Question

A person has a bunch of $n$ keys, only one of which can open a lock. The person tries the keys at random rejecting those which do not open the lock. The probability that the lock is opened at the kth $(\le n)$ trial is

• A. $\frac{1}{n}$
• B. $\frac{k}{n}$
• C. $\frac{k-1}{n}$
• D. ${\left(1-\frac{1}{n}\right)}^{k-1}\cdot \frac{1}{n}$

Answer 1

Answer: (A)

$P$ (lock is opened at kth trial)
$=P$ (lock is not opened at first $(k-1)$ trials) $\times P$ (lock is opened at kth trial)
$=\left(1-\frac{1}{n}\right)\left(1-\frac{1}{n-1}\right)\left(1-\frac{1}{n-2}\right)\dots \left(1-\frac{1}{n-(k-2)}\right)\left(\frac{1}{n-(k-1)}\right)$
$=\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n-1}\right)\left(\frac{n-3}{n-2}\right)\dots \left(\frac{n-k+1}{n-k+2}\right)\left(\frac{1}{n-k+1}\right)=\frac{1}{n}$