Compute probabilities of balls not being drawn from urns Independent events $A_1$ through $A_n$, where $A_n$ is event that nth ball is not drawn in a random draw from urn full of $n^2$ balls labeled $1, 2, \dots, n^2$

What is $\lim_{n \to \infty} P(A_n)$?

What is the $P(\bigcap_{n=2}^{\infty} A_n)$?
What does continuity have to do with this?

Anyway


$P(A_n) = \frac{n^2-1}{n^2}$.


Thus $\lim_{n \to \infty} P(A_n) = \lim_{n \to \infty} \frac{n^2-1}{n^2}$


$P(\bigcap_{n=1}^{\infty} A_n) = \prod_{n=1}^{\infty} P(A_n)$ by independence


$\to \prod_{n=1}^{\infty} P(A_n) = P(A_1)P(A_2)P(A_3) \dots = 0P(A_2)P(A_3) \dots = 0$

$P(\bigcap_{n=2}^{\infty} A_n) = \prod_{n=2}^{\infty} P(A_n) = \prod_{n=2}^{\infty} \frac{n^2-1}{n^2}$

Can you take it from here?
What does continuity have to do with this?

Anyway


$P(A_n) = \frac{n^2-1}{n^2}$.


Thus $\lim_{n \to \infty} P(A_n) = \lim_{n \to \infty} \frac{n^2-1}{n^2}$


$P(\bigcap_{n=1}^{\infty} A_n) = \prod_{n=1}^{\infty} P(A_n)$ by independence


$\to \prod_{n=1}^{\infty} P(A_n) = P(A_1)P(A_2)P(A_3) \dots = 0P(A_2)P(A_3) \dots = 0$

$P(\bigcap_{n=2}^{\infty} A_n) = \prod_{n=2}^{\infty} P(A_n) = \prod_{n=2}^{\infty} \frac{n^2-1}{n^2}$

Can you take it from here?