This is Problem 1.59 from Tsitsiklis, Bertsekas, Introduction to Probability, 2nd edition.
A parking lot contains 100 cars, k of which happen to be lemons. We
select m of these cars at random and take them for a test drive. Find
the probability that n of the cars tested turn out to be lemons.
I worked towards a solution and got stuck after I draw a series of branches for the different possible selects and didn't show how to express them mathematically.
The solution book says
Clearly if n > m, or n > k, or m − n > 100 − k, the probability must
be zero. If n ≤ m, n ≤ k, and m − n ≤ 100 − k, then we can find the
probability that the test drive found n of the 100 cars defective by
counting the total number of size m subsets, and then the number of
size m subsets that contain n lemons. Clearly, there are 100 m
different subsets of size m. To count the number of size m subsets
with n lemons, we first choose n lemons from the k available lemons,
and then choose m − n good cars from the 100 − k available good cars.
Thus, the number of ways to choose a subset of size m from 100 cars,
and get n lemons is$${k \choose n} {{100 – k} \choose {m – n}}$$
And the desired probability is
$${ {k \choose n} {{100 – k} \choose {m – n}} \over {100 \choose m}}$$
This answer really doesn't make much sense to me at all.
One possible explanation is that they are treating the solution as a sequence where the first stage is select a lemon and the second stage is select a non-lemon and they are saying that there are ${k \choose n}$ ways to select a lemon and, with the sequential rule, there are ${100 – k} \choose {m – n}$ ways to choose a non-lemon and so to get the total we multiply them together.
Best Answer
The problem asks for the probability that $n$ cars among $m$ chosen cars are lemons.
$\binom{k}{n}\binom{100-k}{m-n}$ is the number of ways to choose $m$ cars so that $n$ of them are lemons.