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An insurance company has a paper record and an electronic

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An insurance company has a paper record and an electronic [#permalink] New post 13 Sep 2004, 22:43
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An insurance company has a paper record and an electronic record for every claim. For an inaccurate paper record, 60% chances that the electronic record is inaccurate. For an inaccurate electronic record, 75% chances that the paper record is inaccurate. 3% of all the claims are inaccurate both in paper record and in electronic record. Pick one claim randomly, what are the chances that it is both accurate in paper record and in electronic record?
A.97% B.94% C.68% D.65% E.35%
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Re: PS - insurance company [#permalink] New post 14 Sep 2004, 01:16
Hi,

I am unable to get the answer. However, could you correct me pointing out the flaw in my approach given below?

Symbols are ~ = complement, ^ = intersecton of the sets, v = union of the sets.

P- accurate paper, E - accurate electronic stuff

p(~P) = 0.6, p(~E) = 0.75 p(~P^~E) = 0.03
p(PvE) = 1-p(~P^~E) = 0.97
Through ven diagrams, it is clear that p(E)-p(P^E) = p(~P)-p(~P^~E)
= 0.6-0.03 = 0.57

Similarly p(P)-p(P^E) = p(~E)-p(~P^~E)
= 0.75-0.03 = 0.72

But, p(PvE) = [p(P)-p(P^E)]+[p(E)-p(P^E)]+[p(P^E)]
0.97 = 0.72+0.57+ [p (P^E)]

=> the required p(P^E) = 0.97-1.29 = -0.32

There is no negative probaility. So, I am wrong. however, I am unable to find fault with my approach. Please spend sometime in debugging.

Anyway, I am choosing E as the answer is the closest,though negative.

Thanks in advance.

mba4me wrote:
An insurance company has a paper record and an electronic record for every claim. For an inaccurate paper record, 60% chances that the electronic record is inaccurate. For an inaccurate electronic record, 75% chances that the paper record is inaccurate. 3% of all the claims are inaccurate both in paper record and in electronic record. Pick one claim randomly, what are the chances that it is both accurate in paper record and in electronic record?
A.97% B.94% C.68% D.65% E.35%

_________________

Awaiting response,

Thnx & Rgds,
Chandra

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 [#permalink] New post 16 Sep 2004, 23:21
Hi mallelac,
I only know the answer and not the approach to this question . However i posted this question in testmagic forum and this is the response i got.

P(P)=accurate paper records
P(E)=accurate electronic records
P(E'/P')=0.60
P(P'/E')=0.75
P(P'E')=0.03

P(EP)=?

P(E'/P')=P(E'P')/P(P') => P(P')=0.03/0.60=0.05

P(P'/E')=P(E'P')/P(E') => P(E')=0.03/0.75=0.04

P(E'P')=P((EUP)')=0.03 => P(EUP)=1-0.03=0.97

P(EUP)=P(E)+P(P)-P(EP) => P(EP)=0.96+0.95-0.97

Thus,

P(EP)=0.94

This doesn't look like simple probability. Will try to do some research and post some more explanations to this question.


Thanks.
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 [#permalink] New post 17 Sep 2004, 01:36
Hi mba4me,

Thank you very much. I got the solution very clearly. It is conditional probability. I think I am NOT reading the query properly. The query is clearly phrased suggesting conditional probability(however, it is tough to envision such a scenario of Electronic and paper mistakes). I could not get it when I read it first time.

Thanks a lot. It is a very good problem. Thanks to you and the guy who has given the solution.

Hope I start reading the queries better. 8-) 8-) 8-)


mba4me wrote:
Hi mallelac,
I only know the answer and not the approach to this question . However i posted this question in testmagic forum and this is the response i got.

P(P)=accurate paper records
P(E)=accurate electronic records
P(E'/P')=0.60
P(P'/E')=0.75
P(P'E')=0.03

P(EP)=?

P(E'/P')=P(E'P')/P(P') => P(P')=0.03/0.60=0.05

P(P'/E')=P(E'P')/P(E') => P(E')=0.03/0.75=0.04

P(E'P')=P((EUP)')=0.03 => P(EUP)=1-0.03=0.97

P(EUP)=P(E)+P(P)-P(EP) => P(EP)=0.96+0.95-0.97

Thus,

P(EP)=0.94

This doesn't look like simple probability. Will try to do some research and post some more explanations to this question.

Thanks.

_________________

Awaiting response,

Thnx & Rgds,
Chandra

  [#permalink] 17 Sep 2004, 01:36
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