TITLE: Probability of getting off at the bus stop
QUESTION [5 upvotes]: 6 stops on the line, and a car with 4 passengers. Assume they are equally likely to get off at any stop. What is the probability that A $2$ passengers get off stop 2, and $2$ get off stop 4? B $2$ get off stop 1, and $2$ get of at another stop (but they get off at the same stop as each other?)

A
I get $1/81$ which is wrong?
B I get $5/6^4$ which is also wrong?
Why are both of these wrong?

REPLY [3 votes]: For both the parts, the total number of possible ways by which the passengers can exit is $6^4$.  
For part A,
the ways by which $2$ passengers can be grouped is $\binom{4}{2}$ (and the other $2$ are automatically grouped so there's nothing else to do). So the number of ways in which the exiting of passengers can be done is $\binom{4}{2}$.
So the required probability in part A is $$\frac{\binom{4}{2}}{6^4}$$
For part B,
the number of ways of grouping remains same:  $\binom{4}{2}$; but now the exit stop can be assigned in $5$ ways (one group exits at stop $1$ and other can exit on remaining $5$ stops). So the required probability is $$\frac{5 \times \binom{4}{2}}{6^4}$$