[Guest]

You are tested for a rare flesh-eating virus that afflicts 1% of the population. If you have the virus, the test will always come back positive, but if you do not, there is a 20% chance of a false positive. You are told that your test came back positive. What is the probability that you have the disease?

[curr3nt]

nevermind...I should sleep

Edited April 19, 2011 by curr3nt

[Guest]

OMG, I did all this math and realized it has to be 1% since only 1% of the population is afflicted with the virus.

Edited April 19, 2011 by liliquoi_moon

[Guest]

Out of every 100 people, 1 will have the virus and will therefore receive a positive result. Additionally, 20% of the remaining 99 people will receive positive results, albeit falsely. That means that 21 people will receive positive results but only 1 out of the 21 will actually have the virus. So, if you receive a positive result the probability that you have the disease is 1 in 21 or 4.75% (rounded).

[Guest]

The 1% of population is a red herring. Of those getting tested, 80% really have the disease and 20% do not. Hence the answer is an 80% probability of having the disease.

[Guest]

Easy. Bayes' Theorem

P(H) = 0.01

P(~H) = 0.99

P(E|H) = 1.00

P(E|~H) = 0.20

We are asked to find P(H|E).

P(H|E) = (P(H)P(E|H))/(P(H)P(E|H)+P(~H)P(E|~H))

Plugging t3h numberz we get:

P(H|E) = (0.01*1.00)/(0.01*1.00+0.99*0.20) = 0.0480769

[Guest]

Since the 1% that have the test and have the disease, and out of the other 99% that do not have the disease, 20% or 19.8 out of the other 99 will test incorrectly. So 20.8% will test positive while only 1% will have the disease, your probability is 1 out of 20.8 or less than 5% if you have a positive test.

[Guest]

To keep the simple math simple, n=100 was used and numbers are rounded to whole people.

Of 100 people, 1 true positive and 20 false positive results are reported.

A total of 21 positives are reported and thus, 1/21 positives are true positives.

1/21 x 100% = 5%

That is the answer, of the 21 positive results, he has a 5% chance of being one of the "lucky" 20; I'd rather be one of the 100-20-1, or the lucky 79.

The question:

You are tested for a rare flesh-eating virus that afflicts 1% of the population. If you have the virus, the test will always come back positive, but if you do not, there is a 20% chance of a false positive. You are told that your test came back positive. What is the probability that you have the disease?

Edited April 19, 2011 by DDN

[Akriti]

i was about to do very long calculation. then i realised that the answer is only one percent

Edited April 19, 2011 by Akriti

[Akriti]

guys i need help!!!!!

Can anyone tell me how to lock a topic that's over?

plz....

thank u....

[Guest]

Out of every 100 people, 1 will have the virus and will therefore receive a positive result. Additionally, 20% of the remaining 99 people will receive positive results, albeit falsely. That means that 21 people will receive positive results but only 1 out of the 21 will actually have the virus. So, if you receive a positive result the probability that you have the disease is 1 in 21 or 4.75% (rounded).

I guess I should have kept doing the math! I was one step away from your answer when I stopped.

[Guest]

Thoughtful Fellow above... the riddle says that the probability of a false positive is 20%. You get a positive. Basic arithmetic - 100%-20%=80% probability. The 1% part is just noise.

[Guest]

It won't let me edit my first post, but I realized I made a significant error!

Ok, so 1 out of 100 will be infected. Of the other 99, 20% will test positive falsely, leaving 19.8. Therefore 20.8 will test positive with only one positive result. That gives me a probability of 4.81%. Not sure if my math is right though.

[Guest]

^ It is.

I got:

0.0480769

[Guest]

Psychosmurf is correct (so were several other people) and gets bonus points for using Bayes' Theorem!

[Guest]

Psychosmurf is correct (so were several other people) and gets bonus points for using Bayes' Theorem!

yay! Bonus points!