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

Question

1. In an insurance company, each policy has a paper record and an 
electric record. For those policies having incorrect paper record, 
60% also having 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..

MartinMag · Answer

Took me like 5 minutes!!

A=Incorrect Paper Record
B=Incorrect electronic Record
P(A/B)=means probability of ocurring A given that B already happened

Info from the question:
P(A AND B)=0.03
P(A/B)=0.75
P(B/A)=0.6

We want to know 1-P(AUB)

Calculations:

P(B) = 0.03/0.75= 1/25 = 0.04
P(A) = 0.03/0.6 = 1/20 = 0.05

P(AUB)=P(A)+P(B)-P(A AND B)=0.05+0.04-0.03=0.06

1-P(AUB)=0.94


Did I convince anyone?

Praetorian · Answer

Hey martin/rich

It cant be this complex..seriously!

Let me wait a while before i tell you the answer..would like to see others participate

Thanks 
Praetorian

rich28 · Answer

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

Praetorian · Answer

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

rich28 · Answer

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

Vicky · Answer

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

Praetorian · Answer

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

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

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