91Ó°ÊÓ

First, we need to find the probability that a person chosen at random will have no accidents in a year. We can use the Law of Total Probability for this: \(P(X=0) = P(X=0 \mid \lambda = 2)P(\lambda = 2) + P(X=0 \mid \lambda = 3)P(\lambda = 3)\) We know that \(P(\lambda = 2) = 0.6\) and \(P(\lambda = 3) = 0.4\). Using the Poisson distribution formula to calculate probabilities for a given mean: \(P(X=k \mid \lambda) = \frac{e^{-\lambda} \cdot \lambda^k}{k!}\) Substituting the values, we get: \(P(X=0) = \frac{e^{-2} \cdot 2^0}{0!} (0.6) + \frac{e^{-3} \cdot 3^0}{0!} (0.4) = e^{-2}(0.6) + e^{-3}(0.4)\)

Now, we will do the same for the situation when a person has exactly 3 accidents in a year: \(P(X=3) = P(X=3 \mid \lambda = 2)P(\lambda = 2) + P(X=3 \mid \lambda = 3)P(\lambda = 3)\) Using the Poisson distribution formula again, we get: \(P(X=3) = \frac{e^{-2} \cdot 2^3}{3!} (0.6) + \frac{e^{-3} \cdot 3^3}{3!} (0.4) = e^{-2}\frac{8}{6}(0.6) + e^{-3}\frac{27}{6}(0.4)\)

To calculate the conditional probability \(P(X=3 \mid X_1=0)\), where \(X_1\) represents the number of accidents in the previous year, we can use Bayes' theorem: \(P(X=3 \mid X_1=0) = \frac{P(X_1=0 \mid X=3)P(X=3)}{P(X_1=0)}\) We know the values of \(P(X=3)\) and \(P(X_1=0)\) from the previous steps. To calculate \(P(X_1=0 \mid X=3)\), we consider that the events of having 0 accidents in the previous year and having 3 accidents this year are independent, since the Poisson distribution shows no memory property: \(P(X_1=0 \mid X=3) = P(X_1=0)\) Now, we can substitute all the values and calculate the conditional probability: \(P(X=3 \mid X_1=0) = \frac{P(X_1=0)P(X=3)}{P(X_1=0)} = P(X=3) = e^{-2}\frac{8}{6}(0.6) + e^{-3}\frac{27}{6}(0.4)\) #Final Answer, Probabilities: (a) 0 accidents, (b) 3 accidents, and Conditional Probability for 3 Accidents given 0 Accidents last year# \(P(X=0) = e^{-2}(0.6) + e^{-3}(0.4)\) \(P(X=3) = e^{-2}\frac{8}{6}(0.6) + e^{-3}\frac{27}{6}(0.4)\) \(P(X=3 \mid X_1=0) = e^{-2}\frac{8}{6}(0.6) + e^{-3}\frac{27}{6}(0.4)\)