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 $k t h$ trial, then the value of $1000 p$ is

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Solution

Let $A$ denotes the event that the stranger succeeds at $k t h$ trial.
$p = \frac{999}{1000} \times \frac{998}{999} \dots \frac{1000 - k + 1}{1000 - k + 2} \times \frac{1}{1000 - k + 1}$
$p = \frac{1}{1000}$
$∴ 1000 p = 1$

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