Kth trial locker

The locker opens in kth trial.Hence we should have (k-1) unsuccessful attempts and exactly at the kth attempt it should be successful. P(failing (k-1) times continuously)=(1- $\frac{1}{1000}$ )x(1- $\frac{1}{999}$ )x(1- $\frac{1}{998}$ )x(1- $\frac{1}{997}$ )x....(1- $\frac{1}{1000-(k-1)+1}$ )(As each time we use a particular combination, we discard it.) = $\frac{999}{1000}$ x $\frac{998}{999}$ x $\frac{997}{998}$ x $\frac{996}{997}$ ...... $\frac{1000-(k-1)}{1000-(k-1)+1}$ then successful at the next attempt, P(successful at kth time)= $\frac{1}{1000-(k-1)}$ Since both these events occur simultaneously, by multiplication theorem,
P(k)= $\frac{999}{1000}$ x $\frac{998}{999}$ x $\frac{997}{998}$ x $\frac{996}{997}$ ...... $\frac{1000-(k-1)}{1000-(k-1)+1}$ x $\frac{1}{1000-(k-1)}$ = $\frac{1}{1000}$

Using the numbers (0-9) and allowing repeatation of digits we can make 10×10×10=1000 3-digited codes . Among them only one code will open the locker . So the probability that the 'k'th code will open the locker is (1/1000).