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An insurance company has a paper record and an electronic

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Director
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An insurance company has a paper record and an electronic [#permalink] New post 06 Oct 2004, 03:41
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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. If one claim is picked randomly, what are the chances that it is both accurate in paper record and in electronic record?
Director
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 [#permalink] New post 06 Oct 2004, 04:04
Live draft :

4 cases : a = accurate, w = wrong

Ew+Pw = 3%
Ea+Pa = X%
Ew+Pa = 3%.(1-75%)/100 = 0,75%
Ea+Pw = 3%.(1-60%)/100 = 1,2%

P (Ea+Pa) = 1-3%-0,75%-1,2%

Any other opinion ?
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 [#permalink] New post 06 Oct 2004, 04:20
Twixt , can u plz explain with words please.
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Re: PS Probability Accuracy!! [#permalink] New post 06 Oct 2004, 10:56
saurya_s 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. If one claim is picked randomly, what are the chances that it is both accurate in paper record and in electronic record?


Abbrevs: Inaccurate Electronic Record = IE, Inaccurate Paper = IP
Pick #s... 100 IPs
Then:
100 * 60% = 60 both IE and IP

x IEs * 75% = 60 both IE and IP

60/0.75 = 80 IEs

3% * TOTAL = 60 IE & IP
TOTAL = 60/0.03 = 2000

TOTAL - IE - IP = # Records that are accurate in both formats.

2000 - 80 - 60 = 1860

P(accurate in both formats) = 1860 / 2000 = 93%
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 [#permalink] New post 06 Oct 2004, 14:11
what is the official anwer to this question
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 [#permalink] New post 06 Oct 2004, 21:34
I got 94%

Paper Inacc:
(60/100)P = 3/100
P = 5%

Elec Inacc:
(75/100)Q = 3/100
Q = 4%

Both acc = 100 - (5+4-3) = 94%
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 [#permalink] New post 07 Oct 2004, 02:38
my answer is 91.2%

Note product rule: P(A given B) = P(A and B) / P(B)

Use it and answer will follow through easily
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 [#permalink] New post 07 Oct 2004, 07:07
keyV wrote:
my answer is 91.2%

Note product rule: P(A given B) = P(A and B) / P(B)

Use it and answer will follow through easily


KeyV, in the interest of of tthose who can't figure out what is the different probabilities here, why don't you use figures in solving this for the forum. That would be a good service.
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 [#permalink] New post 07 Oct 2004, 23:45
Quote:
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. If one claim is picked randomly, what are the chances that it is both accurate in paper record and in electronic record?


Let prob that elec record is inaccurate given paper rec is inacc be P(e/p)
Let prob that paper record is inaccurate given elec rec is inacc be P(p/e)
Let P(e) and P(p) be prob that elec and paper records respectively are inaccurate.
Let P(p and e) be prob that both are inaccurate.

IT is given that:
P(e/p) = .6
P(p/e) = .75
P(p and e) = .03
We are asked for the value of [1-P(p)] x [1-P(e)] ie no error....clear so far?

Unleash product rule:
P(e/p) = P(p and e) / P(p)
.6 = .03 / P(p)

Similarly,
P(p/e) = P(p and e)/P(e)
.75 = .03/P(e)

Now just solve for P(p) and P(e).....
Dont forget to calculate [1-P(p)] * [1-P(e)].....as that is what Q asks for.

Hope this helps...
Cheers
  [#permalink] 07 Oct 2004, 23:45
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