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)}](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D)
In which
P(B|A) is the probability of event B happening, given that A happened.
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}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
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)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
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](https://tex.z-dn.net/?f=P%28A%29%20%3D%200.01%2A0.6%20%2B%200.05%2A0.3%20%2B%200.09%2A0.1%20%3D%200.03)
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](https://tex.z-dn.net/?f=P%28A%20%5Ccap%20B%29%20%3D%200.05%2A0.3%20%3D%200.015)
Desired probability:
![P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.015}{0.03} = 0.5](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D%20%3D%20%5Cfrac%7B0.015%7D%7B0.03%7D%20%3D%200.5)
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](https://tex.z-dn.net/?f=P%28A%20%5Ccap%20B%29%20%3D%200.1%2A0.09%20%3D%200.009)
The probability is
![P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.009}{0.03} = 0.3](https://tex.z-dn.net/?f=P%28B%7CA%29%20%3D%20%5Cfrac%7BP%28A%20%5Ccap%20B%29%7D%7BP%28A%29%7D%20%3D%20%5Cfrac%7B0.009%7D%7B0.03%7D%20%3D%200.3)
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](https://tex.z-dn.net/?f=p%20%3D%200.3)
Two accidents mean that ![n = 2](https://tex.z-dn.net/?f=n%20%3D%202)
This probability is:
![P(X \geq 1) = 1 - P(X = 0)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%201%29%20%3D%201%20-%20P%28X%20%3D%200%29)
In which
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{2,0}.(0.3)^{0}.(0.7)^{2} = 0.49](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B2%2C0%7D.%280.3%29%5E%7B0%7D.%280.7%29%5E%7B2%7D%20%3D%200.49)
![P(X \geq 1) = 1 - P(X = 0) = 1 - 0.49 = 0.51](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%201%29%20%3D%201%20-%20P%28X%20%3D%200%29%20%3D%201%20-%200.49%20%3D%200.51)
0.51 = 51% probability that at least one of them has been classified as high-risk.