# Probability number chosen at random from S will be odd

The question is:

Let $S = \{1,2,3,\ldots\}$. Let $P$ be a probability measure defined by $P(\{n\}) = 2(\frac{1}{3})^n$ for all $n$ in $S$. What is the probability that a number chosen at random from $S$ will be odd?

The way that I solved this question is by taking an infinite sum $\sum_{n=1}^\infty 2(\frac{1}{3})^n$. Because the sum starts at $1$ instead of $0$, when using the $1/(1-r)$ formula, I subtracted out $1$ (the first term in the sequence). So I got $1/(2/3) - 1$, which is $1/2$.

It would make sense to me that the probability of selecting an odd number from an infinite set is about half. But I chose to compute the sum sort of arbitrarily (mainly because I realized I could do a sum here). So I guess I'm confused as to whether my answer is correct, and why would taking the infinite sum of the probability measure yield the probability of taking odd numbers only, not both odd and even numbers.

## 1 Answer

I think the correct sum is $$P(1)+P(3)+P(5)+\cdots=\sum_{k=0}^\infty 2\left(\frac{1}{3}\right)^{2k+1}=\frac{2}{3} \sum_{k=0}^\infty\left(\frac{1}{9}\right)^k=\frac{3}{4}~~,$$ because in that way you sum in the odd numbers only.

It is not useful to think you are selecting a number "at random" from $S$ because you have the probabilities explicitly: they are telling you that the way numbers are selected at random from $S$ is such that you get a "1" in two thirds of the time (on average) and a "2" in two out of nine cases, etc...

• Thanks, that makes sense! I figured there was a way to isolate the odd numbers but I couldn't figure out how to do it. – Sveinn May 1 '16 at 20:01