# Probability proving the series of a pmf converges to a probability.

1. Sep 24, 2009

### ryanj123

1. The problem statement, all variables and given/known data

There are two separate series I'm having trouble with, although they're related.

The scenario: Roll a fair die until a six comes up.

pmf = (5/6)^(x-1) * (1/6)

So first, show the sum from 1 to infinity of p(x) =1

Next, determine P(X=1,3,5,7,...) that it will appear on an odd roll.

2. Relevant equations

I'm trying to prove 1=sum (5/6)^(x-1) * (1/6)

and sum (5/6)^(2x-1) * (1/6) = ?

3. The attempt at a solution

For the first, I tried induction but got lost along the way.

The second, the back of the book states = 6/11 but getting there sends me back to computing the first series.

2. Sep 24, 2009

### Dick

Your first series is a geometric series. What's the sum of 1+a+a^2+a^3+... for a<1? The second series will be a geometric series as well with a different ratio.

3. Sep 24, 2009

### ryanj123

I'm getting sum (1/5)*(5/6)^n

So,

1 = sum (1/5)*(5/6)^n
1 = (1/5) sum (5/6)^n
5 = sum (5/6)^n
5 = (5/6) + (5/6)^2 + (5/6)^3 + ... + (5/6)^n
5 = (5/6)[1 + (5/6) + (5/6)^2 +...+ (5/6)^(n-1)]

Now, I need to show this for n+1.

5 = (5/6)[1 + (5/6) + (5/6)^2 +...+ (5/6)^(n-1)]+(5/6)*(5/6)^(n-1+1)
5 = (5/6)[1 + (5/6) + (5/6)^2 +...+ (5/6)^(n-1)]+(5/6)*(5/6)^n
= 5 + (5/6)(5/6)^n
= 5 + (5/6)^(n+1)
so true for all n.

I'm not sure if that was correct? But I need a derivation instead of the geometric assumption.

4. Sep 24, 2009

### Dick

You don't need induction. Just sum the geometric series. It's not an assumption. It's a fact. Can you write something a little clearer next time?