# Homework Help: Partition Theorem

1. Jan 10, 2012

### Philip Wong

1. The problem statement, all variables and given/known data
Assume that it is appropriate to transfer the probabilities IP(F|L) and IP(F|T) from the police context to the insurance context.
Define the following new events for the insurance context:
L = “insurance claimant is lying”;
T = “insurance claimant is truthful”;
F = “insurance claimant failed lie-detector test on phone”;
P = “insurance claimant passed lie-detector test on phone”.
An insurance company finds that a massive 52.5% of claimants fail the liedetector
test on the phone. What is the probability that a claimant is actually
lying?

IP(F) = 0.525 IP(P) = 1-0.525=0.475
IP(F|L)=0.38 IP(F|T)=0.23
IP(P|L)=0.14 IP(P|T)=0.25

2. Relevant equations
Bayesian TheoremP(B |A) = P(A|B)P(B)/ P(A)
IP(L) = IP((L|F) $\cap$ (L|P))

3. The attempt at a solution
IP(L|F)=(0.38*0.525) / (0.38*0.525+0.23*0.525)=0.1995/0.32025=0.62
IP(L|P)=(0.14*0.475) / (0.14*0.475+0.25*0.475)=0.0665/0.18525 = 0.36

IP(L) = IP((L|F) $\cap$ (L|P))
= IP(L|F) * IP(L|P)
= 0.62 * 0.36 = 0.2232

is my workings right? I'm kind of worried that I used the wrong formula to work out IP(L), so it would be nice if someone could double check that part too
thanks

2. Jan 10, 2012

### lanedance

Though I haven't checked numbers, your use of Bayes to get IP(L|F) and IP(L|P) is on the right track, and given the results show someone who fails the test is likely to be lying (62%) or someone who passes the test is unlikely to be lying (36%) is encouraging

I don't understand what you've done in the last step however. I would notice (or assume) that the events P&F span the entire probabilty universe, and use the following:

IP(L) = IP(L|F)IP(F) + IP(L|P)IP(P)

Last edited: Jan 10, 2012
3. Jan 10, 2012

### Philip Wong

you are right, I just figured it out about half an hour ago. I knew I had implied some of the formula wrongly.

Thanks for confirming that with me!