On Texas Avenue between University Drive and George Bush Drive, accidents occur according to a Poisson process at a rate of three accidents per week. (a) What is the probability that a given day has no accidents? (b) Whenever there is an accident, the risk is 1 in 8 that personal injury will result. What is the probability that February (which has four weeks) has at least one accident with a personal injury?

Answer:(a) The probability is 0.6514(b) The probability is 0.7769Step-by-step explanation:If the number of accidents occur according to a poisson process, the probability that x accidents occurs on a given day is:[tex]P(x)=\frac{e^{-at}*(at)^{x} }{x!}[/tex]Where a is the mean number of accidents per day and t is the number of days. So, for part (a), a is equal to 3/7 and t is equal to 1 day, because there is a rate of 3 accidents every 7 days.Then, the probability that a given day has no accidents is calculated as:[tex]P(x)=\frac{e^{-3/7}*(3/7)^{x}}{x!}[/tex][tex]P(0)=\frac{e^{-3/7}*(3/7)^{0}}{0!}=0.6514[/tex]On the other hand the probability that February has at least one accident with a personal injury is calculated as:P(x≥1)=1 - P(0) Where P(0) is calculated as:[tex]P(x)=\frac{e^{-at}*(at)^{x} }{x!}[/tex]Where a is equivalent to (3/7)(1/8) because that is the mean number of accidents with personal injury per day, and t is equal to 28 because 4 weeks has 28 days, so:[tex]P(x)=\frac{e^{-(3/7)(1/8)(28)}*((3/7)(1/8)(28))^{x}}{x!}[/tex][tex]P(0)=\frac{e^{-(3/7)(1/8)(28)}*((3/7)(1/8)(28))^{0}}{0!}=0.2231[/tex]Finally, P(x≥1) is:P(x≥1) = 1 - 0.2231 = 0.7769