# Thread: Bayes Theorem-Probabilty Question

1. ## Bayes Theorem-Probabilty Question

Hello Hi everybody!
Here is a new thread I faced difficult in solving please give me the way out.

Question;
Michael Schofield, credit manager for Lincoln Burrows knows that the company uses three methods to encourage collection of delinquent accounts. From the past collection records, he learn that 70% of the accounts are called on personally, 20% are phoned, and 10% are sent a letter. The probabilities of collecting an overdue amount from an account of the three methods are 0.75, 0.60 and 0.65 respectively. Scofield has just received payment from the past due account what is the probability that this account;

(a) Was called on personally
(b) Received a phone call
(c) Received a letter
**end**
I look forward to getting a good start!!1

2. I'll denote

P(p) = prior probability of someone being contacted personally
P(ph) = prior probability of someone being contacted by phone
P(l) = prior probability of someone being contacted by letter

And

P(c) = probability of collecting an overdue ammount

We know:
P(p) = 0.7
P(ph) = 0.2
P(l) = 0.1
P(c | p) = 0.75
P(c | ph) = 0.60
P(c | l) = 0.65

And we want to find the posterior probabilities:

p(p | c) = ?
p(ph | c) = ?
p(l | c) = ?

From Bayes' Theorem we have:

p(p | c) = p(c | p) * p(p) / p(c) (1)
p(ph | c) = p(c | ph) * p(ph) / p(c) (2)
p(l | c) = p(c | l) * p(l) / p(c) (3)

All we have to compute is p(c), which is:

p(c) = p(c & p) + p(c & ph) + p(c & l) = p(c | p) * p(p) + p(c | ph) * p(ph) + p(c | l) * p(l)

Now we have all values and can substitute in (1), (2), (3).

EDIT: Bayes' theorem the other way around =X

3. Originally Posted by pedrosorio
From Bayes' Theorem we have:

p(p | c) = p(c | p) * p(c) / p(p) (1)
p(ph | c) = p(c | ph) * p(c) / p(ph) (2)
p(l | c) = p(c | l) * p(c) / p(l) (3)
Bayes' theorem is:

$\displaystyle Pr(A|B)=\frac{Pr(B|A)Pr(A)}{Pr(B)}$

which is not what you have here.

CB

4. Originally Posted by CaptainBlack
Bayes' theorem is:

$\displaystyle Pr(A|B)=\frac{Pr(B|A)Pr(A)}{Pr(B)}$

which is not what you have here.

CB
Terrible mistake