Setting $P\left(\{k\}\right)=\frac{1}{2^k}$ when picking number from the naturals

Question

We randomly peak a number from the natural numbers $\mathbb{N}$. What is the probability to get $13$? My prof explained that if for every $k$, the probability is $0$ then we get: $$ 1=P\left(\mathbb{N}\right)=P\left(\bigcup_{k=1}^{\infty} \{k\}\right)=\sum_{k=1}^{\infty} P\left(\{k\}\right)=0 $$ Then it's impossible to have $p=0$. Also it's impossible to have $p>0$ because then you get $1=P\left(\mathbb{N}\right)=\infty$.

I understand this but then he said that for $P\left(\{k\}\right)=\frac{1}{2^k}$ (sequence so the sum is 1) it works. I don't understand this part. What should work? Why can we choose it to be $P\left(\{k\}\right)=\frac{1}{2^k}$? I also looked at Probability of picking a random natural number but didn't see something that explain this part.

Answer 1

In the latter case , we have countable infinite many events from which exactly one occurs.

In this case, the probability that any event occurs is the sum of the probabilities, so we need a sequence of countable infinite many entries summing up to $1$.

The given sequence does the job, hence is a suitable probability distribution.