praetorian123 wrote:

In an insurance company, each policy has a paper record and an electric record. For those policies having incorrect paper record, 60% also have incorrect electric record; For those policies having incorrect electric record, 75% also having incorrect paper record. 3% of all policies have both incorrect paper and incorrect electric

records. If we randomly pick out one policy, what's the probability

that the one having both correct paper and correct electric records?

Please explain steps..

ok..heres the answer

we use one of the most important probability concepts

P( correct paper and correct electric) = 1- P ( incorrect paper and incorrect electric)

Let T be total policies

Let x be total incorrect paper policies

Let y be total incorrect electric policies

It follows from the problem statement that

0.03T is the number of both incorrect electric and paper policies

0.6x is the number of both incorrect electric and paper ..

0.75y is the number of both incorrect electric and paper...

So now that all are the same, we have

0.03 T = 0.6 x = 0.75 y

x = 0.03 T/0.6 = 5% * T

y= 0.03 T/0.75 = 4 % * T

Vicky , x + y double counts the "BOTH" incorrect part.

So total incorrect paper OR total incorrect electric = x +y - both incorrect

5% T + 4% T - 3% T = 6%

Required Prob = 1 - 0.06 = 0.94

Answer 0.94

Thanks

Praetorian