91Ó°ÊÓ

4 guests check their hats when they arrive at a restaurant, and when they leave hats are returned to them in a random order.

The probability of event E, that no guest will receive proper hat is,

\(\begin{aligned}{}P\left( E \right) &= 1 - P\left( {{\rm{at}}\;{\rm{least}}\;{\rm{one}}\;{\rm{guest}}\;{\rm{would}}\;{\rm{recieve}}\;{\rm{proper}}\;{\rm{hat}}} \right)\\ &= 1 - P\left( {{E^c}} \right)\\ &= 1 - P\left( {\bigcup\nolimits_{i = 1}^4 {{A_i}} } \right)\end{aligned}\)

Where, \({A_i}\)is the event that i-th person receives proper hat.

Expanding the expression in the above equation,

\(P\left( {\bigcup\nolimits_{i = 1}^4 {{A_i}} } \right) = \sum\limits_{i = 1}^4 {P\left( {{A_i}} \right)} - \sum\limits_{i < j}^4 {P\left( {{A_i} \cap {A_j}} \right)} + \sum\limits_{i < j < k}^4 {P\left( {{A_i} \cap {A_j} \cap {A_k}} \right)} - \sum\limits_{i < j < k < l}^{} {P\left( {{A_i} \cap {A_j} \cap {A_k} \cap {A_l}} \right)} \)

Probability that one person received proper hat is,

\(\begin{array}{}P\left( {{A_i}} \right) = \frac{1}{4}\\ \Rightarrow \sum\limits_i {P\left( {{A_i}} \right)} = 4\left( {\frac{1}{4}} \right)\\ = 1\end{array}\)

Similarly,

Any two people received proper hat is,

\(\begin{aligned}{}P\left( {{A_i} \cap {A_j}} \right) &= \frac{1}{4} \times \frac{1}{3}\\\sum\limits_{i < j} {P\left( {{A_i} \cap {A_j}} \right)} { = ^4}{C_2}\left( {\frac{1}{4} \times \frac{1}{3}} \right)\\ &= \frac{{4!}}{{2!2!}}\left( {\frac{1}{4} \times \frac{1}{3}} \right)\\ &= \frac{1}{{2!}}\end{aligned}\)

Similarly,

\(\begin{array}{c}\sum\limits_{i < j < k} {P\left( {{A_i} \cap {A_j} \cap {A_k}} \right) = \frac{1}{{3!}}} \\\sum\limits_{i < j < k < l} {P\left( {{A_i} \cap {A_j} \cap {A_k} \cap {A_l}} \right) = \frac{1}{{4!}}} \end{array}\)

Therefore, substituting the value in the above equation,

\(\begin{aligned}{}P\left( E \right) &= 1 - \left( {1 - \frac{1}{{2!}} + \frac{1}{{3!}} - \frac{1}{{4!}}} \right)\\ &= \frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}}\\ &= \frac{1}{2} - \frac{1}{6} + \frac{1}{{24}}\\ &= \frac{9}{{24}}\\ &= \frac{3}{8}\end{aligned}\)

The probability that none of the guest received proper hat is \(\frac{3}{8}\)