Answer:
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
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