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I'm trying to solve the following problem:

Ten people are sitting around a round table. Three of them are chosen at random to give a presentation. What is the probability that the three chosen people were sitting in consecutive seats?

I got the wrong answer but cannot see the error in my reasoning. This is how I see it:

1) the selection of the first person is unconstrained.

2) the next person must be selected from the 2 spots adjacent to the first. So this choice is limited to 2/9 of the possible choices.

3) the third choice must be taken from the one free spot next to the first person chosen, or the one free spot next to the 2nd person chosen. So this choice is limited to 2/8 of the possible choices.

4) multiplying these we get:

2/9 * 2/8 = 1/18

However, the official answer is:

Let's count as our outcomes the ways to select 3 people without regard to order. There are $\binom{10}{3} = 120$ ways to select any 3 people. The number of successful outcomes is the number of ways to select 3 consecutive people. There are only 10 ways to do this -- think of first selecting the middle person, then we take his or her two neighbors. Therefore, the probability is $\frac{10}{120} = > \boxed{\frac{1}{12}}$.

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You forgot the possibility that the second person can be chosen to sit two seats away from the first, and then the third person is chosen to be the one between the first and the second. This gives an additional $\frac29\cdot \frac18 = \frac1{36}$, bringing the total up to $\frac1{18}+\frac1{36} = \frac1{12}$.

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  • $\begingroup$ great answer, thanks! $\endgroup$ – David J. Feb 8 at 9:00
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For your mistake see the answer of Arthur.

A bit more concise solution:

If the first person has been chosen then yet $2$ out of $9$ must be chosen.

In $3$ of these cases the three chosen persons will sit consecutively so the probability on that is:$$\frac3{\binom92}=\frac1{12}$$

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  • $\begingroup$ A nice middle ground between the OP's approach and the given official answer. $\endgroup$ – Arthur Feb 8 at 10:10

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