# From a set of infinite positive integers a number is picked at random. What is the probability that the integer the person picked is greater than 100. [duplicate]

I tried this question and obtained the probability to be $$1$$

The total no. of outcomes are $$∞$$ and the no. of outcomes greater than $$100$$ is also $$∞$$ so the probability should be $$∞/∞$$ which turns out to be undefined

Another approach is to denote the probability of picking a positive integer greater than 100 as $$P(X > 100)$$. We can express this probability as a limit:

$$P(x>100) = \lim_{n \to \infty} \frac{n-100}{n}$$

As n approaches infinity, the number of positive integers greater than $$100$$ in the first $$n$$ positive integers will also approach infinity. Therefore, the limit of this ratio will be $$1$$.

$$P(X > 100) = 1$$

Hence the probability comes out to be $$1$$, but this is impossible. What are your thoughts on the question? (Sorry if I made any mistake as I am new to this site)

• You cannot pick a value uniformly from all the positive integers. You can do so from a finite number of integers, but there is no limiting distribution as the number of integers increases. Commented Nov 13, 2023 at 12:17
• Does this answer your question? The linked question isn't exactly the same, but the answer there answers here too. Given two randomly chosen natural numbers, what is the probability that the second is greater than the first? Commented Nov 13, 2023 at 12:26
• “a set of infinite positive integers” – so, the empty set? There are no infinite positive integers. Unless we are working in a nonstandard model… Commented Nov 13, 2023 at 21:23

Infinity is not a number, but is instead a symbol of unbounded growth.

To resolve the paradox, re-state the problem as letting $$~p(n)~$$ denote the probability that the number chosen is less than or equal to $$~100,~$$ assuming that there are $$~n~$$ positive integers to choose from.

Then, you have that $$~\displaystyle \lim_{n \to \infty} p(n) = 0,~$$ despite the fact that $$~p(n)~$$ is not equal to $$~0,~$$ for any finite value of $$~n.$$

Now, define $$~q(n)~$$ to be the probability of the complementary event, that among $$~n~$$ positive integers, the number chosen is $$~> 100.~$$

Then:

• $$~p(n) + q(n) = 1, ~$$ for all $$~n \in \Bbb{Z^+}.$$

• $$\displaystyle \lim_{n\to \infty} q(n) = 1,~$$ despite the fact that $$~q(n)~$$ is not equal to $$~1,~$$ for any finite value of $$~n.$$

Similarly, define the sequence $$~\langle a_n \rangle ~$$ by $$~a_n = \frac{1}{n}.~$$

Then $$~\displaystyle \lim_{n \to \infty} a_n = 0,~$$ despite the fact that $$~a_n \neq 0,~$$ for any finite value of $$~n.$$

You could then define the sequence $$~\langle b_n \rangle ~$$ by $$~b_n = 1 - \frac{1}{n}.~$$

Then $$~\displaystyle \lim_{n \to \infty} b_n = 1~$$ despite the fact that $$~b_n \neq 1,~$$ for any finite value of $$~n.$$

• The OP asks for the probability of the complement event. So it might be confusing having you say the probability tends to $0$. Commented Nov 13, 2023 at 12:32
• @SahajSatishSharma Good point. I edited my answer accordingly. Commented Nov 13, 2023 at 12:37