What is the probability of a customer leaving a workshop happy?

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Discussion Overview

The discussion revolves around calculating the probability of a customer leaving a workshop happy based on conditional probabilities related to the timeliness and quality of repairs performed by a mechanic. Participants explore the concepts of conditional probability, particularly in the context of a specific scenario involving customer satisfaction.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about conditional probability and seeks guidance on solving the problem.
  • Another participant suggests defining the events involved and writing down the known probabilities, specifically P(A|B) and P(B).
  • A participant initially considers the use of Bayes' rule but later questions its applicability to the problem.
  • Clarifications are made regarding the meaning of conditional probabilities and the distinction between intersection and union of events.
  • One participant calculates that if 77% of jobs are done on time and 85% of those are satisfactory, approximately 65.45 customers out of 100 will leave happy, leading to a probability of 0.6545.
  • A later participant expresses satisfaction with the problem, indicating it seems straightforward after the discussion.

Areas of Agreement / Disagreement

Participants generally agree on the approach to solving the problem using conditional probabilities, but there are some misunderstandings about terminology and concepts, particularly regarding the definitions of intersection and union in probability.

Contextual Notes

Some participants express uncertainty about the definitions and applications of conditional probability, and there are unresolved questions about the correct interpretation of certain terms used in the discussion.

Who May Find This Useful

Individuals interested in learning about conditional probability, particularly in practical applications related to customer satisfaction and service quality in workshops.

morry
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Ok guys, I don't really understand conditional probability, can you guys tell me how to go about solving this?

To please customers, repairs need to be done satisfactorily and completed on time. For one mechanic, if the job is done on time, he has a 85% chance that it was also done satisfactorily.
He completes 77% of his jobs on time.

Whats the probability that a customer leaves the workshop happy?

Cheers guys.
 
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Probability questions can often be confusing (and deceptively easy).
Best thing to do is write out the given and the unknown.

Let's say the event A is the event that the repair was done satisfactorily and B is the event that it was done on time. You are given P(A|B) and P(B). The unknown is P(A and B). Writing this down is the starting point for any problem.
 
Thanks for the start Galileo!

I was initially thinking that this would be a problem involving Bayes rule. I now reckon that it isnt.

I have no formulas that include both the union and A/B. Whats the next step?
 
P(A|B) is the probability of A conditioned on B, i.e. the probability A occurs given that B has occurred. It's simply the way conditional probabilities are written. It has nothing to do with setminuses.
 
Galileo said:
It has nothing to do with setminuses.

Umm, what are setminuses?

Ill have another go and see if I can get it out.
 
I thought that's what you meant by A/B (Those elements of A that are not in B), since you mentioned it in the same sentence as the union.

I didn't mean the union, by (A and B) I mean the intersection of A and B (both satisfactory AND on time).
Anyway, look up the definition of conditional probability and figure out the given and unknowns for yourself. After that, it's just plug and chug.
 
I presume you mean that if the job is not done on time, a customer is not happy. If the job is done on time but not satisfactorily, a customer is not happy. So in order for a customer to be happy, the job must be done on time and satisfactorily. Imagine 100 customers whose jobs are done by this mechanic. 77 of them will be done on time. Of those 77 jobs that are done on time, 77*0.85= 65.45 are done satisfactorily: 65.45 of the customers will be happy which is 0.6545 probability that a given customer will leave happy.
That's how "conditional probability" is defined: Using P(A|B) to mean "probability of A given B" (which is what I think you meant by P(A/B)) then P(A)= P(A|B)*P(B).
 
Thanks for the help guys. That question was too easy! Seems so obvious now.
 

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