Problem 50 - A First Course in Probability

Question

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 \(1997,\) what is the probability that he or she is a good or average risk?

Step-by-Step Solution

Answer

The proportion of people who have accidents in a fixed year is 0.175 or 17.5%. The probability that policyholder $A$ is either a good or average risk given they had no accidents in 1997 is approximately 0.7455 or 74.55%.

Step 1: Identify the probabilities given

We are given the following probabilities: - Probability of having an accident for a good risk: P(Accident | Good) = 0.05 - Probability of having an accident for an average risk: P(Accident | Average) = 0.15 - Probability of having an accident for a bad risk: P(Accident | Bad) = 0.30 - Probability of a person being a good risk: P(Good) = 0.20 - Probability of a person being an average risk: P(Average) = 0.50 - Probability of a person being a bad risk: P(Bad) = 0.30

Step 2: Apply the law of total probability

We will use the law of total probability to find the proportion of people who have accidents in a fixed year. P(Accident) = P(Accident | Good) * P(Good) + P(Accident | Average) * P(Average) + P(Accident | Bad) * P(Bad)

Step 3: Calculate P(Accident)

Now, we can plug the values into the formula to find the probability of having an accident in a fixed year. P(Accident) = (0.05 * 0.20) + (0.15 * 0.50) + (0.30 * 0.30) P(Accident) = 0.01 + 0.075 + 0.09 P(Accident) = 0.175 So, the proportion of people who have accidents in a fixed year is 0.175 or 17.5%.

Step 4: Calculate probabilities of no accidents within risk groups

We need to find the probabilities of not having an accident for good, average, and bad risks: - P(No Accident | Good) = 0.95 (since 1 - 0.05 = 0.95) - P(No Accident | Average) = 0.85 (since 1 - 0.15 = 0.85) - P(No Accident | Bad) = 0.70 (since 1 - 0.30 = 0.70)

Step 5: Use Bayes' theorem

We want to find the probability that policyholder A is a good or average risk given that they had no accidents in 1997: P(Good or Average | No Accident). Since "Good or Average" can be treated as two separate events, we can find the probabilities separately and then add them together: - P(Good | No Accident) = (P(No Accident | Good) * P(Good)) / P(No Accident) - P(Average | No Accident) = (P(No Accident | Average) * P(Average)) / P(No Accident) First, we need to find the probability of not having an accident: P(No Accident). We can use the law of total probability again: P(No Accident) = P(No Accident | Good) * P(Good) + P(No Accident | Average) * P(Average) + P(No Accident | Bad) * P(Bad)

Step 6: Calculate P(No Accident)

Plugging the values into the formula, we get: P(No Accident) = (0.95 * 0.20) + (0.85 * 0.50) + (0.70 * 0.30) P(No Accident) = 0.19 + 0.425 + 0.21 P(No Accident) = 0.825

Step 7: Calculate P(Good | No Accident) and P(Average | No Accident)

Now we can find the probabilities: P(Good | No Accident) = (0.95 * 0.20) / 0.825 ≈ 0.2303 P(Average | No Accident) = (0.85 * 0.50) / 0.825 ≈ 0.5152

Step 8: Calculate P(Good or Average | No Accident)

Finally, we add the probabilities together: P(Good or Average | No Accident) = P(Good | No Accident) + P(Average | No Accident) ≈ 0.2303 + 0.5152 ≈ 0.7455 So, the probability that policyholder A is either a good or average risk given they had no accidents in 1997 is ≈ 0.7455 or 74.55%.

Key Concepts

Law of Total Probability

The Law of Total Probability is a powerful and essential theorem in probability theory. It helps us calculate the overall probability of an event by considering all possible ways that event can be achieved. For instance, in the context of insurance risk assessment, we are interested in finding the proportion of people expected to have accidents in a year.
To do this, we need to look at different risk categories: "good," "average," and "bad" risks.
Each category has a different probability of being involved in an accident.

• "Good" risk individuals have a probability of 0.05.
• "Average" risk individuals have a probability of 0.15.
• "Bad" risk individuals have a probability of 0.30.

Additionally, each group makes up a specific portion of the population: 20%, 50%, and 30% respectively.
The Law of Total Probability states that to find the probability of an accident occurring in the general population, you combine all the individual probabilities from each risk group weighted by their respective population proportions. This is formulated as:
\[P(Accident) = \left[ P(Accident | Good) \times P(Good) \right] + \left[ P(Accident | Average) \times P(Average) \right] + \left[ P(Accident | Bad) \times P(Bad) \right]\]
In this scenario, we calculated it to be 17.5%, meaning that 17.5% of the overall population is expected to have an accident within a year.

Conditional Probability

Conditional probability measures the likelihood of an event occurring given that another event has already occurred. This is where Bayes' Theorem comes handy, allowing us to update our understanding given new evidence.
Imagine you are interested in figuring out policyholder A's risk status, knowing they didn’t have any accidents in a particular year.
Conditional probability lets us reevaluate A's situation compared to the general population by considering any prior information about their group risk probabilities as well as the new "no accident" information.

• We defined individual probabilities of not having an accident: 0.95 for "good," 0.85 for "average," and 0.70 for "bad" risks.

Next, Bayes' Theorem is applied to determine the probability that A is a "good" or "average" risk given they had no accidents:
\[P(Good | No\ Accident) = \frac{P(No\ Accident | Good) \times P(Good)}{P(No\ Accident)}\]
and similarly for "average" risks. This results in a much clearer picture as the information gets updated based on the fact of having no accident in 1997, resulting in probabilities for A being 23.03% good and 51.52% average, totaling a 74.55% chance A is either a good or average risk.

Risk Assessment

Risk assessment is a methodical way to analyze potential risks by calculating the probability of an event, like an accident, occurring.
In the context of the insurance industry, it allows for categorizing individuals based on their likelihood of having an accident.
For instance, using known probabilities:

• People classified as "good" risks have a 5% chance of an accident.
• Those considered "average" have a 15% chance.
• "Bad" risks face a 30% chance.

With these classifications, insurance companies like the hypothetical one in our example can effectively assess and manage risks.
Risk assessment becomes crucial when determining policy premiums or evaluating the profitability of various insurance plans.
Here, it also involves continuously updating class assignments as more data becomes available, such as through conditional probability and Bayes' Theorem.
This reclassification helps insurers make more informed decisions, aligning policies with actual risk levels. It's a bit like tuning in to a radio station; you adjust based on the signal strength to get the clearest sound possible. The clearer your signal (or risk understanding), the better your decision making!