# Probability problem in textbook

There's this problem in my textbook I don't really know how to do, and the answer key doesn't give steps:

"There are 6 stops left on a subway line and 4 passengers on a train. Assume they are each equally likely to get off at any stop. What is the probability that 2 get off at one stop and 2 at another stop?"

The correct answer is supposed to be 5/72. How should I proceed with the problem?

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Since each passenger has equal probability of choosing any of the stops, we need only take the number of desired configurations over the number of possible configurations.

There are $4$ choices to make in order for all passengers to exit the train (each must choose some stop), and $6$ options for each, so there are $6^4$ total arrangements in which the passengers can exit the train. If you like, think of it as each of the $4$ passengers rolling a distinctive $6$-sided die, and getting off at the stop that their die indicates. There are $6^4$ ways to roll $4$ distinct $6$-sided dice.

There are ${}_4C_2=6$ possible ways we can choose the first pair to exit (and the first pair determines the second pair, so there's nothing else to do. There are ${}_6C_2=15$ ways that we can choose two stops for our pairs, so there are $6\cdot 15=90$ ways that the passengers can exit the train in two separate pairs.

Hence, the probability is $$\frac{90}{6^4}=\frac{2\cdot 3^2\cdot 5}{2^4\cdot 3^4}=\frac5{2^3\cdot 3^2}=\frac5{72}.$$

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Imagine that the passengers each toss a fair die to decide which stop to get off at. List their results as a string of length $4$ made up of the symbols $1$ to $6$. (The order is determined by their student ID.)

There are $6^4$ such strings, all equally likely. We now count the number of strings that have two pairs.

The two numbers that we have two each of can be chosen in $\binom{6}{2}$ ways. For each such choice, the $2$ locations of the higher of the two numbers can be chosen in $\binom{4}{2}$ ways, for a total of $\binom{6}{2}\binom{4}{2}$. So our probability is $$\frac{\binom{6}{2}\binom{4}{2}}{6^4}.$$

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Probably the easiest way would be to count the number of ways that they can get off total and the number of ways they can get off in pairs. There are 3 possible pairs and they can get off at $\binom{6}{2}$ pairs of stops. Also there are a total of $6^4$ ways the people can get off. So the probability is $$\frac{3\cdot\binom{6}{2}}{6^4}=\frac{5}{72}$$

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