A health insurer sells policies to residents of territory X and territory Y. Past claims experience indicates the following:

1. 20% of the total policyholders from territory X and territory Y combined filed no claims.
2. 15% of the policyholders from territory X filed no claims.
3. 40% of the policyholders from territory Y filed no claims.

Calculate the probability that a randomly selected policyholder was a resident of territory X, given that the policyholder filed no claims.

• 0.09
• 0.27
• 0.50
• 0.60
• 0.80

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

Solution: D

Define the following 3 events:

A: policyholder is in territory A

B: policyholder is in territory. B

N: policyholder does not file claim.

Then using Bayes Theorem, we want

[[math]] \operatorname{P}(A | N) = \frac{\operatorname{P}( A) \operatorname{P}( N | A)}{\operatorname{P}( A) \operatorname{P}( N | A) + \operatorname{P}( B) \operatorname{P}( N | B)} = \frac{\operatorname{P}( A) [ 0.15]}{\operatorname{P}( A) [ 0.15] + \operatorname{P}( B) [ 0.40]} [[/math]]

Now let [math]\operatorname{P}( A) = p[/math] . Then [math]\operatorname{P}( B) = 1 − p[/math] . To find [math]p[/math], note

[[math]] \begin{align*} 0.2 &= \operatorname{P}( N ) = \operatorname{P}( A ∩ N ) + \operatorname{P}( B ∩ N ) = \operatorname{P}( A) \operatorname{P}( N | A) + \operatorname{P}( B) \operatorname{P}( N | B) \\ &=( p )(0.15) + (1 − p )(0.4) = 0.4 − 0.25 p. \end{align*} [[/math]]

Thus [math]0.25p = 0.2[/math], and [math]p = \frac{0.2}{0.25} = 0.8 [/math]. So

[[math]] \operatorname{P}(A | N) = \frac{(0.8)(0.15)}{( 0.8)( 0.15) + ( 0.2 )( 0.4 )} = \frac{0.12}{0.20} = 0.6. [[/math]]

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