An actuary has done an analysis of all policies that cover two cars. 70% of the policies are of type A for both cars, and 30% of the policies are of type B for both cars. The number of claims on different cars across all policies are mutually independent. The distributions of the number of claims on a car are given in the following table.

Number of Claims	Type A	Type B
0	40%	25%
1	30%	25%
2	20%	25%
3	10%	25%

Calculate the probability that exactly one of the four policies has the same number of claims on both covered cars.

• 0.104
• 0.250
• 0.285
• 0.417
• 0.739

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

Step 1: Probability of Equal Claims for Type A Policies

The problem states that the number of claims on different cars across all policies are mutually independent. This means if a policy is of Type A, the probability that both cars have [math]k[/math] claims is [math]P(\text{Car 1 has } k \text{ claims}) \times P(\text{Car 2 has } k \text{ claims})[/math]. The probability that a Type A policy has the same number of claims on both cars is the sum of probabilities for each possible number of claims ([math]0, 1, 2, 3[/math]) where both cars match. We denote the probability of [math]k[/math] claims for a car under a Type A policy as [math]P(C_A=k)[/math]. From the given table, these probabilities are:

• [math]P(C_A=0) = 0.40[/math]
• [math]P(C_A=1) = 0.30[/math]
• [math]P(C_A=2) = 0.20[/math]
• [math]P(C_A=3) = 0.10[/math]

The probability of equal claims for a Type A policy is:

[[math]]P(\text{Equal Claims } | \text{ Type A}) = P(C_A=0)^2 + P(C_A=1)^2 + P(C_A=2)^2 + P(C_A=3)^2[[/math]]

[[math]]= (0.40)^2 + (0.30)^2 + (0.20)^2 + (0.10)^2[[/math]]

[[math]]= 0.16 + 0.09 + 0.04 + 0.01 = 0.30[[/math]]

Step 2: Probability of Equal Claims for Type B Policies

Similarly, for a Type B policy, we use the probabilities for claims under Type B. We denote the probability of [math]k[/math] claims for a car under a Type B policy as [math]P(C_B=k)[/math]. From the given table, these probabilities are:

• [math]P(C_B=0) = 0.25[/math]
• [math]P(C_B=1) = 0.25[/math]
• [math]P(C_B=2) = 0.25[/math]
• [math]P(C_B=3) = 0.25[/math]

The probability of equal claims for a Type B policy is:

[[math]]P(\text{Equal Claims } | \text{ Type B}) = P(C_B=0)^2 + P(C_B=1)^2 + P(C_B=2)^2 + P(C_B=3)^2[[/math]]

[[math]]= (0.25)^2 + (0.25)^2 + (0.25)^2 + (0.25)^2[[/math]]

[[math]]= 4 \times (0.25)^2 = 4 \times 0.0625 = 0.25[[/math]]

Step 3: Overall Probability of a Randomly Selected Policy Having Equal Claims

We are given the proportion of policies of each type:

• 70% of policies are Type A ([math]P(\text{Type A}) = 0.70[/math])
• 30% of policies are Type B ([math]P(\text{Type B}) = 0.30[/math])

To find the overall probability that a randomly selected policy has the same number of claims on both cars, we use the Law of Total Probability:

[[math]]P(\text{Equal Claims}) = P(\text{Equal Claims } | \text{ Type A}) P(\text{Type A}) + P(\text{Equal Claims } | \text{ Type B}) P(\text{Type B})[[/math]]

Using the results from Step 1 and Step 2:

[[math]]P(\text{Equal Claims}) = (0.30)(0.70) + (0.25)(0.30)[[/math]]

[[math]]= 0.21 + 0.075 = 0.285[[/math]]

Let [math]p = 0.285[/math] be the probability that a single, randomly selected policy has the same number of claims on both covered cars.

Step 4: Probability of Exactly One of Four Policies Having Equal Claims

We are asked to calculate the probability that exactly one of four policies has the same number of claims on both covered cars. This is a binomial probability problem. Let [math]X[/math] be the number of policies (out of four) that have an equal number of claims on both cars. We have:

• Number of trials [math]n = 4[/math] (four policies selected).
• Probability of success (a policy having equal claims) [math]p = 0.285[/math] (from Step 3).
• Number of successes desired [math]k = 1[/math].

The probability mass function for a binomial distribution is given by:

[[math]]P(X=k) = {n \choose k} p^k (1-p)^{n-k}[[/math]]

For [math]n=4[/math], [math]k=1[/math], and [math]p=0.285[/math]:

[[math]]P(X=1) = {4 \choose 1} (0.285)^1 (1 - 0.285)^{4-1}[[/math]]

[[math]]P(X=1) = 4 \times (0.285) \times (0.715)^3[[/math]]

First, calculate [math](0.715)^3[/math]:

[[math]](0.715)^3 \approx 0.365228875[[/math]]

Now, substitute this value back into the equation:

[[math]]P(X=1) = 4 \times 0.285 \times 0.365228875[[/math]]

[[math]]P(X=1) \approx 0.416805096[[/math]]

Rounding to three decimal places, the probability is approximately [math]0.417[/math].

"'Key Insights"'

• Independence Principle: When events (like claims on two different cars) are independent, their joint probability is the product of their individual probabilities. This is crucial for calculating [math]P(C_1=C_2)[/math].
• Law of Total Probability: This fundamental rule allows calculation of the overall probability of an event by considering different mutually exclusive scenarios (e.g., policy types) and weighting their conditional probabilities by their prior probabilities.
• Binomial Distribution Application: Situations involving a fixed number of independent trials (policies), each with two possible outcomes (equal claims or not), and a constant probability of success, are modeled using the binomial distribution. Key parameters are the number of trials ([math]n[/math]), the number of successes ([math]k[/math]), and the probability of success ([math]p[/math]).
• Decomposition of "Same Number of Claims": To find the probability that two cars have the "same number of claims," one must sum the probabilities of all specific matching outcomes (e.g., both 0 claims, or both 1 claim, etc.).

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