**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