An insurance company has a paper record and an electronic record for

Question

An insurance company has a paper record and an electronic record for every claim. For an inaccurate paper record, the probability is 60% that the electronic record is inaccurate. For an inaccurate electronic record, the probability is 75% 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 is the probability 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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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?

imeanup · Answer

Product Rule

P(X ∩ Y)= P(X|Y)P(Y)

Given;  P(X|Y) = 60  Whereas, Y= inaccurate paper record and X= inaccurate electronic record

P(Y)=\frac{P(X ∩ Y)}{ P(X|Y)}=\frac{3}{60}=\frac{1}{20}

P(X)=\frac{P(X ∩ Y)}{ P(Y|X)}=\frac{3}{75}=\frac{1}{25}

P(X ∪ Y)= P(X) + P(Y)-P(X ∩ Y) = \frac{1}{25}+\frac{1}{20}-\frac{3}{100}=\frac{6}{100}

Therefore, Required Probability:

P`=1-P=1-\frac{6}{100}=\frac{94}{100}

Ans B

rsrighosh · Answer

Seems like this is a dependent event. Is this a GMAT type question?

Regor60 · Answer

Inacc P Inacc P Acc P Acc P
Inacc E Acc E Inac E Acc E
_______ _______ ______ _____

a b c d

a/a+b = .6.

a/a+c = .75

a=.03

Find d

.03=.18+.6b. b=.02

.03=(.09/4)+.75c. c=.01

a+b+c+d=1

d=1-.03_.02-.01 = .94

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pappal · Answer

incorrect paper 
 correct paper

incorrect electronic 
 0.6x=0.75y=3% 
 0.25y 
 y% let 
 
  correct electronic 
 0.4x 
 z=? let

x% let 
  
 100%

as per given conditions above grid can be formed
so 0.6x=3% so x=5% and 0.75y=3% so y=4%
if x=5% so the box below z=? will read as 95%
and if y=4% so 0.25y=1%
hence required probability=z=95-1= 94% both entries correct

okHedwig · Answer

IMO B
I tried to solve it like this:
Let Ei be the event of electronic inaccurate & Pi be the event of Paper inaccurate.
P(Ei/Pi) = 0.6 = P(Ei intersection Pi) / P(Pi) = 0.03/P(Pi), solving P(Pi) = 0.04
similarly P(Pi/Ei) = 0.75 = P( Ei intersection Pi)/P(Ei) , solving P(Ei) = 0.05
So , P(Pi U Ei) = 0.04+0.05-0.03 = 0.06
Let Pa = paper acurate, Ea= Electronic acurate
Therefore: P(Pa inetrsection Ea) = 1-0.06 = 0.94

Please let me know if my logic is correct?

An insurance company has a paper record and an electronic record for

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	incorrect paper	correct paper
incorrect electronic	0.6x=0.75y=3%	0.25y	y% let
correct electronic	0.4x	z=? let
	x% let		100%

GMAT Problem Solving (PS) Questions

An insurance company has a paper record and an electronic record for

Prep Toolkit

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