David W. answered • 09/27/16
Tutor
4.7
(90)
Experienced Prof
A salesperson plans to visit Annapolis, Bloomington, Carmel, Detroit, and Elksville on his business trip. Since Annapolis is not in the Midwest, he wants to visit that city last. If his boss randomly selects the order of all five cities to be visited, what is the probability that Annapolis is last?
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This problem offers a great lesson in understanding [1] word (story) problems, [2] combinations and probabilities, and [3] theoretical probability.
First, a key to solving word (story) problems is to read and re-read the problem until you understand it and can put it into your own words. This helps to identify TMI (Too Much Information). For this problem, it does not matter what the salesperson wants. Also, note the nice first letters of the city names. Here is a re-wording:
“Five numbers (1,2,3,4,5) are randomly ordered without replacement. What is the probability that 1 is first?” [note: any specific position will do]
Second, without replacement, there are 5! [that is, 5*4*3*2*1 = 120] ways to order the numbers.
Third, the Theoretical Probability of 1 being first (or last) is defined as P=(number of successes)/(number of possibilities). So, P=S/120 where S is the number of times that “1” is first (or, you could say last). You could enumerate them (and for beginner problems with only a couple variables, this is very instructive):
1 2 3 4 5 *
1 2 3 5 4 *
1 2 4 3 5 *
1 2 4 5 3 *
1 2 5 3 4 *
1 2 5 4 3 *
1 3 2 4 5 *
1 3 2 5 4 *
1 3 4 2 5 *
1 3 4 5 2 *
1 3 5 2 4 *
1 3 5 4 2 *
1 4 2 3 5 *
1 4 2 5 3 *
1 4 3 2 5 *
1 4 3 5 2 *
1 4 5 2 3 *
1 4 5 3 2 *
1 5 2 3 4 *
1 5 2 4 3 *
1 5 3 2 4 *
1 5 3 4 2 *
1 5 4 2 3 *
1 5 4 3 2 *
2 1 3 4 5
2 1 3 5 4
2 1 4 3 5
2 1 4 5 3
2 1 5 3 4
2 1 5 4 3
2 3 1 4 5
2 3 1 5 4
2 3 4 1 5
2 3 4 5 1
2 3 5 1 4
2 3 5 4 1
2 4 1 3 5
2 4 1 5 3
2 4 3 1 5
2 4 3 5 1
2 4 5 1 3
2 4 5 3 1
2 5 1 3 4
2 5 1 4 3
2 5 3 1 4
2 5 3 4 1
2 5 4 1 3
2 5 4 3 1
3 1 2 4 5
3 1 2 5 4
3 1 4 2 5
3 1 4 5 2
3 1 5 2 4
3 1 5 4 2
3 2 1 4 5
3 2 1 5 4
3 2 4 1 5
3 2 4 5 1
3 2 5 1 4
3 2 5 4 1
3 4 1 2 5
3 4 1 5 2
3 4 2 1 5
3 4 2 5 1
3 4 5 1 2
3 4 5 2 1
3 5 1 2 4
3 5 1 4 2
3 5 2 1 4
3 5 2 4 1
3 5 4 1 2
3 5 4 2 1
4 1 2 3 5
4 1 2 5 3
4 1 3 2 5
4 1 3 5 2
4 1 5 2 3
4 1 5 3 2
4 2 1 3 5
4 2 1 5 3
4 2 3 1 5
4 2 3 5 1
4 2 5 1 3
4 2 5 3 1
4 3 1 2 5
4 3 1 5 2
4 3 2 1 5
4 3 2 5 1
4 3 5 1 2
4 3 5 2 1
4 5 1 2 3
4 5 1 3 2
4 5 2 1 3
4 5 2 3 1
4 5 3 1 2
4 5 3 2 1
5 1 2 3 4
5 1 2 4 3
5 1 3 2 4
5 1 3 4 2
5 1 4 2 3
5 1 4 3 2
5 2 1 3 4
5 2 1 4 3
5 2 3 1 4
5 2 3 4 1
5 2 4 1 3
5 2 4 3 1
5 3 1 2 4
5 3 1 4 2
5 3 2 1 4
5 3 2 4 1
5 3 4 1 2
5 3 4 2 1
5 4 1 2 3
5 4 1 3 2
5 4 2 1 3
5 4 2 3 1
5 4 3 1 2
5 4 3 2 1
Or, you could select a number (or letter or city) and then ask, “How many ways can four numbers/letters/cities be arranged after (or before) it?” Well, it is 4*3*2*1 = 24.
Now, we have P=24/120.
This assumes that the probability of any number/letter/city being in each of the 5 positions is equally-likely (for some events, we say “fair”).
