Expected value and variance of a discrete random variable

Question

Problem I'm working on:

On average, 4% of Americans have blood type AB. Suppose we are screening a group of 100 people for this blood type, and we decide to pool blood samples in 20 groups of 5 samples each in order to speed up the testing. If any pool shows evidence of type AB, then we’ll go on to test the 5 individuals in the pool separately. We can assume that blood types of different individuals are independent.

What is the expected number of blood tests to be performed? (Compare this to the 100 we’d do without pooling.)

What is the variance in the number of tests to be performed?

It's taking a lot of work for me to add up 21 terms for the expected value. I feel that there's a way to turn this into a series, but I'm having trouble doing so.

For the variance, I would use the fact that var(x)=E(x^2)-E(x)^2, so I'm having the same problem there.

Answer 1

If the groups of five are all independent of each other, then find the mean and variance for one group of five and then multiply them by $20$.

With one group of five, you do one test, and your probability of NOT having to do five others is $0.96^5\approx0.815.$ So the number of tests $$ \begin{cases} 1 & \text{with probability approximately }0.815, \\ 6 & \text{with probability approximately }0.185. \end{cases} $$ The variance is $(6-1)^2$ times the variance of a Bernoulli distribution (whose possible values are $0$ and $1$). I'll let you find the expected value.