91Ó°ÊÓ

Given that\(n\)letters are placed at random in\(n\)envelopes.

Also,\({q_n}\)denote the probability that no letter is placed in the correct envelope.

Also, let \({q_{n - 1}}\) denote the probability that exactly one letter is placed.

Let \({A_i}\) be the event that the letter is placed in the correct envelope \(\left( {i = 1, \ldots ,n} \right)\). Then \({q_n}\)is given by \({q_n} = pr\left( {\bigcup\nolimits_{i = 1}^n {{A_i}} } \right)\) .

Here we use this theorem to solve is

\(pr\left( {\bigcup\nolimits_{i = 1}^n {{A_i}} } \right) = \sum\limits_{i = 1}^n {pr\left( {{A_i}} \right) - \sum\limits_{i < j} {pr\left( {{A_i} \cap {A_j}} \right) + \sum\limits_{i < j < k} {pr\left( {{A_i} \cap {A_j} \cap {A_k}} \right) + \ldots + {{\left( { - 1} \right)}^{n + 1}}pr\left( {{A_1} \cap {A_2}, \ldots , \cap {A_n}} \right)} } } \)

Since the letters are placed in the envelopes correctly at random, the probability\(pr\left( {{A_i}} \right)\)that any particular letter will be placed in the correct envelope is\(\frac{1}{n}\).

Then exactly one letter is placed in the correct envelope of probability is given by

\(\begin{aligned}{c}\sum\limits_{i = 1}^n {pr\left( {{A_i}} \right) = {q_{n - 1}}} \\{\rm{ = n \times }}\frac{{\rm{1}}}{{\rm{n}}}\\{\rm{ = 1}}\end{aligned}\)

Hence proving that exactly one letter is placed in the correct envelope is \({q_{n - 1}}\)