Question

The average amount of time until a car accident on a particular 60 mile stretch of road is 30 minutes. Assume (unreasonably) that car accidents are independent, and that two accidents cannot occur at the same time. You may assume X is distributed negative binomial.

(a) What is the probability of a car accident occurring in the first hour? 

(b) What is the probability of a car accident occurring between 15 and 45 minutes? 

(c) What is the variance of the time until a car accident occurs? 

(d) If a car accident has not happened 2 hours, what. is the probability it will happen in the next hour?

Answer 1

Here average amount for car to have an accident on a particular on a particular 60 mile stretch = 30 minutes

so the poisson parameter = 2 accidents per hour

(a) Here expected number of a car accident occuring in the first hour = 1 - Pr(No car accident occuring in first hour)

= 1 - POISSON (X = 0 ; 2)

= 1- e^-2 2⁰ /0!

= 1 - 0.1353 = 0.8647

(b) Here expected number of accidents to occur in 15 to 45 minutes (or 30 minutes ) = 1 accident

= 1 - Pr(x = 0 ; = 1) = 1 - e^-1 = 1 - 0.3679 = 0.6321

(c) Variance of time until a car accident occurs = 1/30² = 1/900 minutes²

(d) Here the probability it will happen in the next hour will not depend on the what happend in previous times.

so,

Pr(x >0 ; in next one hour l no accident in last 2 hours) = Pr(x > 0 ; 2) = 1 - Pr(x = 0 ; 2) = 1 - e^-2 = 1- 0.1353 = 0.8647