MATHEMATICS COLLEGE

An insurance company writes policies for a large number of newly-licensed drivers each year. Suppose 40% of these are low-risk drivers, 40% are moderate risk, and 20% are high risk. The company has no way to know which group any individual driver falls in when it writes the policies. None of the low-risk drivers will have an at-fault accident in the next year, but 10% of the moderate-risk and 20% of the high-risk drivers will have such an accident. If a driver has an at-fault accident in the next year, what is the probability that he or she is high-risk

Answers

Answer 1

Answer:

Answer:

The probability that he or she is high-risk is 0.50

Step-by-step explanation:

P(Low risk) = 40% = 0.40

P( Moderate risk) = 40% = 0.40

P(High risk) = 20% = 0.20

P(At - fault accident | Low risk) = 0% = 0

P(At-fault accident | Moderate risk) = 10% = 0.10

P(At-fault accident | High risk) = 20% = 0.20

If a driver has an at-fault accident in the next year, what is the probability that he or she is high-risk. Hence, We need to calculate P( High risk | at-fault accident) = ?

Using Bayes' conditional probability theorem

P( High risk | at-fault accident) = ( P( High risk) * P(At-fault accident | High risk) ) / { P( Low risk) * P(At-fault accident | Low risk) +P( Moderate risk) * P(At-fault accident | Moderate risk) + P( High risk) * P(At-fault accident | High risk) }

P( High risk | at-fault accident)= (0.20 * 0.20) / ( 0.40 * 0 + 0.40 * 0.10 + 0.20 * 0.20 )

P( High risk | at-fault accident) = 0.04 / 0 + 0.04 + 0.04

P( High risk | at-fault accident) = 0.04 / 0.08

P( High risk | at-fault accident) = 0.50.

Answer 2

Answer:

Final answer:

The probability that a driver is high-risk given that they had an at-fault accident can be found using Bayes' theorem. Given the probabilities provided in the question, the probability is approximately 0.3333 or 33.33%.

Explanation:

To find the probability that a driver is high-risk given that they had an at-fault accident, we can use Bayes' theorem. Let's define the events:

A: Driver is high-risk
B: Driver has an at-fault accident

We are given the following probabilities:

P(A) = 0.20 (probability of a driver being high-risk)
P(B|A) = 0.20 (probability of an at-fault accident given that they are high-risk)
P(~A) = 0.80 (probability of a driver not being high-risk)
P(B|~A) = 0.10 (probability of an at-fault accident given that they are not high-risk)

Using Bayes' theorem, the probability of a driver being high-risk given that they had an at-fault accident is:

P(A|B) = (P(A) * P(B|A)) / (P(A) * P(B|A) + P(~A) * P(B|~A))

Substituting the given probabilities:

P(A|B) = (0.20 * 0.20) / (0.20 * 0.20 + 0.80 * 0.10) = 0.04 / (0.04 + 0.08) = 0.04 / 0.12 = 0.3333.

Therefore, the probability that a driver is high-risk given that they had an at-fault accident in the next year is approximately 0.3333 or 33.33%.

Learn more about probability here:

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Answers

Volume of sphere is given by:
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Answer:

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Step-by-step explanation:

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Suppose we have two bags with the numbers. Each bag has a total of 100 numbers. In the first bag there are 31 lucky numbers, in the second bag there are 18 lucky numbers. We want to add one more bag with 100 numbers to decrease the probability that a randomly selected number from a random bag is the lucky number. How many lucky numbers should be in the third bag?

Answers

Answer:

There should be at most 24 lucky numbers in the third bag.

Step-by-step explanation:

Initially, there are 200 numbers. Two bags with 100 each. There are 31+18 = 49 lucky numbers. So there is a 49/200 = 0.245 probability that a randomly selected number from a random bag is the lucky number.

Now with 300 numbers, we want this probability to be lower than 24.5%. So we should solve the following rule of three:

200 - 49

300 - x

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$x = 73.5$

With the third bag, the probability will be the same if 73.5-49 = 24.5 lucky numbers are added. So there should be at most 24 lucky numbers in the third bag.

A formal power series over R is a formal infinite sum f = X[infinity] n=0 anxn, where the coefficients an ∈ R. We add power series term-by-term, and two power series are the same if all their coefficients are the same. (We don’t plug numbers in for x, because we don’t want to worry about issues with convergence of the sum.) There is a vector space V whose elements are the formal power series over R. There is a derivative operator D ∈ L(V ) defined by taking the derivative term-by-term: D X[infinity] n=0 anxn ! = X[infinity] n=0 (n + 1)an+1xn What are the eigenvalues of D? For each eigenvalue λ, give a basis of the eigenspace E(D, λ). (Hint: construct eigenvectors by solving the equation Df = λf term-by-term.)

Answers

Answer:

Check the explanation

Step-by-step explanation:

where the letter D is the diagonal matrix with diagonal entries λ1,…,λn. Now let's assume V is invertible, that is, this particular given eigenvectors are linearly independent, you get M=VDV−1.

Kindly check the attached image below to see the step by step explanation to the question above.

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Answers

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