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Finally they multiplied it by .4.

I don't know what formula they use but obviously it is not Bayes rule.

- Thread starter torquerotates
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- #1

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Finally they multiplied it by .4.

I don't know what formula they use but obviously it is not Bayes rule.

- #2

Stephen Tashi

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What you describe seems to be calculation for P( (X + Y > 8000) and X > 2000) rather than the calculation for P(X + Y > 8000 | X > 2000). Which did the book intend to calculate?Finally they multiplied it by .4.

(It isn't clear what you mean when you say "some textbooks just assume A|B works". How do you interpret "A|B"? What does it mean to say that that it "works"?)

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the problem states that there are two employees are covered by a policy. The policy will reimburse no more than one loss per employee. The policy reimburses the full amount of the loss up to a company maximum of 8000. The probability of an employee incurring a loss is 40%. The probability of an employee incurring a loss independent of any other employee incuring a loss. The amount of each loss is uniformly distributed on [1000, 5000]. Given that one of the employees incurs a loss, find the probability that losses will exceed reimbursements.

For the solution, they let X be the claim of the employee that has incurred a loss and Y be the claim of the other employee. I just don't know why they can do this. They assumed that one of the employees already incurred a loss,calculate the probability and then say, oh well the probability of this employee incurring a loss in the first place is 40% so they multiply it by .4. It just don't understand how they could have not accounted for the other case Y. You usually have to account for that due to the law of total probability.

- #4

Stephen Tashi

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I think the .4 represents the probability that the second employee incurs a loss given the first one has already incurred a loss. Their notation must be that X is the employee who is known to have a loss and Y is the second employee.They assumed that one of the employees already incurred a loss,calculate the probability and then say, oh well the probability of this employee incurring a loss in the first place is 40% so they multiply it by .4.

If you want X and Y to denote two employees without saying which one is known to occur a loss then you would indeed have to deal with the possibility that X alone or Y alone or both X and Y incur losses. However, you'd be working in a different probability space.

Bayes rule relates probabilities in two different probability spaces. As a set there is no difference between [itex] A \cup B [/itex] and [itex] A | B [/itex] but there is a difference in what we are to consider the space of all possible events. It's sometimes possible to compute probabilities in the space of possible events corresponding to [itex] A | B [/itex] by working entirely in that space.

It has been said that "All probabilities are conditional probabilities". For example, if a probability problem is given in a test book and it involves P(A) then this is really P(A| E) where E are the conditions given the in the problem. Those conditions define the space of all possible events. If you work the textbook problem without using Bayes rule and find P(A), you've actually found P(A|E). Conditional probabilities can be sometimes be found without using Bayes.

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HallsofIvy

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