Insurance company examines its pool of auto insurance customers and gathers the following information:

1. All customers insure at least one car.
2. 64% of the customers insure more than one car.
3. 20% of the customers insure a sports car.
4. Of those customers who insure more than one car, 15% insure a sports car.

Calculate the probability that a randomly selected customer insures exactly one car, and that the car is not a sports car.

• 0.16
• 0.19
• 0.26
• 0.29
• 0.31

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

Consider the following events about a randomly selected auto insurance customer:

A = customer insures more than one car

B = customer insures a sports car

We want to find the probability of the complement of A intersecting the complement of B (exactly one car, non-sports). We have [math]\operatorname{P}(A^c \cap B) = 1 - \operatorname{P}(A \cup B). [/math] By the additive Law,

[[math]] \operatorname{P}( A ∪ B )= \operatorname{P}( A) + \operatorname{P}( B) − \operatorname{P}( A ∩ B) [[/math]]

By the Multiplicative Law,

[[math]] \operatorname{P}( A=∩ B ) \operatorname{P}( B | A) \operatorname{P}( A) = (0.15) (0.64) = 0.096. = 0.096 . [[/math]]

Then,

[[math]] \operatorname{P}( A ∪ B ) = 0.64 + 0.20 − 0.096 = 0.744 . [[/math]]

Finally,

[[math]] \operatorname{P}( A^c ∩ B^c ) = 1 − 0.744 = 0.256. [[/math]]

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