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September 23rd, 2007, 10:13 AM  #1 
Newbie Joined: Aug 2007 From: University of Waterloo Posts: 5 Thanks: 0  Stats 230 Problem
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 a)they all get off at different stops? I know how to do this : →Sample space = 6^4 = 1296 →They all get off at different stops = (6 choose 1)(5 choose 1)(4 choose 1)(3 choose 1) = 360 [How i thought was each person chooses 1 stop to get off and next person must not get off at the same stop. Like each bracket represents each person's decision] ●So the probability = 360 / 1296 = 5/18 b)if 2 get off at one stop and 2 at another stop? →Same sample space = 6^4 = 1296 →The probability (I think and am NOT sure) = (6 choose 1)(1 choose 1)(5 choose 1)(1 choose 1) = 6*1*5*1 = 30 [It is the way I go : Each bracket represents each person's decision. Since a person chooses 1 stop to go , the one follows so who can only have one choice] But the answer is not matched to the answer. TT.. Please help, Thank you, Yeonbi 
September 24th, 2007, 06:27 PM  #2 
Member Joined: Mar 2007 Posts: 57 Thanks: 0 
a) You are correct. The first person has 6 stops to choose from, the second has 5 left to choose from, the third has 4, and the fourth has 3. The sample space is 6^4 = 1296. The required probability is (6*5*4*3) / (6^4) = 360/1296. b) Label the passengers 1,2,3,4. Arrange the 4 people into two pairs, A and B: there are 12,34 13,24 14,23 = three different ways of choosing A and B. There are 6 stops A can choose from, and B chooses from the remaining 5. Therefore the required probability is (6*5*3) / 1296 = 90 / 1296. 

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