Question

rolde An individual who has automobile insurance from a company is randomly selected. Let Y moving violations for which the individual was cited during the last 3 years. The pmf of Y is be the number of 2 p(y) 0.6 0.25 0.10 0.05 (c) What is the probability that three randomly chosen individuals like that (ie., with automobile insur of them more than 1 violations? d) What is the probability that of three randomly chosen individuals with automobile insurance at least one of them has more than one violations

Answer 1

Here probability that any random individual will have more than 1 violation = P(x = 2) + P(x = 3) = 0.10 + 0.05= 0.15

so here we have three persons

so

Pr(All three have more than 1 violation) = 0.15 * 0.15 * 0.15 = 0.003375

(d) Here Probability that any random individual will have more than 1 violation = P(x = 2) + P(x = 3) = 0.10 + 0.05= 0.15

Now there are 3 persons

Pr(At least one of them has more than one violation) = 1 - Pr(None of them has more than one violations)

Pr(None of them has more than one violations) = (1 - 0.15)³= 0.614125

Pr(At least one of them has more than one violation) = 1 - Pr(None of them has more than one violations)

= 1 - 0.614125 = 0.3859