Five men go to a restaurant together and each of them orders a dish that is different from the dishes ordered by the other members of the group. However the waiter serves the dishes randomly. Then the probability that exactly one of them gets the dish he ordered equals__________________ (round off to 2 decimal places)

Let us denotes the (1,,3,4,5) as the dishes ordered by the 5 persons A,B,C,D,E. Then, the probability that the only the first person gets the dish he ordered will be:

Thus the probability that exactly one of them gets the dish he ordered equals 5*0.05= 0.25

Suppose the event that the 5 persons get their ordered right is denoted by A_{1},A_{2},A_{3},A_{4},A_{5}. Then,

P(A_5 \cap A_1^C \cap A_2^C \cap A_3^C \cap A_4^C ) = P(A_5) *P(A_1^C \cap A_2^C \cap A_3^C \cap A_4^C / A_5)

Now,

P(A_5) = \frac{1}{5} \\ P(A_1^C \cap A_2^C \cap A_3^C \cap A_4^C / A_5) = P(\left(A_1 \cup A_2 \cup A_3 \cup A_4 \right) ^C / A_5) \\ = 1 - P(\left(A_1 \cup A_2 \cup A_3 \cup A_4 \right) / A_5) \\ = 1- \sum_{i=1}^4P(A_i / A_5)+\sum_{i\neq j=1}^4P(A_i \cap A_j / A_5) \\-\sum_{j\neq i \neq k=1}^4P(A_i \cap A_j\cap A_k / A_5) +\sum_{i \neq j \neq k \neq k =1}^4P(A_i \cap A_j\cap A_k \cap A_l / A_5) \\ = 1 - \binom 4 1 \frac{1}{4} + \binom 4 2 \frac{1}{4.3} - \binom 4 3 \frac{1}{4.3.2} + \frac{1}{4.3.2.1} \\ = 1- \frac{1}{1} + \frac{6}{4.3} - \frac{4}{4.3.2} + \frac{1}{4.3.2.1} = \frac{12-4+1}{24} = \frac{3}{8} = 0.375

Now probability that exactly one of the persons gets the dish he ordered equals,

5* P(A_5 \cap A_1^C \cap A_2^C \cap A_3^C \cap A_4^C ) \\ = 5*P(A_5) *P(A_1^C \cap A_2^C \cap A_3^C \cap A_4^C / A_5) \\ = 5. \frac{1}{5} . 0.375 =0.375

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