Independent solution

How to solve this Expected Value question

Setup

Setup

Partition the policyholders into the mutually exclusive groups with only A, only H, and both coverages.

Pr⁡(A only)=0.65−0.15=0.50\Pr(A\text{ only})=0.65-0.15=0.50
Pr⁡(H only)=0.50−0.15=0.35\Pr(H\text{ only})=0.50-0.15=0.35

Model

Model

Apply the law of total probability by multiplying each group share by its renewal probability and summing the three contributions.

p=0.40(0.50)+0.60(0.35)+0.80(0.15)p=0.40(0.50)+0.60(0.35)+0.80(0.15)

Compute

Compute

The only-A, only-H, and both-coverages contributions are 0.20, 0.21, and 0.12, respectively, for a total renewal probability of 0.53.

p=0.20+0.21+0.12=0.53p=0.20+0.21+0.12=0.53

Answer

Answer

Thus 53% of policyholders renew, corresponding to choice D.

53%(D)\boxed{53\%\quad\text{(D)}}