Guided Practice: Conditional Probability, Independence, and Bayes' Theorem in Actuarial and Health Contexts

Q11. Probability of Neither Collision Nor Disability Coverage

Background

Topic: Conditional Probability, Independence, and Complements

This question tests your understanding of how to use information about independent events, their probabilities, and complements to find the probability that neither of two events occurs.

Key Terms and Formulas

• Let $C$ = collision coverage, $D$ = disability coverage.

• Independence: $P(C \cap D) = P(C)P(D)$

• Complement Rule: $P(C' \cap D') = 1 - P(C \cup D)$

• Addition Rule: $P(C \cup D) = P(C) + P(D) - P(C \cap D)$

Step-by-Step Guidance

1. Let $P(D) = p$. Since collision is twice as likely as disability, $P(C) = 2p$.

2. Since $C$ and $D$ are independent, $P(C \cap D) = P(C)P(D) = 2p^2$.

3. From the problem, $P(C \cap D) = 0.15$. Set $2p^2 = 0.15$ and solve for $p$ (but do not compute the final value yet).

4. Find $P(C)$ using the value of $p$ you found above.

5. Calculate $P(C \cup D)$ using the addition rule.

Try solving on your own before revealing the answer!

Final Answer: 0.67

Using the complement rule, $P(C' \cap D') = 1 - P(C \cup D) = 0.67$.

This means there is a 67% chance an owner purchases neither coverage.

Q13. Probability of No Risk Factors Given Not A

Background

Topic: Conditional Probability and Law of Total Probability

This question involves using given probabilities for combinations of risk factors and applying conditional probability to find the probability that a woman has none of the three risk factors, given she does not have risk factor A.

Key Terms and Formulas

• Let $A$, $B$, $C$ be the three risk factors.

• Conditional Probability: $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$

• Law of Total Probability for $P(A')$ and $P($none$\cap A')$

Step-by-Step Guidance

1. List all possible combinations of risk factors that do not include $A$ (i.e., only $B$, only $C$, $B$ and $C$, or none).

2. Sum the probabilities for all cases where $A$ is not present to find $P(A')$.

3. Identify the probability that a woman has none of the three risk factors (i.e., not $A$, not $B$, not $C$).

4. Apply the conditional probability formula: $P($none$|A') = \frac{P($none$\cap A')}{P(A')}$.

Try solving on your own before revealing the answer!

Final Answer: 0.311

After calculating the relevant probabilities and applying the conditional probability formula, the answer is 0.311.

Q14. Probability of More Than One Claim

Background

Topic: Recurrence Relations and Probability Distributions

This question tests your ability to use a recurrence relation to model probabilities and to sum probabilities for discrete events.

Key Terms and Formulas

• Recurrence relation: $p(n+1) = 0.2p(n)$ for $n \geq 0$

• Probability of more than one claim: $P(N > 1) = 1 - p(0) - p(1)$

Step-by-Step Guidance

1. Express $p(1)$ in terms of $p(0)$ using the recurrence relation.

2. Express $p(2)$ in terms of $p(0)$, and so on, to find a general formula for $p(n)$.

3. Use the fact that the sum of all probabilities must be 1: $\sum_{n=0}^{\infty} p(n) = 1$.

4. Solve for $p(0)$ using the geometric series formula.

5. Find $p(1)$ and then set up $P(N > 1) = 1 - p(0) - p(1)$.

Try solving on your own before revealing the answer!

Final Answer: 0.16

After solving for $p(0)$ and $p(1)$, the probability that a policyholder files more than one claim is 0.16.

Q19. Probability Driver Was Age 16-20 Given Accident

Background

Topic: Bayes' Theorem and Conditional Probability

This question asks you to use Bayes' Theorem to find the probability that a driver was in a certain age group, given that an accident occurred.

Key Terms and Formulas

• Bayes' Theorem: $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$

• Law of Total Probability: $P(B) = \sum_{i} P(B|A_i)P(A_i)$

Step-by-Step Guidance

1. Let $A$ = driver is age 16-20, $B$ = accident occurs.

2. Find $P(B|A)$ and $P(A)$ from the table.

3. Calculate $P(B)$ using the law of total probability, summing over all age groups.

4. Apply Bayes' Theorem to find $P(A|B)$.

Try solving on your own before revealing the answer!