Since probability ranges from 0 to 1, reduce P to 1/5 or 0.2 (Oh, we knew that any of the 5 cities were equally likely to be last/first, didn’t we?)
---------------------------
This problem offers a great lesson in understanding [1] word (story) problems, [2] combinations and probabilities, and [3] theoretical probability.
First, a key to solving word (story) problems is to read and re-read the problem until you understand it and can put it into your own words. This helps to identify TMI (Too Much Information). For this problem, it does not matter what the salesperson wants. Also, note the nice first letters of the city names. Here is a re-wording:
“Five numbers (1,2,3,4,5) are randomly ordered without replacement. What is the probability that 1 is first?” [note: any specific position will do]
Second, without replacement, there are 5! [that is, 5*4*3*2*1 = 120] ways to order the numbers.
Third, the Theoretical Probability of 1 being first (or last) is defined as P=(number of successes)/(number of possibilities). So, P=S/120 where S is the number of times that “1” is first (or, you could say last). You could enumerate them (and for beginner problems with only a couple variables, this is very instructive):
1 2 3 4 5 *
1 2 3 5 4 *
1 2 4 3 5 *
1 2 4 5 3 *
1 2 5 3 4 *
1 2 5 4 3 *
1 3 2 4 5 *
1 3 2 5 4 *
1 3 4 2 5 *
1 3 4 5 2 *
1 3 5 2 4 *
1 3 5 4 2 *
1 4 2 3 5 *
1 4 2 5 3 *
1 4 3 2 5 *
1 4 3 5 2 *
1 4 5 2 3 *
1 4 5 3 2 *
1 5 2 3 4 *
1 5 2 4 3 *
1 5 3 2 4 *
1 5 3 4 2 *
1 5 4 2 3 *
1 5 4 3 2 *
2 1 3 4 5
2 1 3 5 4
2 1 4 3 5
2 1 4 5 3
2 1 5 3 4
2 1 5 4 3
2 3 1 4 5
2 3 1 5 4
2 3 4 1 5
2 3 4 5 1
2 3 5 1 4
2 3 5 4 1
2 4 1 3 5
2 4 1 5 3
2 4 3 1 5
2 4 3 5 1
2 4 5 1 3
2 4 5 3 1
2 5 1 3 4
2 5 1 4 3
2 5 3 1 4
2 5 3 4 1
2 5 4 1 3
2 5 4 3 1
3 1 2 4 5
3 1 2 5 4
3 1 4 2 5
3 1 4 5 2
3 1 5 2 4
3 1 5 4 2
3 2 1 4 5
3 2 1 5 4
3 2 4 1 5
3 2 4 5 1
3 2 5 1 4
3 2 5 4 1
3 4 1 2 5
3 4 1 5 2
3 4 2 1 5
3 4 2 5 1
3 4 5 1 2
3 4 5 2 1
3 5 1 2 4
3 5 1 4 2
3 5 2 1 4
3 5 2 4 1
3 5 4 1 2
3 5 4 2 1
4 1 2 3 5
4 1 2 5 3
4 1 3 2 5
4 1 3 5 2
4 1 5 2 3
4 1 5 3 2
4 2 1 3 5
4 2 1 5 3
4 2 3 1 5
4 2 3 5 1
4 2 5 1 3
4 2 5 3 1
4 3 1 2 5
4 3 1 5 2
4 3 2 1 5
4 3 2 5 1
4 3 5 1 2
4 3 5 2 1
4 5 1 2 3
4 5 1 3 2
4 5 2 1 3
4 5 2 3 1
4 5 3 1 2
4 5 3 2 1
5 1 2 3 4
5 1 2 4 3
5 1 3 2 4
5 1 3 4 2
5 1 4 2 3
5 1 4 3 2
5 2 1 3 4
5 2 1 4 3
5 2 3 1 4
5 2 3 4 1
5 2 4 1 3
5 2 4 3 1
5 3 1 2 4
5 3 1 4 2
5 3 2 1 4
5 3 2 4 1
5 3 4 1 2
5 3 4 2 1
5 4 1 2 3
5 4 1 3 2
5 4 2 1 3
5 4 2 3 1
5 4 3 1 2
5 4 3 2 1
Or, you could select a number (or letter or city) and then ask, “How many ways can four numbers/letters/cities be arranged after (or before) it?” Well, it is 4*3*2*1 = 24.
Now, we have P=24/120.
This assumes that the probability of any number/letter/city being in each of the 5 positions is equally-likely (for some events, we say “fair”).
Since probability ranges from 0 to 1, reduce P to 1/5 or 0.2 (Oh, we knew that any of the 5 cities were equally likely to be last/first, didn’t we?)
