In an insurance company, each policy has a paper record and

Hey Praetorian,

I noticed you are in Houston. How do people in the area view the Rice business school? Where schools do you have your eye on?

Rich

Hey Rich

I am so sorry..though i have been here for two yrs now, i have visited Rice just twice..and i dont know too many students there.

Run a search on Google...may be you can find alumni from Jones...you can contact them...and see how it goes..

My target schools...well..i am applying for the Phd program!
And no , i dont have an MBA.

Best of Luck
Praetorian

Thanks, I'm in D.C. and I Darden is a great school. good luck!

Rich

I will try a shorter and simpler route:
Let a = incorrect Paper records
b = incorrect Electronic recods
c = Total policies.
Now from problem statement => 0.6a = 0.75b (they are same i.e. the ones having both incorrect paper and electric records)

Also, 0.03c = 0.6a = 0.75b --------- eq 1. (all give same data but in a different way)

Therefore, probability for selecting a policy with both correct electric and paper record = (c - b - a)/c
= 0.91 (substituting a,b from above equation -1 )
am i riht praet. I must say its a very tricky problem.
-Vicks

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