29.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%
【确定】B 94%,
For an inaccurate paper record, 60% chances that the electronic record is inaccurate.
x * 60% = 3% x = 5% (inaccurate paper record)
For an inaccurate electronic record, 75% chances that the paper record is inaccurate.
y * 75% = 3% y = 4% (inaccurate electronic record)
(=> 2% inaccurate paper record, accurate electronic record
=> 1% inaccurate electronic record, accurate paper record)
inaccurate paper record + inaccurate electronic record = (5+4-3)=6%
ans:94%
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the explanation is from David thx to him ^^
yea , I've got the answer , but don't have any idea with such a problem ...it's always confusing me
can I think about it this way
set there are 100 incorrect paper record , then the incorrect ele record are 60
and there are 100 incorrect ele record , then the incorrect paper recods are 75
the total 200 records ,3 are both incorrect in ele and paper record ...
hmn ... my idea blocks here ...
would any one go on ? and is this the right way out ??
David emailed me attempting to send my detail explaination , but sth wrong with Yahoo ...