Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 06 Jul 2015, 16:58

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# An insurance company has a paper record and an electronic

Author Message
TAGS:
Manager
Joined: 09 Sep 2004
Posts: 54
Followers: 2

Kudos [?]: 31 [0], given: 0

An insurance company has a paper record and an electronic [#permalink]  13 Sep 2004, 22:43
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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%
Senior Manager
Joined: 22 Jun 2004
Posts: 394
Location: Bangalore, India
Followers: 1

Kudos [?]: 2 [0], given: 0

Re: PS - insurance company [#permalink]  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.

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

Manager
Joined: 09 Sep 2004
Posts: 54
Followers: 2

Kudos [?]: 31 [0], given: 0

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.
Senior Manager
Joined: 22 Jun 2004
Posts: 394
Location: Bangalore, India
Followers: 1

Kudos [?]: 2 [0], given: 0

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.

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

Similar topics Replies Last post
Similar
Topics:
1 A publishing company produced a record high of 180 books tod 3 07 Aug 2012, 04:51
A company has 540 employees, 40% of whom are employed part time. If it 7 19 Oct 2010, 07:06
8 A certain company has records stored with a record storage 6 23 Dec 2009, 17:39
2 Lee Colle insures its students for thefts up to \$1000. The college mak 9 01 Sep 2009, 12:32
paper 3 12 Aug 2009, 06:25
Display posts from previous: Sort by