
Posts: 1401

Re: Probability Word Problem (Not Multiple Choice) [#permalink]
Show Tags
05 Sep 2008, 16:56
freestyla86 wrote: A terrible disease sweeps around the world, luckily only affecting 1 in 10,000, but for that one, the disease is lethal. Shortly after the disease was discovered, scientists developed a test that is 99% accurate regardless of whether you have the disease. In other words, the test yields the correct positive or negative result 99% of the time. You take the test and a week later, you receive the lab report. The outcome of the test is positive.
What is the probability you have the disease? Let me try Probability that one has this disease means that one has to have the disease and the test he/she has taken must be correct positive. P ( one in 10k) X P (test is correct) (1/10,000) X (99/100)



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1346

Re: Probability Word Problem (Not Multiple Choice) [#permalink]
Show Tags
05 Sep 2008, 17:33
freestyla86 wrote: A terrible disease sweeps around the world, luckily only affecting 1 in 10,000, but for that one, the disease is lethal. Shortly after the disease was discovered, scientists developed a test that is 99% accurate regardless of whether you have the disease. In other words, the test yields the correct positive or negative result 99% of the time. You take the test and a week later, you receive the lab report. The outcome of the test is positive.
What is the probability you have the disease? Take a million people, and say they all get tested. 1,000,000 people 100 have the disease 999,900 do not of these: 99 test positive for the disease *and* have the disease 1 tests negative for the disease *and* has the disease 9999 test positive for the disease *and* do not have the disease 989901 test negative for the disease *and* do not have the disease So for every million people, 10,098 will test positive. Only 99 will have the disease. If you test positive, you're one of these 10,098 people, and there is a 99/10,098 = 1/102 chance you actually have it.
_________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com



Manager
Joined: 09 Jul 2007
Posts: 245

Re: Probability Word Problem (Not Multiple Choice) [#permalink]
Show Tags
05 Sep 2008, 17:36
Should be .99
I do not see any catch in this question..I might be missing something.



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1346

Re: Probability Word Problem (Not Multiple Choice) [#permalink]
Show Tags
05 Sep 2008, 19:12
ssandeepan wrote: Should be .99
I do not see any catch in this question..I might be missing something. If you want to see the catch, think of more extreme numbers: Say 1 person in the world has disease ZZZ, and 10 billion people do not. There's a test that is 50% reliable and you take it. If you test "positive", is there a 50% chance that you have the disease? Surely not; billions of people will test positive who don't have ZZZ.
_________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com



Manager
Joined: 03 Jun 2008
Posts: 135
Schools: ISB, Tuck, Michigan (Ross), Darden, MBS

Re: Probability Word Problem (Not Multiple Choice) [#permalink]
Show Tags
05 Sep 2008, 23:21
IanStewart wrote: freestyla86 wrote: A terrible disease sweeps around the world, luckily only affecting 1 in 10,000, but for that one, the disease is lethal. Shortly after the disease was discovered, scientists developed a test that is 99% accurate regardless of whether you have the disease. In other words, the test yields the correct positive or negative result 99% of the time. You take the test and a week later, you receive the lab report. The outcome of the test is positive.
What is the probability you have the disease? Take a million people, and say they all get tested. 1,000,000 people 100 have the disease 999,900 do not of these: 99 test positive for the disease *and* have the disease 1 tests negative for the disease *and* has the disease 9999 test positive for the disease *and* do not have the disease 989901 test negative for the disease *and* do not have the disease So for every million people, 10,098 will test positive. Only 99 will have the disease. If you test positive, you're one of these 10,098 people, and there is a 99/10,098 = 1/102 chance you actually have it. Brilliant!
_________________
 'It's not the ride, it's the rider'



Manager
Joined: 09 Jul 2007
Posts: 245

Re: Probability Word Problem (Not Multiple Choice) [#permalink]
Show Tags
06 Sep 2008, 04:22
I beg to disagree with this..
These two are independent cases,
1 ) affecting 1 in 10,000; so P1= 1/10,000 => so if somebody asks, what is the probability of anybody getting effected , asswer is 1/10000
2 ) test that is 99% accurate => so if somebody is tested, and whatever the result be, the accuracy of the result is =P2=99/100
Here, you take the test and the results are out , so do not need to worry about the 1st case at all. The question is not asking : what is the probability that if a person is tested, the tested person will be +ve, which will be 1P1*P2=19999/10000*99/100



