1. In a college football training session, the defensive coordinator needs to ha
ID: 3073642 • Letter: 1
Question
1. In a college football training session, the defensive coordinator needs to have 23 play standing n a row. Among these 23 players, there are5 freshmansophomore juniors, and 3 seniors. How many different ways can they be arranged in a row if only their class level will be distinguished? (10 points) persn e 2. How many ways are there to select 7 candidates from 12 equally qualified recent graduates for openings in an accounting firm? (Order is not important) (10 points) 3. Orders for a computer are summarized by the optional features that are requested as follows: No Optional Features One Optional Feature More than one Optional Proportion of Orders 0.25 0.65 0.1 Feature (a) What is the probability that an order requests at least one optional feature? (10 points) (b) What is the probability that an order does not request more than one optional feature? (10 points) 11PageExplanation / Answer
(1) This is a permutation problem, according to the theorem,
The number of distinct permutation of n things of which n1 are of one kind, n2 are of second kind, and so on..... nk are of kth kind
is given by n! / (n1! * n2! * .... *nk!)
For this question, N = 23, n1 = 5, n2 = 9, n3 = 6, n4 = 3
Therefore total number of ways = 23! / (5! * 9! * 6! * 3!) = 137425207920
(2) The question says that order is not important, so this is combination question.
7 candidates can be selected from 12 candidates in 12C7 ways = 12! / (7! * (12-7)!)
= 12! / (7! * 5!) = 792
(3) Proportions of each order type based on optional feature is given, the probability will be same as proportions.
Therefore,
Probability that a order requested has "No Optional Feature" = 0.25
Probability that a order requested has "One Optional Feature" = 0.65
Probability that a order requested has "More than one Optional Feature" = 0.10
(a) Probability that an order requests atleast one optional feature.
This condition will be satisfied if either the order requests "One Optional Feature" or "More than one Optional Feature"
Hence, Probability that an order requests atleast one optional feature = Probability that a order requested has "One Optional Feature" + Probability that a order requested has "More than one Optional Feature"
= 0.65 + 0.10 = 0.75
(b) Probability that an order does not requests more than one optional feature. Meaning there should be maximum one optional feature
This condition will be satisfied if either the order requests "No Optional Feature" or "One Optional Feature"
Hence, Probability that an order does not requests more than one optional feature = Probability that a order requested has "No Optional Feature" + Probability that a order requested has "One Optional Feature"
= 0.25 + 0.65 = 0.90
(Alternatively, this part (b) can also be solved by simply subraction Probability that a order requested has "More than one Optional Feature" from the total probability. Total probability is always 1. Therefore answer would be 1 - 0.10 = 0.9)
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