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.

Step 1: Define Events and Identify the Target Probability

We begin by defining the events involved in the problem:

• Event A: A customer insures more than one car.
• Event B: A customer insures a sports car.

The problem asks for the probability that a randomly selected customer insures exactly one car and that the car is not a sports car.

• "Exactly one car" is the complement of "more than one car," which is [math]A^c[/math]. (Note: The problem states "All customers insure at least one car," so "exactly one car" is equivalent to [math]A^c[/math].)
• "Not a sports car" is the complement of "insures a sports car," which is [math]B^c[/math].

Therefore, we need to calculate the probability of the intersection of these two events: [math]\operatorname{P}(A^c \cap B^c)[/math]. Using De Morgan's Laws, we know that [math]A^c \cap B^c = (A \cup B)^c[/math]. Thus, the target probability can be expressed as:

[[math]]\operatorname{P}(A^c \cap B^c) = 1 - \operatorname{P}(A \cup B)[[/math]]

Step 2: List Given Probabilities

The problem provides the following probabilities:

Given Probabilities

Event	Probability
A: Customer insures more than one car	[math]\operatorname{P}(A) = 0.64[/math]
B: Customer insures a sports car	[math]\operatorname{P}(B) = 0.20[/math]
B given A: Customer insures a sports car, given they insure more than one car	[math]\operatorname{P}(B|A) = 0.15[/math]

Step 3: Calculate the Probability of A and B ([math]\operatorname{P}(A \cap B)[/math])

To find [math]\operatorname{P}(A \cap B)[/math] (the probability that a customer insures more than one car AND a sports car), we use the multiplicative law of probability:

[[math]]\operatorname{P}(A \cap B) = \operatorname{P}(B|A) \times \operatorname{P}(A)[[/math]]

Substitute the given values:

[[math]]\operatorname{P}(A \cap B) = 0.15 \times 0.64 = 0.096[[/math]]

Step 4: Calculate the Probability of A or B ([math]\operatorname{P}(A \cup B)[/math])

Next, we calculate the probability that a customer insures more than one car OR a sports car, [math]\operatorname{P}(A \cup B)[/math], using the additive law of probability (inclusion-exclusion principle):

[[math]]\operatorname{P}(A \cup B) = \operatorname{P}(A) + \operatorname{P}(B) - \operatorname{P}(A \cap B)[[/math]]

Substitute the probabilities we have calculated or were given:

[[math]]\operatorname{P}(A \cup B) = 0.64 + 0.20 - 0.096 = 0.84 - 0.096 = 0.744[[/math]]

Step 5: Calculate the Probability of "Exactly One Car and Not a Sports Car"

Finally, we calculate the target probability, [math]\operatorname{P}(A^c \cap B^c)[/math], using the relationship established in Step 1:

[[math]]\operatorname{P}(A^c \cap B^c) = 1 - \operatorname{P}(A \cup B)[[/math]]

Substitute the value of [math]\operatorname{P}(A \cup B)[/math] calculated in Step 4:

[[math]]\operatorname{P}(A^c \cap B^c) = 1 - 0.744 = 0.256[[/math]]

The probability that a randomly selected customer insures exactly one car, and that the car is not a sports car, is 0.256. This corresponds to option C.

Key Insights

• Defining events clearly (e.g., A = "more than one car", B = "sports car") is the first crucial step in solving probability problems.
• Understanding how to translate complex verbal descriptions (e.g., "exactly one car and not a sports car") into set notation (e.g., [math]A^c \cap B^c[/math]) is fundamental.
• De Morgan's Laws ([math]A^c \cap B^c = (A \cup B)^c[/math]) are powerful tools for simplifying probability calculations, especially when dealing with complements of intersections or unions.
• The Additive Law of Probability (Inclusion-Exclusion Principle): [math]\operatorname{P}(A \cup B) = \operatorname{P}(A) + \operatorname{P}(B) - \operatorname{P}(A \cap B)[/math] is essential for calculating the probability of the union of two events.
• The Multiplicative Law of Probability (Conditional Probability): [math]\operatorname{P}(A \cap B) = \operatorname{P}(B|A) \times \operatorname{P}(A)[/math] is vital when information is given in terms of conditional probabilities.
• Systematically breaking down the problem into smaller, manageable steps (defining events, listing givens, calculating intermediate probabilities, then the final target) helps ensure accuracy.

This article was generated by AI and may contain errors. If permitted, please edit the article to improve it.

Solution: C

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.