Consider an engineer who will visit fifteen manufacturing facilities in a single
ID: 3040138 • Letter: C
Question
Consider an engineer who will visit fifteen manufacturing facilities in a single week, and must select a visitation order. Nine facilities are production facilities, five are packaging facilities, and one is a storage facility (facilities of the same type are indistinguishable). (a) How many visitation orders are there? (b) Suppose the visitation order will be chosen randomly (i.e., the next facility to be visited is always chosen randomly among those that have not yet been visited). What is the probability that the storage facility is not the first nor the last facility visited? [HINT: There is more than one approach to compute this.]
Explanation / Answer
a) total numebr of visitation order from multinomial distribution =15!/(9!*5!*1!) =30030
b)number of visitation where storage facility is first =14!/(9!*5!) =2002 =number of visiration where storage facility is last
therefore visitation where storage facility is not the first nor the last facility visited=30030-2002-2002=26026
hence probability that the storage facility is not the first nor the last facility visited =26026/30030 =0.8667