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Suppose that an insurance company classifies people into one of three classes: good risks, average risks, and bad risks. The company's records indicate that the probabilities that good-, average-, and bad-risk persons will be involved in an accident over a 1 -year span are, respectively, .05, .15 and \(30 .\) If 20 percent of the population is a good risk, 50 percent an average risk, and 30 percent a bad risk, what proportion of people have accidents in a fixed year? If policyholder \(A\) had no accidents in \(2012,\) what is the probability that he or she is a good risk? is an average risk?

Short Answer

Expert verified
The proportion of people having accidents in a fixed year is 17.5%. If policyholder A had no accidents in 2012, the probabilities of being in each risk class are: - Good risk: 24.24% - Average risk: 33.33% - Bad risk: 21.82%

Step by step solution

01

Calculate the Proportion of People Having Accidents in a Year

Using the Total Probability Theorem: The proportion of people having accidents in a year can be calculated as: P(Accident) = P(Accident | Good risk) × P(Good risk) + P(Accident | Average risk) × P(Average risk) + P(Accident | Bad risk) × P(Bad risk) Using the given probabilities of accidents for each risk class and the population percentages: P(Accident) = (0.05)(0.20) + (0.15)(0.50) + (0.3)(0.30)
02

Calculate P(Accident) using the obtained expression

The proportion of people having accidents in a year is: P(Accident) = (0.05)(0.20) + (0.15)(0.50) + (0.3)(0.30) P(Accident) = 0.01 + 0.075 + 0.09 P(Accident) = 0.175 So, 17.5% of the population have accidents in a fixed year.
03

Calculate the Probability of Being in a Risk Class Given No Accidents in a Year

Using Bayes' Theorem, the probabilities of being in each risk class given no accidents in a year are: P(Good risk | No accident) = P( No accident | Good risk) × P(Good risk) / P(No accident) P(Average risk | No accident) = P( No accident | Average risk) × P(Average risk) / P(No accident) P(Bad risk | No accident) = P( No accident | Bad risk) × P(Bad risk) / P(No accident) We already know P(Accident) from step 2. We can calculate P(No accident) as: P(No accident) = 1 - P(Accident) = 1 - 0.175 = 0.825
04

Calculate the Required Probabilities of being in a Risk Class

Using the expression we obtained in Step 3, we get: P(Good risk | No accident) = (1 - 0.05) × 0.20 / 0.825 ≈ 0.2424 P(Average risk | No accident) = (1 - 0.15) × 0.50 / 0.825 ≈ 0.3333 P(Bad risk | No accident) = (1 - 0.3) × 0.30 / 0.825 ≈ 0.2182 Therefore, the probabilities of policyholder A being in each risk class given no accidents in 2012 are: - Good risk: 24.24% - Average risk: 33.33% - Bad risk: 21.82%

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bayes' Theorem
In the context of insurance risk assessment, Bayes' Theorem provides us with a powerful tool for updating the probability estimates based on new information. This theorem relates the conditional and marginal probabilities of random events, allowing us to revise our beliefs when presented with evidence.

For example, if we want to know the probability that a person is a good risk, given that they had no accidents, Bayes' Theorem comes into play. We start by considering the initial probability of a person being a good risk, and how likely a good risk is to avoid accidents. Bayes' Theorem combines this information with the overall rate of accidents in the population to refine our estimate.

Specifically, the theorem states that the probability of an event B given A is equal to the likelihood of A given B times the initial likelihood of B, all divided by the likelihood of A. Mathematically, this is expressed as:
\[ P(B | A) = \frac{P(A | B) \times P(B)}{P(A)} \]
By applying this formula, we can update our beliefs about the risk category of an individual based on their accident record, and therefore make more informed insurance pricing or policy decisions.
Probability Calculation
Probability calculation is fundamental to assessing risk and making predictions in various industries, including insurance. In our exercise scenario, different risk classes (good, average, bad) have different probabilities of getting into an accident. To calculate the overall probability of an accident happening within a year, we use the Total Probability Theorem.

