On the table, there are eight $100 bills, three of which are forged. We may take three of them. What is the probabiliy that we take at least one forged bill?
Let's split the whole problem into subproblems. We could take one, two or three forged bills. There are $3 \cdot\binom{5}{2}$ sets with one forged bill, $\binom{3}{2} \cdot 5$ sets with two of them and $1$ set with three of them.
This adds up to 46.
There are $\binom{8}{3}$ = 56 possible sets of three, and so the probability is
$p(x) = \frac{23}{28}$
Is this the right solution? If so, is there an easier way to solve this?
The easiest way is finding the probability of have no forged bill. Because all chances add up to one, so $$1-P(0forged bill)=P(atleast1forged bills)$$
$$P(0forgedbills)=\frac{5}{8}\cdot \frac{4}{7}\cdot \frac{3}{6}$$
$$P(atleast1forged bills) = 1 - \frac{5}{8}\cdot \frac{4}{7}\cdot \frac{3}{6} = \frac{23}{28}$$