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22 November, 02:50

A car insurance company insures four types of drivers:

• Good drivers who have a 3% chance of getting in an accident each year

• Mediocre drivers have a 7% chance of getting in an accident in each year

• Atrocious drivers who have a 20% chance of getting in an accident in each year

• Ludicrously bad drivers who have a 75% chance of getting in an accident each year 1

70% of drivers are good, 20% are mediocre, 7% are atrocious, and 3% are ludicrously bad. A driver they insure gets in an accident, what is the likelihood he is a good, mediocre, atrocious, or ludicrously bad driver?

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  1. 22 November, 02:56
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    A driver they insure gets in an accident, the likelihood he is

    A good driver, P (G|GA) = 0.294

    A mediocre driver, P (M|GA) = 0.196

    A atrocious driver, P (A|GA) = 0.196

    A ludicrously bad driver, P (L|GA) = 0.315

    Step-by-step explanation:

    Let G represent good drivers, P (G) = 0.70

    M represent mediocre drivers, P (M) = 0.20

    A represent Atrocious drivers, P (A) = 0.07

    L represent Ludicrously bad drivers, P (L) = 0.03

    If we represent the probability of getting involved in an accident as P (GA),

    We're given in the question that the probability of getting in an accident, given one is a good driver, P (GA|G) = 0.03

    P (GA|M) = 0.07

    P (GA|A) = 0.20

    P (GA|L) = 0.75

    And we're told to find the probability that given an accident has occurred, the driver is a good driver, P (G|GA), moderate driver, P (M|GA), atrocious driver, P (A|GA) and ludicrously bad driver, P (L|GA).

    To do this, we will require the total probability of getting in an accident P (GA) which is the total of probability of each type of driver getting in an accident.

    P (GA) = P (G n GA) + P (M n GA) + P (A n GA) + P (L n GA)

    To obtain P (G n GA)

    Conditional probability P (GA|G) given in the question is represented mathematically as P (GA n G) / P (G).

    So, P (GA n G) = P (G n GA) = P (GA|G) * P (G) = 0.7 * 0.03 = 0.021

    P (GA n M) = P (M n GA) = P (GA|M) * P (M) = 0.07 * 0.2 = 0.014

    P (GA n A) = P (A n GA) = P (GA|A) * P (M) = 0.20 * 0.07 = 0.014

    P (GA n L) = P (L n GA) = P (GA|L) * P (L) = 0.75 * 0.03 = 0.0225

    P (GA) = 0.021 + 0.014 + 0.014 + 0.0225 = 0.0715

    So, our required probability now,

    P (G|GA) = P (G n GA) / P (GA) = 0.021/0.0715 = 0.294

    P (M|GA) = P (M n GA) / P (GA) = 0.014/0.0715 = 0.196

    P (A|GA) = P (A n GA) / P (GA) = 0.014/0.0715 = 0.196

    P (L|GA) = P (L n GA) / P (GA) = 0.0225/0.0715 = 0.315
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