The following information is given about a group of high-risk borrowers.

1. Of all these borrowers, 30% defaulted on at least one student loan.
2. Of the borrowers who defaulted on at least one car loan, 40% defaulted on at least one student loan.
3. Of the borrowers who did not default on any student loans, 28% defaulted on at least one car loan.

A statistician randomly selects a borrower from this group and observes that the selected borrower defaulted on at least one student loan.

Calculate the probability that the selected borrower defaulted on at least one car loan.

• 0.33
• 0.40
• 0.44
• 0.65
• 0.72

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: C

Let S represent the event that the selected borrower defaulted on at least one student loan.

Let C represent the event that the selected borrower defaulted on at least one car loan.

We need to find [math] \operatorname{P}(C | S) = \frac{\operatorname{P}(C \cap S)}{\operatorname{P}(S)}[/math].

We are given

[[math]] \operatorname{P}(S) = 0.3, \, \operatorname{P}(S | C) = \frac{\operatorname{P}(C \cap S)}{\operatorname{P}(C)} = 0.4, \, \operatorname{P}(C | S^c) = \frac{\operatorname{P}(C \cap S^c)}{\operatorname{P}(S^c)} = 0.28. [[/math]]

Then

[[math]] \operatorname{P}(C \cap S^c) = 0.28 \operatorname{P}(S^c) = 0.28(1-0.3) = 0.196. [[/math]]

Because [math] \operatorname{P}(C) = \operatorname{P}(C \cap S) + \operatorname{P}(C \cap S^c)[/math] and [math]\operatorname{P}(C) = \operatorname{P}(C \cap S)/0.4 [/math] we have

[[math]] \operatorname{P}(C ∩ S ) / 0.4= \operatorname{P}(C ∩ S ) + 0.196 ⇒ \operatorname{P}(C ∩ S )= 0.196 /1.5= 0.13067. [[/math]]

Therefore

[[math]] \operatorname{P}(C | S) = \frac{\operatorname{P}(C \cap S)}{\operatorname{P}(S)} = \frac{0.13067}{0.3} = 0.4356. [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.