Probability questions.

Question

In studying the relationship between loan delinquency and borrower characteristics, a bank
found the following information:
• 80% of borrowers have been employed at their present job for 3 or more years,
• Of the borrowers who have been employed at their present job for 3 or more years, 5%
are behind in their payments
• Of the borrowers who have been employed at their present job for less than 3 years,
15% are behind in their payments

If a loan account is randomly selected, what is the probability that the borrower has been
employed at their present job for less than 3 years AND is behind in their payments?
(a) 0.17
(b) 0.03
(c) 0.20
(d) 0.014
(e) 0.24

If a loan account is randomly selected, what is the probability that the borrower has been
employed at their present job for 3 or more years OR is not behind in their payments?
(a) 0.76
(b) 0.744
(c) 0.97
(d) 0.83
(e) 0.96

If a loan account is randomly selected from a borrower who has been employed at their
present job for 3 or more years, what is the probability that the borrower is behind in their
payments?
(a) 0.5714
(b) 0.05
(c) 0.04
(d) 0.056
(e) 0.07

Answer 1

Hello Max,

 

1.  First, since 80% have been employed for 3 or more years, 20% have been employed less than 3 years, since those who are and aren't have to add up to 100%.  Then, we know that 15% of this 20% are behind.  The key word "of" often means to multiply, so we have.

15%*20%

=.15*.20

=.03

 

2.  In probabilities, the word "or" tells us that we can generally add the probabilities of these two together as long as there is no overlap.

First we know that p(3 or more years) is simply 80% or .8.

Next we consider those who are behind.  There are 5% who are behind and 3 or more years, but since I already counted all of the 3 or more years group, I cannot count them again.  This is an area of overlap, also known as non-mutually exclusive.  Therefore I can ignore this set.

Third, I know that there are 15% of the less than 3 years group that I need to include.  We already figured this amount is .3 from the last question.

So I add those two probabilities.

 .80+.03=.83

 

3. Since it is provided that the 3 or more years group has already been selected, we only have to consider the probability within this set.  This is 5%, or .05