Probability that two people Mike and Joe sit next to each in a random circular seating arrangement of eight seats.

probability

1,684

Solution 1

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.

Solution 2

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 ... ?

1,684

John

Updated on August 01, 2022

Comments

John 7 months

Is the sample space 8 choose 2? Why? What would the answer to this be?
- Barbara Osofsky about 10 years
  
  $\binom 8 2$ is a number, not a sample space. Moreover, where did 8 come from. I do not see it anywhere else in the question.
John about 10 years

Sorry forgot to mention circular seating arrangement
Brian M. Scott about 10 years

@Koy: It makes only a small change: you get one extra pair of adjacent seats.
John about 10 years

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?
Brian M. Scott about 10 years

@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 about 10 years

... them next to each other.