Question

A locker can be opened by dialing a fixed three-digit code (between $000$ and $999$). A stranger, who does not know the code, tries to open the locker by dialing three digits at random. If $p$ is the probability that the stranger succeeds at the $kth$ trial, then the value of $1000p$ is

Answer 1

Let $A$ denotes the event that the stranger succeeds at $kth$ trial.
$p=\frac{999}{1000}\times \frac{998}{999}\cdots \frac{1000-k+1}{1000-k+2}\times \frac{1}{1000-k+1}$
$p=\frac{1}{1000}$
$\therefore 1000p=1$