Daniela M.

asked • 02/04/14

When selecting two people, find the probability that one works in an office and the other one drives a car?

I have a probability question from my intro to stats class. I can't figure out if it's dependent or independent. Overall, I really don't know how to figure it out. There are 2000 people in total from a given chart; but the most important are the office workers 1050 people, and the car drivers 1150 people. I hope that's enough information. Please help!

Kenneth G.

Note that the probability will be different depending on the number of people who are both office workers and also drive.  Given your description of the problem, the number of people who can have both characteristics is between 200 and 1050.  The more people who have both characteristics, the more people who don't have either.   So the probability of picking the two people with one a driver and the other an office worker will be lower in this case since more people will have neither of the characteristics.
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02/05/14

Kenneth G.

Correction:  should say "between 100 and 1050 people both drive and work in the office" above.
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02/05/14

Kenneth G.

Another Correction:  I changed my mind -   200 is the correct minimum, not 100.  The minimum intersection is 200.  Since 1150 + 1050 = 2200,  200 objects need to be in common.  
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02/06/14

Kenneth G.

Correction:  Changed my mind again.  The correct minimum intersection is 200 items since 1150 + 1050 = 2200 but there are only 2000 items.  If there were only 100 in the intersection of these sets then 1050 + 950 + 100 would have to equal 2000, which is not true.
 
I will be revising my solution again shortly.
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02/06/14

Kenneth G.

Correction:  I've changed my mind.  200 is the minimum intersection of these two sets.  Because 1150 + 1050 = 2200, if the intersection were only 100 then 1050 + 100 + 950 would have to equal 2000, which is not true.
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02/06/14

Kenneth G.

I'm happy with my answer now.
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02/06/14

3 Answers By Expert Tutors

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Kenneth G. answered • 02/05/14

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Experienced Tutor of Mathematics and Statistics

Alex S. answered • 02/05/14

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Retired, Looking to Share My Love of Mathematics and Computers

Kenneth G.

Thanks; you're right the minimum is 100 people who both drive and work in the office, not 200 as I said.  I will correct that in a minute.
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02/05/14

Kenneth G.

Alex,
 
I'm glad to see someone else took on this question also.
 
I'm not sure I agree with you that "works in the office" is independent of "is a driver" because office workers may get paid less and not be able to afford a car for example, or because drivers can get better jobs elsewhere.  However, I also don't see how whether these are dependent or independent events impacts the problem at all. The problem is about the selection of two items from 2000. 
 
By the way, you didn't compute a probability for your "put the numbers in context" example. I got 60.2% for the case of 100 people who are both drivers and office workers.   Do you agree with my computation? Let me know if you find any problem.
 
Thanks.
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02/05/14

Alex S.

Ken,
I think I actually agree with you about the "indepedent" and, likely, about your calculations.  The sum of the probabilities must equal 1;  mine do not ... I did not account for the probability of the 100 who both drive and work in office ... I think that is expressed a P(A|B).  My forte is NOT statistics, forgot how to address this component, so my answer is, in fact, wrong.
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02/05/14

Kenneth G.

Alex,
 
Thanks for posting.  You definitely helped me correct the error in the minimum intersection of the two sets.  I expected more posts on this.  I guess the problem must have scared off a lot of people.
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02/05/14

Alex S.

On further thought, I now think the question (one drives, one works in office) is ambiguous, is it: one drives, but does not work in office; and, the other  works in office, but does not drive?  Or, is it just simply, independent probability of finding one person drives (whether or not person works in office) and another person works in office (whether or not person drives).  Since this is "Intro to Statistics," I will bet on the later, simple, interpretation.  Hence, think my original response, 1150/2000 (57.5%) and 1050/2000 (52.5%) are correct ... BUT I FREELY ADMIT THAT STATISTICS IS NOT ONE OF MY AREAS OF EXPERTISE.
 
Please take my response "with a grain of salt," as intended.  I will now bow out!
 
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02/05/14

Kenneth G.

Thanks for posting. I posted my first solution to this problem several days ago, but it keeps disappearing for days because of WyzAnt's review process - even if I change one word it disappears.

Some questions for you.

