Question

A key bunch contain $m$ keys exactly one of which opens the door. A man tries to open the door by choosing a key at random discarding the wrong key after each trial. The probability that the door opens at the $r$th trial $(1\le r\le m)$, is

• A. $1/m$
• B. $1/r$
• C. $r/m$
• D. $(r-1)/m$

Answer 1

Answer: (A)

$1/m$

Probability that it will open in the first try = $\frac{1}{m}$

Probability that it will open in the second try = $\frac{m-1}{m}\times \frac{1}{m-1}=\frac{1}{m}$Probability that it will open in the third try = $\frac{m-1}{m}\times \frac{m-2}{m-1}\times \frac{1}{m-2}=\frac{1}{m}$

Thus, the probability of opening is always $\frac{1}{m}$ irrespective of the number of trials.