Final Answer: 0.13

After applying Bayes' Theorem, the probability is 0.13.

Q20. Probability Deceased Policyholder Was Ultra-Preferred

Background

Topic: Bayes' Theorem and Conditional Probability

This question requires you to use Bayes' Theorem to find the probability that a deceased policyholder was ultra-preferred, given the probabilities and proportions for each policy type.

Key Terms and Formulas

• Bayes' Theorem: $P(U|D) = \frac{P(D|U)P(U)}{P(D)}$

• Law of Total Probability: $P(D) = \sum_{i} P(D|T_i)P(T_i)$

Step-by-Step Guidance

1. Let $U$ = ultra-preferred, $D$ = died in the next year.

2. Find $P(D|U)$ and $P(U)$ from the problem statement.

3. Calculate $P(D)$ using the law of total probability, summing over all policy types.

4. Apply Bayes' Theorem to find $P(U|D)$.

Try solving on your own before revealing the answer!

Final Answer: 0.0141

After applying Bayes' Theorem, the probability is 0.0141.

Q21. Probability Patient Was Serious Given Survival

Background

Topic: Conditional Probability and Law of Total Probability

This question asks you to use conditional probability to find the probability that a patient was categorized as serious, given that they survived.

Key Terms and Formulas

• Let $S$ = serious, $V$ = survived.

• Conditional Probability: $P(S|V) = \frac{P(S \cap V)}{P(V)}$

• Law of Total Probability for $P(V)$

Step-by-Step Guidance

1. Find the probability that a patient was serious and survived: $P(S \cap V) = P(S) \times P($survived$|S)$.

2. Calculate $P(V)$ by summing the probabilities of survival for all categories (critical, serious, stable).

3. Apply the conditional probability formula: $P(S|V) = \frac{P(S \cap V)}{P(V)}$.

Try solving on your own before revealing the answer!

Final Answer: 0.29

After calculating the relevant probabilities, the answer is 0.29.

Q22. Probability Participant Was Heavy Smoker Given Death

Background

Topic: Bayes' Theorem and Conditional Probability

This question involves using Bayes' Theorem to find the probability that a participant was a heavy smoker, given that they died during the study period.

Key Terms and Formulas

• Bayes' Theorem: $P(H|D) = \frac{P(D|H)P(H)}{P(D)}$

• Law of Total Probability: $P(D) = \sum_{i} P(D|G_i)P(G_i)$

Step-by-Step Guidance

1. Let $H$ = heavy smoker, $D$ = died during the study.

2. Assign probabilities for each group (heavy, light, nonsmoker) and their death rates based on the information given.

3. Calculate $P(D)$ using the law of total probability.

4. Apply Bayes' Theorem to find $P(H|D)$.

Try solving on your own before revealing the answer!

Final Answer: 0.42

After applying Bayes' Theorem, the probability is 0.42.

Q27. Probability Automobile Was Model Year 2014 Given Accident

Background

Topic: Bayes' Theorem and Conditional Probability

This question asks you to use Bayes' Theorem to find the probability that a car involved in an accident was a 2014 model, given the proportions and accident probabilities for each model year.

Key Terms and Formulas

• Bayes' Theorem: $P(2014|A) = \frac{P(A|2014)P(2014)}{P(A)}$

• Law of Total Probability: $P(A) = \sum_{i} P(A|Y_i)P(Y_i)$

Step-by-Step Guidance

1. Let $A$ = accident, $Y_i$ = model year $i$.

2. Find $P(A|2014)$ and $P(2014)$ from the table.

3. Calculate $P(A)$ using the law of total probability, summing over all model years.

4. Apply Bayes' Theorem to find $P(2014|A)$.

Try solving on your own before revealing the answer!

Final Answer: 0.22

After applying Bayes' Theorem, the probability is 0.22.