You are diagnosed with an uncommon disease. You know that there is only a 1% chance of getting it. Use the letter D for the event "you have the disease" and T for "the test says so." It is known that the test is imperfect:: P(T | D) = 0.98 and P(!T | !D) = 0.95 (Here I'm using ! to mean the complement of event that follows the !)
A. Given that you test positive, what is the probability that you really have the disease?
B. You obtain a second opinion: an independent repetition of the test. You test positive again. Given this, what is the probability that you really have the disease?
So, first question I have is this: If P(!T | !D) = 0.95, then does P(T | !D) = 1 - 0.95 = 0.05?
This would mean that P(T) = P(T | D) + P (T | !D) = 0.98 + .05 = 1.03 ...? That can't be right...
Then, here's my strategy for solving part A:
Looking for P(D | T)
P(D | T) = P(T intersect D) / P(T) = [P(T | D) * P(D)] / P(T)
... But P(T) can't be 1.03!
Is there an error in this question?
UPDATE: Here is a hint from the professor:
During office hours today, the following hint came up that I thought would be good to share with the entire class. This is the problem that asks about the probability of having the disease given two tests that were positive.
Let's define two events T (first test positive) and S (second test positive). When you use Bayes' rule, you are going to need to figure out how to compute the total probability P(T intersect S). To do this, you should assume that these two tests are independent, and therefore you will get:
P(T intersect S | D) = P(T | D) * P(S | D)
P(T intersect S | D^c) = P(T | D^c) * P(S | D^c)
(I'm using "D^c" to denote complement of the event D.) A point to remember here is that the rules of probability stay the same for conditional probability if the event on the right of the "|" stays constant. For instance, the complement rule looks like this:
P(A^c | B) = 1 - P(A | B)
The other rules we have learned work out similarly.