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.
Thanks in advance.
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
close
