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

[morry]

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.

 

[Galileo]

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.

 

[morry]

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?

 

[Galileo]

P(A|B) is the probability of A conditioned on B, i.e. the probability A occurs given that B has occured. It's simply the way conditional probabilities are written. It has nothing to do with setminuses.

 

[morry]

It has nothing to do with setminuses.

Umm, what are setminuses?

Ill have another go and see if I can get it out.

 

[Galileo]

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.

 

[HallsofIvy]

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).

 

[morry]

Thanks for the help guys. That question was too easy! Seems so obvious now.