I'm doing this problem from "The Probability Tutoring Book" by Carol Ash.
The probability that a well is polluted is .1. Suppose 100 wells are tested as follows.
Divide the wells into 5 groups of 20. For each group, water samples from each of the 20 wells are pooled and one test is performed. If a test is negative, all the wells in that group are safe. If a test is positive each one of the 20 wells must be tested.
Find the expected number of tests we need to perform under this procedure.
I'm going to attempt this problem by using the Law of Total Expectation.
$$E[X] = E[E[X|i\text{ groups positive}]] = \sum_{i = 0}^{5}{E[X|i\text{ groups positive}]}{P(i\text{ groups positive})}$$
Now let's find all the $P(i\text{ groups positive})$ and $E[X|i\text{ groups positive}]$ $$E[X|0\text{ groups positive}] = 5 \text{ and}$$ $$E[X|i\text{ groups positive}] = 20i \text{ for i = 1, 2, ..., 5}$$
$$P(i\text{ groups positive}) = {5 \choose i}{\left({1 - .99^{20}}\right)^i}\left(.99^{20}\right)^{5-i}$$
If we want exactly $i$ of them to be polluted then then the probability of that would be ${5 \choose i}{\left({1 - .99^{20}}\right)^i}\left(.99^{20}\right)^{5-i}$. Any feedback on the probability here would be greatly appreciated :)
So now that I've found all the terms in the sum for E[X] I can just add them together.
Is this solution correct? If not, constructive feedback is happily welcome!
EDIT: I've made edits to the above problem thanks to be brilliant people in the comments right below. However, now I'm having a different problem.
So I computed the above summation via wolframalpha and got an expected number of tests $\approx 20$, but the answer at the back of the book claims that the answer should be $\approx 83$