Probability that a random individual is a carrier of a disease given that at least one of two independent blood samples test positive I have a quick question about a two part problem that I am trying to solve. I can get a numeric answer, but I was wondering how I would get an answer if the probabilities were unknown.
The question:
1% of all individuals in a certain population are carriers of a particular disease. A diagnostic test for this disease has a 98% detection rate for carriers and a 3% detection rate for noncarriers. Suppose the test for this disease is applied independently to two different blood samples from the same randomly selected individual.
a) What is the probability that at least one test is positive ?
b) What is the probability that the selected individual is a carrier if at least one test is positive ?
To solve the problem, I defined $C$ as the event that the individual is a carrier, $T$ as the event that the individual tested positive on a random blood sample, and $X$ as the event that the individual tested negative on two independent blood samples (so $X'$ is the event that at least one test was positive). From the problem, $P(C) = 0.01$, $P(T|C) = 0.98$, and $P(T|C') = 0.03$.
Here's how I did the first part:
$$P(T) = P(T \cap C) + P(T \cap C') = P(T|C)P(C) + P(T|C')P(C') = 0.0395$$
$$P(X) = P(T')^2 = 0.92256025$$
$$P(X') = 0.07743975$$
For the second part, I can get a numeric answer, but I am having trouble creating an equation for it. I'm stuck on the first step.
$$P(C|X') = \frac {P(X'|C)P(C)} {P(X')}$$
To get the numerical answer, I can read $P(X'|C)$ as "the probability that an individual tests positive on at least one of two independent blood samples given that the individual is a carrier", so 
$$P(X'|C) = 1 - (1 - P(T|C))^2 = 0.9996$$
After that,
$$P(C|X') = 0.12908099522$$
First off, is this answer correct? If it is, how can I solve the problem without doing what I did?
 A: You have the right idea in general, but unfortunately there's a slip in your equation
$\ P(X) = P(T')^2\ .$ The problem here is that the two tests are not absolutely indepenent, only conditionally independent given that the individual tested is a carrier or given he or she is not a carrier.
Thus, while you can assume $\ P(X|C) = P(T'|C)^2\ $, and $\ P(X|C') = P(T'|C')^2\ $, when you combine these together you get $ P(X) = P(T'|C)^2P(C)+ P(T'|C')^2P(C') \ne P(T')^2\ $.  My calculations give me $\ 0.931495\ $ for $\ P(X)\ $, or $\ 0.068505\ $ for $\ P(X')\ $.
Your solution to the second part looks perfectly fine to me (apart from the use of an incorrect value for $\ P(X')\ $), so I'm not sure what more you're looking for.  If you want to express $\ P(C|X')\ $ in terms of a single expression, you could simply substitute your expression for $\ P(X'|C)\ $  into the one for $\ P(C|X')\ $ to get
$$P(C|X') = \frac {(1 - (1 - P(T|C))^2)\,P(C)} {P(X')}\ ,$$
but I don't see why you would need to do this to answer the questions posed.
