# Statistics Binomial Probability

Suppose that in a group of 10 people, each person has a 0.1 probability of having a certain disease, independently from person to person. Now consider a testing plan where the group is split into two groups of 5 people where in each group blood samples are taken from each person, then their blood samples are combined and tested together (so there is one test per group of 5). If a group tests positive for the disease then each person in that group is tested individually for the disease, if a group does not test positive, then the testing is concluded for that group. Let the random variable X represent the total number of tests required with this plan.

Find the probability distribution of X.

I know that there are 2, 7, 12 possible values of X but I don't know what formula to use to find P(X). I'm assuming its the binomial probability formula with nCk(p^k)(q^n-k)

but how do you use 2, 7 and 12 in that? what is n and k?

• That's a crazy test plan. $\ddot\smile$ – Graham Kemp Oct 27 '16 at 0:32

Well, then you've figured out that the number of tests will be $2,7,12$ depending on how many groups test positive.
What is the probability that a group of five tests positive?   Call that $p_5$.   Then the expected value is evidently :
$$\mathsf E(X) = 2\cdot (1-p_5)^2 + 7\cdot 2 p_5 (1-p_5) + 12\cdot p_5^2$$
Evaluate $p_5$ is and you are done.
Hint: What is the probability of the complement event: that $0$ from a group of $5$ is tests positive, if the population test rate is $0.1$?