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Is the sample space 8 choose 2? Why? What would the answer to this be?

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$\binom 8 2$ is a number, not a sample space. Moreover, where did 8 come from. I do not see it anywhere else in the question. –  Barbara Osofsky Feb 6 '13 at 1:50
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2 Answers 2

up vote 3 down vote accepted

Wherever Mike is seated, there are $7$ unoccupied seats, and Joe is equally likely to be at any of them. Of these $7$ seats, $2$ are next to Mike, so the probability Joe is next to Mike is $2/7$.

Remark: There can be more than one sample space that is useful for calculating a probability. If we are going to solve the problem by counting, the most important thing is to make sure that we use a sample space of equaly likely outcomes.

One possible sample space is the set of pairs of seats. That sample space has $\binom{8}{2}$ elements. Or else it may be useful to think of Mike as sitting down first, and then Joe. The set of ordered pairs of seat choices is then a good sample space. That sample space has $(8)(7)$ elements. Or else we can exploit the symmetry, and use as our sample space the set of choices Joe can make, after Mike is seated. That gives a nice small sample space with $7$ equally likely elements.

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HINT: If they’re seated in a circle, there are indeed $\binom82$ possible pairs of seats in which Mike and Joe might sit. There are $8$ pairs of adjacent seats, so the probability that they sit together is ... ?

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Sorry forgot to mention circular seating arrangement –  John Feb 6 '13 at 1:42
    
@Koy: It makes only a small change: you get one extra pair of adjacent seats. –  Brian M. Scott Feb 6 '13 at 1:44
    
Is it 8/28? But if we place Mike in one seat, aren't there 2 options of Joe? Wouldn't that make it 16? –  John Feb 6 '13 at 1:53
    
@Koy: $\frac8{28}=\frac27$ is correct. You can also work it the way you’re thinking about, but then you have to realize that the denominator is not $\binom82$, but rather $8\cdot7$, because you’re counting ordered pairs of seats: there are $8$ ways for Mike to choose a seat, and once he’s chosen, there are $7$ ways for Joe to choose a seat, for a total of $56$ seat assignments, $16$ of which have them together. In other words, you can look either at how many of the $28$ possible pairs of seats are adjacent, or you can look at how many of the $56$ possible seatings of the two put ... –  Brian M. Scott Feb 6 '13 at 1:56
    
... them next to each other. –  Brian M. Scott Feb 6 '13 at 1:58
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