Independent solution

How to solve this Conditional Probability question

Setup

Setup

Let S denote at least one student-loan default and C at least one car-loan default.

Pr⁡(S)=0.30,Pr⁡(S∣C)=0.40,Pr⁡(C∣Sc)=0.28\Pr(S)=0.30,\qquad \Pr(S\mid C)=0.40,\qquad \Pr(C\mid S^c)=0.28

Model

Model

Write the car-default probability as its contributions inside and outside S.

Pr⁡(C∩Sc)=(0.28)(0.70)=0.196\Pr(C\cap S^c)=(0.28)(0.70)=0.196
Pr⁡(C∩S)=0.40Pr⁡(C)\Pr(C\cap S)=0.40\Pr(C)
Pr⁡(C)=0.40Pr⁡(C)+0.196\Pr(C)=0.40\Pr(C)+0.196

Compute

Compute

Recover the marginal and joint probabilities, then condition on S.

Pr⁡(C)=0.1960.60=0.3266667\Pr(C)=\frac{0.196}{0.60}=0.3266667
Pr⁡(C∩S)=(0.40)(0.3266667)=0.1306667\Pr(C\cap S)=(0.40)(0.3266667)=0.1306667
Pr⁡(C∣S)=0.13066670.30=0.4355556\Pr(C\mid S)=\frac{0.1306667}{0.30}=0.4355556

Answer

Answer

The desired conditional probability rounds to 0.44.

0.44(C)\boxed{0.44\quad\text{(C)}}