Imagine you were on summer holidays on the beautiful island of
Teeceeralia. After returning home, you learn that 1 person in 1000
visitors of Teeceeralia is infected with a strange disease leading to
"Dancing Chicken Syndrome", DCS for short. The good news is, there is
a cure (though painful), and a way to test whether you are infected.

The test detects DCS with 99 % certainty, meaning of 100 people
infected with DCS, 99 are detected by the test and only one is missed
(1 % false negative results).

Conversely, among 100 healthy people, 2 people are wrongly diagnosed
with DCS (2 % false positive results).

You take the test and your test result is positive, meaning the test
says you are infected. Should you worry, and if, how much? What is
that chance that you are in fact infected?

The answer is one of the following:

- 99 %
- 98 %
- 50.5 %
- 33.3 %
- 4.72 %
- 1 %
- 0.1 %

I believe there is no correct answer.

.1 % of people get the disease.

Now if you have the disease, and your test result is positive, the
test is 100% correct.  The fact that the test can bring negative
results even though you are positive does not apply in this question,
because your result is positive. Therefore if you have the disease and
have a positive result, there is a 100% chance you have the disease.

However, if you dont have the disease, yet you have a positive result,
there is a 98% chance you don't have the disease.

Because you don't know for sure if you have the disease or not, the
two answers are mutually exclusive.

However, you could take the average of the two, but again, that would
be a meaningless answer.

While any one person (according to the facts outlined in the Question)
is either infected or not, and each test comes out positive or
negative, the point of the exercise outlined here is perhaps to use
Bayesian reasoning to say what the conditional probability of
infection is for the circumstances specified.  In that sense there is
a "correct answer" other than simply asserting a black and white
dichotomy.

regards, mathtalk-ga

Here is one way to think about it: 

In a group of 100,000 visitors, on the average 100 will be infected
and 99,900 will not.  99 of the infected visitors will test positive
(99% of 100), and 1998 of the health visitors will test positive (2%
of 99,900).  So, on the average, in a group of 100,000 visitors, 99 +
1998 = 2097 visitors will test positive, and of these 2097, 99 will be
infected.  So the chance you're infected, given that you tested
positive, is 99/2097 = 4.72 %.

HW question, right?
Here's the solution:
Let "+" stands for positive diagnosis, "-" stands for negative
diagnosis, "D" for having disease, "ND" for not having disease.
You are given P(D)=0.001 (hence P(ND)=0.999), P(+/D)=0.99, P(+/ND) = 0.02.
You want to find P(D/+).
Now P(D/+)=P(D intersect +)/P(+)
          =(0.99x0.001)/(0.99x0.001+0.02x0.999)
          =0.0472
          =4.72%

i should be woried becaue i have the disease chance of 99% thratically
and 98% practically so ui gor for 99%

What you should use is the bayes rule: P(A|B) = P(A and B)/ P(B).
Probability of A knowing event A occured.

Event "Test" is positive testing, Event "!Test" is negative testing.
Let say P(CVS) = 1/1000 , prob of having the disease,
P(Test|CVS) = 99/100
P(Test|!CVS) = 2/100

So We want P(CVS|Test).
With Bayes Rule:
P(CVS|Test) = P(Test and CVS)/ P(Test)
 = P(CVS) * P (Test|CVS) / P(CVS)        (1)
 
P(Test), prob of having the disease is the prob of having the disease
and being diagnose + Prob of having it and not being diagnose.
So ,
P(Test) = P(CVS and Test) + P(CVS and !Test)
and applying again bayes...
 = P(CVS) * P(Test|CVS) + P(!CVS)*P(Test|!CVS)  (2)

putting back (2) in (1)
we get 
P(CVS|Test) = ( P(CVS) * P(Test|CVS) / ( P(CVS)*P(Test|CVS) +
P(!CVS)*P(Test|!CVS ) )
 = 1/3 = 33.3 %

Is this OK as an answer?

To jsimon:
The question has already been answered by racecar and myself.
Also, there are 2 mistakes in your solution:
Firstly, it is in line (1), it should be:
= P(CVS) * P (Test|CVS) / P(Test)        (1)
Secondly, the evaluated result is 0.0472 or 4.72%. Please see my
working where I have shown the substitution of the values to get
0.0472.

Sorry pkuanko-ga and racecar-ga,
Since my final answer was different from yours, I did not pay
attention and did not notice that the reasonning was quite similar.
I searched a lot for what was the problem with my answer. Finaly, Here
what is wrong.
1) A Typo at line (1), the division should be ( / P(Test) ). 
2) Another Typo, at the line just before (2), but the following line, (2), is OK.
Finaly, I get to the line before the final result, and it is similar to yours:

You put: " =(0.99x0.001)/(0.99x0.001+0.02x0.999)"
I put:      "P(CVS|Test) = ( P(CVS) * P(Test|CVS) / ( P(CVS)*P(Test|CVS) +
P(!CVS)*P(Test|!CVS ) ) "

Wich is the same!

My big error was the final Result, the line I put is not equal to
33.3% but to 4.72% as you say.
Sorry, next time I will double (triple) check.