1. It seems to me the problem doesn't say that the car drivers and the office workers are the only people. It says that there are 2000 people and the most important ones are car drivers and office workers. For example, it seems to me that all the 1050 office workers could actually be car drivers and there could be 850 people of the 200 who do not work in office or drive a car. How did you conclude from the problem statement that this could not be true?

2. When you're figuring probabilities where 200 people both drive cars and work in office, it seems to me that then 1150 - 200 = 950 people only drive cars, and 1050 - 200 = 850 people only work in office. In this case, there are no people who do not drive cars and also do not work in office. In your solution you say, for example, "The probabilities for W and D are (1050/2000)(1150/1999) = .301875." I assume this is a probability for people who only drive and only work in office but do not do both. But there not 1050 people who only work in office and there are not 1150 people who only drive. Can you give your reasoning behind that computation?

3. Your probabilities add up to a probability of .42188 to choose a worker and a driver. At the end you computed the probabilities for the cases "W and W" or "D and D" which make up the remainder of the probability space and got a total of .6059 . Those probabilities, when added to the total you computed for the other cases, .42188, should add to a probability of 1.0 . However .6059 + .42188 = 1.02778 which is more than 100%. Can you explain the extra 2.7%?

Best wishes!
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02/06/14

Kenneth G.

You asked - "If all 1050 workers were also car drivers, wouldn't that lead to multiple answers ?"  
 
Yes, that's exactly my point.  the problem isn't well-defined.    
 
If you read my solution to this problem, I gave exactly 852 different answers to the problem - one answer assuming the driver/office attributes were assigned randomly so that the intersection was random and you didn't need to know what it was, and one for each of the possible fixed intersections N, where N is is between 200 and 1050 inclusive.
 
My answer for N = 200 would correspond to the case where the intersection was exactly 200.  
 
 
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02/07/14

Alex S.

With the proviso, again, that my statistics are "rusty" ... I truly think everyone I'd over complicating the problem and "making a mountain out of molehill"!  This is, after all, a "introduction to statistics" question.  I'm of the mind that this context can not, and does not, entertain unions, intersections, etc (A and B, A not B, B not A, A or B, et al).
 
so, accepting that there are 850 working in office not driving, 950 driving but not working in office, and 200 working in office and driving ...  Representing, in full the  2000 people.  My thinking then is:
 
-  picking each of the two people is an independent event
-  the question simply asks the chance of at least one being a driver and at least one being an office worker (this is probably not accurate or precise, but can't think of how to phrase it ... AND, I DO NOT THINK IT REALLY MATTERS)
- therefore, I think the solution is quite simple ... Calculate total drivers (1150) and calculate total office workers (1050) ... AS GIVEN ... and calculate probabilities as 1150/2000 and 1050/2000
-  the two probabilities add up to more than 1, or 100% ... but this is OK, because the two events of picking are independent.
 
it all comes down in how the question of "one is A, and the other is B" is to be interpreted ... that is the ambiguity!  Is it that the first must be A not B, and the second B not A?  Would an "introduction" question really get into this complexity?
 
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02/07/14

Kenneth G.

Ok.  Thanks for responding. 
 
You asked:  " Would an "introduction" question really get into this complexity?". 
 
First note that the category of this question is not "Introduction to Statistics".  
 
Second I think one of the things that leads students to failure in mathematics is not learning how to carefully examine assumptions in each problem.  In addition, my philosophy is to answer the question which is asked, not the one I guess the student intended to ask. 
 
 
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02/07/14

Arthur D.

tutor
I'm using a different approach. Probability=(# of favorable outcomes)/(total # of equally likely outcomes).
I'm leaving out a lot of computation because I just spent one hour posting all the computations and it disappeared when I hit "save comment".
I'm assuming 850 workers only, 950 drivers only, and 200 people who both work and drive.
Favorable outcomes: each worker can be paired with a driver so we have 850*950=807,500.
The rest of the outcomes are from WW, DD, WD WD, W WD, and D WD.
For WW we have 360,825 using n!/r!(n-r)!
For DD we have 450,775. For WD WD we have 19,900. For W WD we have 850(200)=170,000.
For D WD we have 950(200)=190,000.
Adding all possible outcomes we get 1,999,000
Probability equals 807,500/1,999,000=0.40395=40.395%
 
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02/08/14

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