Question 2.30: It happens that 4 hotels in a certain large city have the sa......
It happens that 4 hotels in a certain large city have the same name, e.g., Grand Hotel. Four persons make an appointment to meet at the Grand Hotel. If each one of the 4 persons chooses the hotel at random, calculate the following probabilities:
(i) All 4 choose the same hotel.
(ii) All 4 choose different hotels.
(i) If ~A=~“all 4 choose the same hotel,” then P(A)\,=\,\frac{n(A)}{n( S)}, where n(A) is the number of sample points in A. Here, n(S)\,=\,4\times4\times4\times4\,=\,4^{4}, by Theorem 7 applied with k\,=\,4 and n_{1}\,=\,n_{2}\,=\,n_{3}\,=\,n_{4}\,=\,4, and n(A)=4, by Theorem 7 again applied with ~k=1~ (the 4 people looked upon as a single unity) and n_{1}=4 (the 4 hotels they can choose). Thus, P(A)={\frac{4}{4^{4}}}={\frac{1}{4^{3}}}={\frac{1}{64}}=0.015625\simeq0.016.
THEOREM 7
(Fundamental Principle of Counting) Suppose a task is completed in k stages by carrying out a number of subtasks in each one of the k stages. If the numbers of these subtasks are n_{1}\ldots n_{ k} for the k stages, respectively, then the total number of different ways the overall task is completed is: n_{1}\times\cdots \times n_{ k}
(ii) If {{B}}=~ “all 4 choose different hotels,” then, by the first part of the corollary to Theorem 7, n(B)\,=\,P_{4,4}\,=\,4!, so that P(B)\,=\,{\textstyle{\frac{\ 4!}{4^{4}}}}\,=\,{\textstyle{\frac{1\times2\times3}{4^{3}}}}\,=\,{\textstyle{\frac{3}{32}}}\,=0.09375\simeq0.094\,.