The Total Probability Theorem states that if you have several mutually exclusive scenarios that cover all possible outcomes, the total probability is the sum of the individual conditional probabilities of each scenario. The formula for our context looks like this:
\[ P(\text{Accident}) = P(\text{Accident | Good risk}) \times P(\text{Good risk}) + P(\text{Accident | Average risk}) \times P(\text{Average risk}) + P(\text{Accident | Bad risk}) \times P(\text{Bad risk}) \]
This allows us to understand the overall risk landscape and is crucial when determining insurance premiums and understanding the distribution of risk across a population.
Risk Assessment
Risk assessment in insurance involves quantifying the likelihood and potential cost of events like car accidents. By combining probability calculations and Bayes' Theorem, insurers can classify individuals into risk categories more accurately.

Each category is assigned a probability of filing a claim based on historical data. In our example, good risks have a lower probability of having an accident compared to bad risks. Insurers use this information to set premiums that are proportional to the level of risk, thus ensuring financial viability.

In practice, insurers collect and analyze large amounts of data on accident occurrences and other risk factors. They then apply probability calculations to this data to estimate risk levels. By performing these assessments, insurance companies strive to minimize their own risk while offering fair prices to policyholders. Such assessments are not only crucial for setting premiums but also for strategic planning and risk management within insurance companies.

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Most popular questions from this chapter

In any given year, a male automobile policyholder will make a claim with probability \(p_{m}\) and a female policyholder will make a claim with probability \(p_{f},\) where \(p_{f} \neq p_{m} .\) The fraction of the policyholders that are male is \(\alpha, 0<\alpha<1 .\) A policyholder is randomly chosen. If \(A_{i}\) denotes the event that this policyholder will make a claim in year \(i,\) show that $$ P\left(A_{2} | A_{1}\right)>P\left(A_{1}\right) $$ Give an intuitive explanation of why the preceding inequality is true.

There is a \(50-50\) chance that the queen carries the gene for hemophilia. If she is a carrier, then each prince has a \(50-50\) chance of having hemophilia. If the queen has had three princes without the disease, what is the probability that the queen is a carrier? If there is a fourth prince, what is the probability that he will have hemophilia?

The color of a person's eyes is determined by a single pair of genes. If they are both blue-eyed genes, then the person will have blue eyes; if they are both brown-eyed genes, then the person will have brown eyes; and if one of them is a blue-eyed gene and the other a brown-eyed gene, then the person will have brown eyes. (Because of the latter fact, we say that the brown-eyed gene is dominant over the blue-eyed one.) A newborn child independently receives one eye gene from each of its parents, and the gene it receives from a parent is equally likely to be either of the two eye genes of that parent. Suppose that Smith and both of his parents have brown eyes, but Smith's sister has blue eyes. (a) What is the probability that Smith possesses a blue eyed gene? (b) Suppose that Smith's wife has blue eyes. What is the probability that their first child will have blue eyes? (c) If their first child has brown eyes, what is the probability that their next child will also have brown eyes?

Consider an unending sequence of independent trials, where each trial is equally likely to result in any of the outcomes \(1,2,\) or \(3 .\) Given that outcome 3 is the last of the three outcomes to occur, find the conditional probability that (a) the first trial results in outcome 1 ; (b) the first two trials both result in outcome \(1 .\)

A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the first actuarial exam in June. If she passes that exam, then she will take the second exam in July, and if she also passes that one, then she will take the third exam in September. If she fails an exam, then she is not allowed to take any others. The probability that she passes the first exam is \(.9 .\) If she passes the first exam, then the conditional probability that she passes the second one is \(.8,\) and if she passes both the first and the second exams, then the conditional probability that she passes the third exam is .7. (a) What is the probability that she passes all three exams? (b) Given that she did not pass all three exams, what is the conditional probability that she failed the second exam?

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