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What is the probability that 2 people getof at the same stop


New member
Oct 1, 2009
Q: there is 6 subway stops left and it has only 4 people in it. Assuming they are each equally likely to get off at any stop. What is the probability that 2 of them get off at the same stop and 2 get off at another.

the given answer is 5/72. My answer is since there are 4 people so " _ _ _ _ ". if put them in paires then in the first paire the first person has a choose of 6 stops to choose from but the second person have to get off at the same stop, so the out come is 6*1. For the second pair, similarly, it's 5*1. So total has 6*1*5*1 = 30 possible out comes, then the P is 30/6^4 which desnot equal to 5/72. I don't understand how they come up with the 5/72 answer.


arthur ohlsten

Full Member
Feb 20, 2005
the number of combinations of 2 of the 4 people get off at a stop = 4C2
The number of combinations of the remaining 2 getting off at a station = 2C2

The number of combinations of these two events happing at any 2 of the 6 stops = 6C2
or 15 possible arrangments of stops that two sets of two can exit.

the total possible events are that the first man can exit at any of 6 stops
the 2nd man at any of 6 stops
the 3rd man at any of 6 stops
the 4th man at any of 6 stops

the probability of the 4 men leaving the train in two sets of 2 at different stops are:
P= { (4C2) (2C2) (6C2) } / [ 6x6x6x6]={ [(4x3)/(1x2)][1][(6x5)/(1x2)] } / [6x6x6x6]

P=5/72 answer