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27 Jan 2021, 14:36
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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%
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
A. 97%
B. 94%
C. 68%
D. 65%
E. 35%
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
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?
29 Jan 2021, 21:26
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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
\(P(X ∩ Y)= P(X|Y)P(Y)\)
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
30 Jan 2021, 07:19
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Seems like this is a dependent event. Is this a GMAT type question?
