Diagnosed with a rare disease, you know that there is only a 1% chance of getting it.
Abbreviate D as the event "you have the disease" and T as "you test positive for the disease."
The test is imperfect: $\Pr(T | D) = 0.98$ and $\Pr(T^C | D^C) = 0.95.$
$\large{A.}$ Given that you test positive, what is the probability that you really have the disease?
$\large{B.}$ You obtain a second opinion: an independent repetition of the test. You test positive again. given two positive tests, what is the probability that you really have the disease?
- So, first question I have is this: Does $\Pr(T | D^C) = 1 - 0.95 = 0.05$?
This would mean that $P(T) = P(T | D) + P (T | D^C) = 0.98 + 0.05 = 1.03$ ...? This can't be right...P(T) can't be 1.03! Is there an error in this question?
- Then, here's my strategy for solving part A: Calculate $\Pr(D | T) = \dfrac{P(T \cap D) }{ P(T) } = \dfrac{ P(T | D) \times P(D) }{ P(T) } $
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 for part B above.
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 $\Pr(T \cap S)$. To do this, you should assume that these two tests are independent, and therefore you will get:
$\Pr(T \cap S | D) = P(T | D) * P(S | D) \quad$ and $\quad \Pr(T \cap S | D^c) = P(T | D^c) * P(S | D^c)$.
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.
Pr(T∩S|D)=P(T|D)∗P(S|D).
– PGupta Aug 22 '17 at 13:37