# Thread: Prob of high risk loans defauting, student text question

1. ## Prob of high risk loans defauting, student text question

Hi, first of all thanks for reading this. I'm 47 and trying to teach myself Stats from an excellent textbook as part of my course in maths and economics which I'm studying independently.
The textbook in question is Statistics for business and economics by Newbold, carson and thorne.

I've been picking my way through probability and have hit a problem which I have a answer but I don't trust my method.

i'm given 3 facts
1 only 15% loans made to high risk customers.
2 5% of loans made are in default.
3 of those in default 40% were made to high risk customers.

I'm taking 3 to be a conditional prob of those being in default being high risk ie (B|A)

I'm asked to find probability of high risk borrower defaulting?

the method I used was to say that P(B|A)= P(AnB)/P(A)

therefore 0.4 x 0.15 = P(AnB) = 6%

Which is higher than the given rate but not by as much as I expect considering only 15 loans made to high risk people and yet they contribute to 40% of defaults.

Am I doing something wrong, any help would be warmly welcomed.

yours

Mark

2. ## Re: Prob of high risk loans defauting, student text question

Originally Posted by Koyunbaba
Hi, first of all thanks for reading this. I'm 47 and trying to teach myself Stats from an excellent textbook as part of my course in maths and economics which I'm studying independently.
Glad to hear you keep on learning!

Originally Posted by Koyunbaba
I've been picking my way through probability and have hit a problem which I have a answer but I don't trust my method.

i'm given 3 facts
1 only 15% loans made to high risk customers.
2 5% of loans made are in default.
3 of those in default 40% were made to high risk customers.
Let's define each of these events in terms of A and B.

Using what you already started with, given that the loan is in default (A), 40% are high risk (B).

So as you noted below, P(B|A) =.4

The "|" means "given".

Originally Posted by Koyunbaba
I'm taking 3 to be a conditional prob of those being in default being high risk ie (B|A)
See just above.

Originally Posted by Koyunbaba
I'm asked to find probability of high risk borrower defaulting?

the method I used was to say that P(B|A)= P(AnB)/P(A)

therefore 0.4 x 0.15 = P(AnB) = 6%
If we stick with the events defined as above, A is default and B is high risk. So, they want to know: given that the borrower is high risk (B), what is the probability he defaults (A); P(A|B).

We know P(A) = .05, P(B)= .15, P(B|A) = .4.

[P(B|A)*P(A)]/P(B) = P(A|B) --> [.4*.05]/.15 = 0.02/.15 = .13

Probability of default GIVEN that the borrower is high risk, P(A|B), is .13 (13%).

Another way to approach problems like this might be with a tree diagram or with natural frequencies (I believe that's the term).

We know that out of every 100 loans:
A) 15 are made to high risk people
B) 5 are in default
A|B) 2 are from high risk people of the 5 total in default, (40% are due to high risk people)

So, 15 are from high risk people, 2 are defaults from high risk people, so 2/15 is probability of default, given that they are high risk loans.

Had to remove a frequency table option due to poor formatting

It's pretty early near me, so if another person could confirm, that would be great. Although, I believe this is correct.

3. ## The Following User Says Thank You to ondansetron For This Useful Post:

Koyunbaba (08-03-2017)

4. ## Re: Prob of high risk loans defauting, student text question

Thank you so much for your reply. That made a lot of sense the way you presented it. I was in the right sort of ballpark, but I don't think I was clear about what I was actually looking for eg a conditional probability. If I could have made that twig, I thing I could have cracked it. Once again thanks for your time. Oh another thing I think I learned from your post was the value of defining clearly what things are. I think that lesson will help me going forward.

5. ## Re: Prob of high risk loans defauting, student text question

Originally Posted by Koyunbaba
Thank you so much for your reply. That made a lot of sense the way you presented it. I was in the right sort of ballpark, but I don't think I was clear about what I was actually looking for eg a conditional probability. If I could have made that twig, I thing I could have cracked it. Once again thanks for your time. Oh another thing I think I learned from your post was the value of defining clearly what things are. I think that lesson will help me going forward.
I considered mentioning that it's quite helpful to define everything, physically write it all out, when approaching a problem...it appears you got there on your own, so great job! It's also useful to think in terms of "X givenY" when you read a sentence. Once you get in these habits I don't think you will have any issues!

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