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 $(\leq n)$ 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)
$=\left(1-\frac{1}{ n }\right)\left(1-\frac{1}{ n -1}\right)\left(1-\frac{1}{ n -2}\right) \ldots\left(1-\frac{1}{ n -( k -2)}\right)\left(\frac{1}{ n -( k -1)}\right)$
$=\left(\frac{ n -1}{ n }\right)\left(\frac{ n -2}{ n -1}\right)\left(\frac{ n -3}{ n -2}\right) \ldots\left(\frac{ n - k +1}{ n - k +2}\right)\left(\frac{1}{ n - k +1}\right)=\frac{1}{ n }$