Accidents occur at a busy intersection at the rate of two per year. What is the probability that it will be at least one year before the next accident at the intersection? Compute the answer using the following two methods: 1. Let X be the number of accidents in the next year. Find the distribution of X and calculate P(X=0). 2. Let T be the amount of time until the next accident. Find the distribution of T and calculate P(T>1).

Answer:1. P(X=0)=0.1352. P(T>1)=0.271Step-by-step explanation:1. Let X be the number of accidents in the next year. Find the distribution of X and calculate P(X=0).The rigth distribution to describe this type of event as number of accidents per unit of time is the Poisson distribution.[tex]P(x=k)\frac{\lambda^ke^{-\lambdat}}{k!}[/tex]In this case k=0 accidents, parameter λ=2 events/year and t=1 year:[tex]P(X=0)=\frac{2^0e^{-2}}{0!}=\frac{1*e^{-2}}{1} = 0.135[/tex]2. Let T be the amount of time until the next accident. Find the distribution of T and calculate P(T>1).In this case, the time between events can be best described by an exponential distribution:[tex]P(X>t)=e^{-\lambda t}[/tex]In this case parameter λ=2 events/year and t=1 year:[tex]P(X>1)= e^{-2}=0.271[/tex]