An insurance company classifies each driver base on his/her driving record, as a

Question

An insurance company classifies each driver base on his/her driving record, as an excellent risk, a good risk, a medium risk, or a poor risk. Of those currently insured by the company, 8% are excellent risk, 32% are good risks, 50% are medium risks, and 10% are poor risks. In a one year period, the probability that a driver will get at least one citation is 0.06 for excellent risks, 0.10 for good risks, 0.14 for medium risks, and 0.20 for poor risks. (i) What is the probability that a randomly selected driver insured by the company will get at least one citation in the year 2008? (ii) Given that a randomly selected driver is given NO citation in the year 2008, what is the probability that the driver is either an excellent or a good risk?

Explanation / Answer

(i)

Let E, G, M, P be the event of insured excellent risk, good risk medium risk and poor risk respectively.

Then, P(E) = 0.08 , P(G) = 0.32, P(M) = 0.5 , P(P) = 0.1

Let the probability that a driver will get at least one citation be P(C)

Then, the conditional probability are

P(C | E) = 0.06 , P(C | G) = 0.1, P(C | M) = 0.14 , P(C | P) = 0.2

By law of total probability,

P(C) = P(C | E) P(E) + P(C | G) P(G) + P(C | M) P(M) + P(C | P) P(P)

= 0.06 * 0.08 + 0.1 * 0.32 + 0.14 * 0.5 + 0.2 * 0.1

= 0.1268

Hence, the probability that a driver will get at least one citation is 0.1268

(ii)

We need to calculate the conditional probabilities P(E | ~C) and P(G | ~C)

where ~C is the event that the driver has not received the citation.

P(~C) = 1 - P(C) = 1 - 0.1268 = 0.8732

and  P(~C | E) = 1 - P(C | E) = 1 - 0.06 = 0.94

and P(~C | G) = 1 - P(C | G) = 1 - 0.1 = 0.9

By Bayes rule,

P(E | ~C) = P(~C | E) P(E) / P(~C)

and P(G | ~C) = P(~C | G) P(G) / P(~C)

P(E | ~C) = 0.94 * 0.08 / 0.8732 = 0.0861

P(G | ~C) = 0.9 * 0.32 / 0.8732 = 0.3298

As, the events E and G are disjoint (An insured can neither be excellent or good at the same time), so the probability that a randomly selected driver is given NO citation in the year 2008 is either an excellent or a good risk is

P(E | ~C)  + P(G | ~C)  = 0.0861 + 0.3298 = 0.4159

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