In a 6-car metro train, there are 4 drunk passengers among the travelers. What is the probability that at most two cars have drunk passengers?
Solution. The 4 patients can independently choose which of the 6 cars they will travel in. Thus, the total number of possibilities is: $6^{4}=1296$.

The favorable cases are as follows:

a) 2-2 overbooked passengers get into two different cars. The two passengers can be chosen in $\binom{4}{2}$ ways, and the two cars can be chosen in $\binom{6}{2}$ ways. The number of possibilities is: $\binom{4}{2}\binom{6}{2}=90$

b) One car has 3 passengers, and another car has 1 passenger. The 3 overbooked passengers can be chosen in $\binom{4}{3}$ ways, and they can choose from 6 cars; the remaining 1 passenger can choose from 5 cars. The number of possibilities is: $\binom{4}{3} \cdot 6 \cdot 5=120$.

c) If the 4 overbooked passengers all go into one car, they have 6 choices.

The cases a), b), and c) are mutually exclusive, so the total number of favorable cases is: $90+120+6=216$. The probability is:

$$
\frac{216}{1296}=\frac{1}{6}
$$