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Sagot :
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
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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
[tex]P(A) = 0.01*0.6 + 0.05*0.3 + 0.09*0.1 = 0.03[/tex]
Probability of having an accident and being a medium risk driver:
0.05 of 0.3. So
[tex]P(A \cap B) = 0.05*0.3 = 0.015[/tex]
Desired probability:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.015}{0.03} = 0.5[/tex]
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
[tex]P(A \cap B) = 0.1*0.09 = 0.009[/tex]
The probability is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.009}{0.03} = 0.3[/tex]
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 [tex]p = 0.3[/tex]
Two accidents mean that [tex]n = 2[/tex]
This probability is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{2,0}.(0.3)^{0}.(0.7)^{2} = 0.49[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.49 = 0.51[/tex]
0.51 = 51% probability that at least one of them has been classified as high-risk.
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