I have a question that I am overcomplicating.
Here is a summary:
The probability of having a disease is $p$.
Tests are pooled into groups of $k$ people. (The samples of $k$ people are mixed together and tested at the same time.)
Whether people have the disease is independent.
If the pooled test is positive, give an expression for the probability that exactly $y$ of the $k$ people in the group are infected, for $y = 1, 2, . . . .k$.
I can't decide if this is a Binomial Distribution or not, given the given in the question. If it was binomial, should I use $p$ as the probability, or should I use $\frac{1}{k}$? All cases in a binomial distribution would not have a sum of 1.
I also thought about using Bayes' Theorem, but I wasn't sure how to apply it when more than one person had the disease in the pool.
Any help will be greatly appreciated.
Thanks in advance!
With the given parameters, the probability that exactly $y$ people among a group of $k$ people are infected is of Binomial distribution, that is, $P(y)=C_k^yp^y(1-p)^{k-y}$, where $y=0, 1, 2, ..., k$.
On the other hand, under the condition that at least one people is infected, the conditional probability that exactly $y$ people are infected is
$P(y|at\ least\ one\ people\ is\ infected)=\frac{P(y)}{1-P(0)}=\frac{C_k^yp^y(1-p)^{k-y}}{1-C_k^0p^0(1-p)^{k-0}}=\frac{C_k^yp^y(1-p)^{k-y}}{1-(1-p)^k}$,
where $y=1, 2, ..., k$.