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 stranger succeeds at the kth, find 1000 p. (Assume that the stranger does not repeat unsuccessful combinations.)

Answer 1

Let A denote the event that the stranger succeeds at the kth trial. Then
$P\left(A\right)=\frac{999}{1000}\times \frac{998}{999}\times ....\times \frac{1000-k+1}{1000-k+2}\times \frac{1}{1000-k+1}$
$\Rightarrow p=\frac{1}{1000}\Rightarrow 1000p=1$.