Question

In: Statistics and Probability

An insurance company divided its clients into two categories, amateur and professional. The last year accident...

An insurance company divided its clients into two categories, amateur and professional. The last year accident report shows 10% of amateur drivers and 2.5% of professional drivers are involved in an accident. If 15% of drivers are amateur license and 85% are professional, find the following probabilities. For simplicity, assume year to year accident accidents for any driver are independent.

a) What is the probability of accident for a driver?
b) What is the probability that an amateur driver is involved in at least one accident during 3 years?
c) What is the probability of not having an accident for a professional driver during 3 years?
d) The annual insurance premium is computed according to being involved to at least one accident in 3 years. If a professional driver pay $1000 premium, how much the premium of amateur drivers should be?

Expert Solution

here let say

Amateur Driver = A

so, P(A) = 0.15

Professional Drive = B

so, P(B) = 0.85

Now if Getting accident done is Cthen

it is also given

P(Amateur Driver do accident) = P(C l A) = 0.10

P(Professional Driver do accident) = P(B l A) = 0.025

(a) P(Accident of a driver) = P(C) = P(C l A) * P(A) + P(C l B) * P(B)

= 0.10 * 0.15 + 0.025 * 0.85 = 0.03625

(b) P(An amateur driver is involved in at least one accident during 3 years) = 1 - P(No accident in 3 years)

=1 - (1 - 0.10)³ = 0.271

(c) P(Not having an accident for a professional driver during 3 years) = (1 - 0.025)³ = 0.9269

(d) Here annual insurance premium is computed according to being involved to at least one accident in 3 years

as we know

P(Amateur driver had at least one accident in 3 years) = 0.271

P(Professional Driver has at least one accident in 3 years) = 1 - 0.9269 = 0.07314

so here more probability of accident means more premium. Let say the premium of Amteur driver is X.

so here value of the premium is proportion to probability of accident/ (1 - probability of accident)

X/1000 = [(0.271/(1 - 0.271)/(0.07314/(1 -0.07314)]

X = $ 4711

orchestra answered 2 years ago

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