Intern
Joined: 03 Sep 2008
Posts: 22

Re: Probability Word Problem (Not Multiple Choice) [#permalink]
Show Tags
06 Sep 2008, 05:28
This is an example of Bayes Theorem (see Wikipedia)
P(A and B) = P(A)*P(AB)
so
P(AB) = P(A and B)/P(A)
The events are not independent since the the probability you test positive is 99% if you have the disease and 1% if you don't have the disease.



Manager
Joined: 09 Jul 2007
Posts: 245

Re: Probability Word Problem (Not Multiple Choice) [#permalink]
Show Tags
06 Sep 2008, 06:10
My take is :
The outcome of the experiment independent of whether the person has the disease or not . SO the probalility of the the experiment being correct is independent of the probabality of a person being effected by the desase.
so P1 and P2 are independent.



Senior Manager
Joined: 04 Jan 2006
Posts: 276

Re: Probability Word Problem (Not Multiple Choice, No OA) [#permalink]
Show Tags
06 Sep 2008, 06:54
freestyla86 wrote: A terrible disease sweeps around the world, luckily only affecting 1 in 10,000, but for that one, the disease is lethal. Shortly after the disease was discovered, scientists developed a test that is 99% accurate regardless of whether you have the disease. In other words, the test yields the correct positive or negative result 99% of the time. You take the test and a week later, you receive the lab report. The outcome of the test is positive.
What is the probability you have the disease? I will try to solve this one in a long way. Assuming 1 million people in the city. People who have disease = (1/10,000) x (1,000,000) = 100 persons People who do not have disease = 1,000,000  100 = 999,900 persons For people with disease, the lab can identify correctly = (99/100)*100 = 99 persons the lab will make a mistake on = (1/100)*100 = 1 person For people with no disease, the lab can identify correctly = (99/100) * 999,900 = 989,901 person the labe will make a mistake on = (1/100)*999,900 = 9,999 person Now, Prob(you have disease based on the fact that the lab say so) = (99)/(99 + 9,999) = 99 / 99*(1+101) = 1/102 = 0.9804%
Last edited by devilmirror on 06 Sep 2008, 12:00, edited 4 times in total.



Intern
Joined: 03 Sep 2008
Posts: 22

Re: Probability Word Problem (Not Multiple Choice, No OA) [#permalink]
Show Tags
06 Sep 2008, 07:23
The events are not independent
A= test is positive
B= person has the disease
P(AB) = 0.99
P(AB') = 0.01
The probability of A depends on the status of the person being tested.
This is a classical problem in statistics/probability so much so that it has been given a name, Bayes Theorem
The wikipedia article explains it well and as did a previous poster in this thread.



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1346

Re: Probability Word Problem (Not Multiple Choice, No OA) [#permalink]
Show Tags
06 Sep 2008, 11:04
There is a profound difference between the following two questions: If you have the disease, what is the probability you test positive? (the answer is 99%) If you test positive, what is the probability you have the disease? (the answer is not 99%) The original post asks the second question, not the first.
_________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com



Manager
Joined: 15 Jul 2008
Posts: 205

Re: Probability Word Problem (Not Multiple Choice, No OA) [#permalink]
Show Tags
06 Sep 2008, 12:01
Yes this is a conditional probability problem which needs to be solved using Bayes' theorem. Getting a little more technical, we know the priory P(D) and and marginal probability P(p) (all possibilities of getting positive test result). We need to find the posterior P(Dp). D=having disease and ~D=not having disease p=testing positive and n= testing negative The question asks probability P(having disease given positive result) P(Dp) = [P(pD) * P(D) ] / P(p) where P(p) = P(pD)*P(D) + P(p~D)*P(~D) Here we know P(pD)=99/100 P(D)=1/10k P(p~D)=1/100 and P(~D)=9999/10k Plug in the values... and voila P(Dp) = 1/102




Re: Probability Word Problem (Not Multiple Choice, No OA)
[#permalink]
06 Sep 2008, 12:01






