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fatcarl
11-06-2005, 11:49 PM
I'm not sure if this is the right place to post this....

Five people enter an elevator at the ground floor af an 8 story builiding. Assuming that each person independenly of the others, is equally likely to get off at any one of the 7 flooors above ground, what is the probablity that exactly 2 people will get off at the same floor? (ie 2 people exit on one of the floors and the other 3 people exit alone, each on a different floor)?

Any ideas?

soroban
11-07-2005, 01:28 AM
Hello, fatcarl!

Five people enter an elevator at the ground floor af an 8 story builiding.
Assuming that each person is equally likely to get off at any one of the 7 floors above ground,
what is the probablity that exactly 2 people will get off at the same floor?
(ie 2 people exit on one of the floors and the other 3 people exit alone, each on a different floor)
I'll assume that the five people are distinguishable (i.e., have different names).

Person "A" has 7 choices of floors, person "B" has 7 choices of floors, person "C" has 7 choices, etc.
. . Hence, there are 7^5\,=\,16,807 possible ways they could exit.

First, select the two that will exit together. .There are C(5,2)\,=\,\frac{5!}{2!3!}\,=\,10 possible pairs.
. . Duct-tape the pair together. .Then we have 4 "people" to consider.
. . These 'four' can exit in: .P(7,4)\,=\,\frac{7!}{3!}\,=\,840 ways.

Hence, they can exit in: 10\,\times\,840\:=\:8400 ways.

The probability is: .\frac{8400}{16807}\:=\:\frac{1200}{2401}\:\approx \:50\%

fatcarl
11-07-2005, 02:04 AM
ok great, thank you
that makes sense

so you take the possible pairs of people C(5,2) and multiply it by the number of ways the other people can exit P(7,4) and divide that by the total exit possibilities...

thank you