Question

An insurance company classifies drivers as low risk, medium risk, high risk. Of those insured, 60% are low-risk, 30% are medium-risk, and 10% are high risk. After a study, the company finds that during a1-year period, 1% of the low risk drivers had an accident, 5% of the medium risk drivers had an accident, and 9% of the high-risk drivers had an accident.
Required:
a. If a driver had an accident during the year, find the probability that the driver is selected as a medium-risk driver.
b. If a driver who had an accident during the I-year period is selected, what is the probability that he has been classified as high-risk?
c. If two drivers who had an accident during the I -year period are selected, what is the probability that at least one of them has been classified as high-risk?

Answer 1

Answer:

a. 0.5 = 50% probability that the driver is selected as a medium-risk driver.

b. 0.3 = 30% probability that he has been classified as high-risk

c. 0.51 = 51% probability that at least one of them has been classified as high-risk.

Step-by-step explanation:

To solve this question, we need to understand conditional probability, for items a and b, and the binomial distribution, for item c.

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

a. If a driver had an accident during the year, find the probability that the driver is selected as a medium-risk driver.

Event A: Had an accident

Event B: Medium-risk driver

Probability of having an accident:

0.01 of 0.6(low risk)

0.05 of 0.3(medium risk)

0.09 of 0.1(high risk)

So

$P(A) = 0.01*0.6 + 0.05*0.3 + 0.09*0.1 = 0.03$

Probability of having an accident and being a medium risk driver:

0.05 of 0.3. So

$P(A \cap B) = 0.05*0.3 = 0.015$

Desired probability:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.015}{0.03} = 0.5$

0.5 = 50% probability that the driver is selected as a medium-risk driver.

b. If a driver who had an accident during the I-year period is selected, what is the probability that he has been classified as high-risk?

Event A: Had an accident

Event B: High risk driver.

From the previous item, we already know that P(A) = 0.03.

Probability of having an accident and being a high risk driver is 0.09 of 0.1. So

$P(A \cap B) = 0.1*0.09 = 0.009$

The probability is

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.009}{0.03} = 0.3$

0.3 = 30% probability that he has been classified as high-risk

c. If two drivers who had an accident during the I -year period are selected, what is the probability that at least one of them has been classified as high-risk?

0.3 are classified as high risk, which means that $p = 0.3$

Two accidents mean that $n = 2$

This probability is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.3)^{0}.(0.7)^{2} = 0.49$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.49 = 0.51$

0.51 = 51% probability that at least one of them has been classified as high-risk.