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Relative frequency assesment method

Posted on 2016-07-18
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Can you tell me if my calculations are right?  Thanks.

A hotel conducted a survey of its guests.​ Sixty-two surveys were completed. Based on the data from the​ survey, shown in the accompanying​ table, determine the following probabilities using the relative frequency  assessment method.      

Relative frequency assessment method---> P(E1) = Number of times E1 occurs / N


                                           Any Problems                  
      Type of Trip Problems No problems
                  
Business            24      9      15
Pleasure            23      3      20
Combination      15      1      14
Grand Total      62      13      49

                                        Alternative of Staffs                  
            No Res-      Pass      Fail
            Ponse            
Business            1      20      3
Pleasure            1      20      2
Combination      0      10      5
Grand Total      2      50      10

                                 Phone Number                  
            No Res-      In-State      Out-State
            Ponse            
Business            3      3      18
Pleasure            9      2      12
Combination      1      1      13
Grand Total      13      6      43


a. Of two customers​ selected, what is the probability that both will be on a business​ trip?
A. 24+9+15+1+20+3+3+3+18/62+13+49+2+50+10+13+6+43

b. What is the probability that a customer will be on a business trip or will experience a hotel problem
during a stay at the​ hotel?
B. 24+9/62+13

c. What is the probability that a customer is strictly on business and has an​ in-state area code phone​ number?
C. 3/6
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Question by:mustish1
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sdstuber earned 2000 total points
ID: 41718063
no, your answers are not correct.

to find the probability of two events both occurring you multiply their individual probabilities.
to find the probability of either of two events occuring you add their individual probabilities.

both of the above statements assume the events are mutually exclusive.

I have no idea what you were attempting with your a, b, or c answers.  

Also, for A,  do your combination people count as business, or is A asking only about strictly business.

If it's strictly business then your probability of one person is the number of business people out of the total number of people.

If combination counts then the probability of one persion is the number business and combination people out of the total number of people.

So, what would those be?

Then, the odds of the second person would be one less than what the first was (because the first person "uses up" one event) but out of the same total.

Then you multiply those two together.
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Author Closing Comment

by:mustish1
ID: 41718073
Thanks.
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