We use optional analytics, advertising and affiliate technologies to
measure usage, attribute advertising and improve EverythingQuant.
You can accept or reject non-essential tracking.
An insurance company classifies insured policyholders into accident prone or non-accident prone. Their current risk model works with the following probabilities:
- The probability that an accident prone insured has an accident within a year is 0.4, and
- The probability that a non-accident prone insured has an accident within a year is 0.2.
If 30% of the population is accident prone, what is the probability that a policy holder will have an accidental within a year?
Further, suppose now that the policy holder has had an accident within one year. What is the probability that they are accident prone?
Solution
\[
A = \text{policyholder is accident prone}
\]
\[
N = \text{policyholder is non-accident prone}
\]
\[
C = \text{policyholder has an accident within a year}.
\]
The given probabilities are
\[
P(A) = 0.3, \qquad P(N) = 0.7
\]
\[
P(C \mid A) = 0.4, \qquad P(C \mid N) = 0.2.
\]
First find the overall probability that a policyholder has an accident.
Using the law of total probability,
\[
P(C) = P(C \mid A)P(A) + P(C \mid N)P(N).
\]
\[
P(C) = (0.4)(0.3) + (0.2)(0.7).
\]
\[
P(C) = 0.26.
\]
Now suppose we know that the policyholder has had an accident.
We want the probability they are accident prone:
\[
P(A \mid C).
\]
Using Bayes' theorem,
\[
P(A \mid C) =
\frac{P(C \mid A)P(A)}{P(C)}.
\]
\[
P(A \mid C) =
\frac{(0.4)(0.3)}{0.26}.
\]
\[
P(A \mid C) \approx 0.462.
\]
Thus,
\[
P(\text{accident within a year}) = 0.26
\]
and the probability that a policyholder who had an accident is accident prone is approximately
\[
0.462.
\]