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.

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