- #1

lizarton

- 14

- 0

## Homework Statement

Consider a dial having a pointer that is equally likely to point to each of [itex]n[/itex] region numbered [itex]1,2,...,n.[/itex] When we spin the dial three times, what is the probability that the sum of the selected numbers is [itex]n[/itex]?

## Homework Equations

A Theorem I believe is relevant:

With repetition allowed, there are [itex]\left(\stackrel{n+k-1}{k - 1}\right)[/itex] ways to select [itex]n[/itex] objects from [itex]k[/itex] types. This also equals the number of nonnegative integer solutions to [itex]x_{1} + \ldots + x_{k} = n.[/itex]

My problem is identifying [itex]n[/itex] and [itex]k[/itex] in these problems. Any help would be greatly appreciated!

## The Attempt at a Solution

I have to use summations, and I'm sure binomial coefficients. I believe that this is a selection; it seems to imitate rolling an [itex]n[/itex]-sided die three times, but I even have trouble computing that problem.

The total number of outcomes is [itex]n^3[/itex] (I think)

I started counting ordered triples of some n terms...

[itex]n=3:[/itex] There is only one way, [itex]{(1,1,1)}[/itex]

[itex]n=4:[/itex] There are 3 ways, [itex]{(1,2,1),(1,1,2),(2,1,1)}[/itex]

[itex]n=5:[/itex] There are 6 ways, [itex]{(1,1,3),(1,2,2),(1,3,1),(2,1,2,),(2,2,1),(3,1,1)}[/itex]

[itex]n=6:[/itex] There are 10 ways...

[itex]n=7:[/itex] 15 ways...

For arbitrary n, you can start making ordered triples...

[itex](1,1,n-2)

(1,2,n-3)

(1,3,n-4)

\ldots

(1,n-2,1)

(1,n-3,2)

\ldots

[/itex]