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September 23rd, 2007, 11:13 AM   #1
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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. T-T..


Please help,

Thank you,
Yeonbi
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September 24th, 2007, 07:27 PM   #2
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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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