Q:

Government approval for a nuclear power plant on the California coast requires a hazard evaluation. Included is a probability analysis of various potentially damaging accidents or natural disasters. Compute the probability of at least one occurrence, (1) in a single year and (2) sometime in the next 100 years, from each of the following potentially damaging events, all of which arise from independent Poisson processes:(a) impact from an airplane crash, presumed to occur in the vicinity of the generator site at a mean annual rate of .0000011) P(at least 1 in a year) =2) P(at least 1 in a 100 years) =b) being hit by a large tsunami (tidal wave), known to occur once every 1,000 years with a further chance of 1/500 of hitting a particular location the width of the generator site1) P(at least 1 in a year) =2) P(at least 1 in a 100 years) =(c) an earthquake causing rupture in the reactor cooling system. This could be only from a Richter-8 or greater shock whose epicenter falls near the generator site. This event is judged to have a mean rate of .00001 per year.1) P(at least 1 in a year) =2) P(at least 1 in a 100 years) =

Accepted Solution

A:
Answer:See the step by step calculations below.Step-by-step explanation:(a) (1) P(at least 1 in a year) impliesThe provided mean is λ=0.000001.We need to compute Pr(X≥1). Therefore, the following is obtained:Pr(X≥1)=1−Pr(X<1)=1−Pr(X≤0)= 1 - 1=0which completes the calculation.(2) P(at least 1 in a 100 years) implies mean =100X0.000001=0.0001The provided mean is λ=0.0001.We need to compute Pr(X≥1). Therefore, the following is obtained:Pr(X≥1)=1−Pr(X<1)=1−Pr(X≤0)=1−0.9999 =0.0001which completes the calculation.(b) (1) P(at least 1 in a year) mean=1/500=0.002The provided mean is λ=0.002.We need to compute Pr(X≥1). Therefore, the following is obtained:Pr(X≥1)=1−Pr(X<1)=1−Pr(X≤0)=1−0.998 =0.002which completes the calculation.(2) P(at least 1 in a 100 years) impliesmean=100 X 0.002=0.2The provided mean is λ=0.2.We need to compute Pr(X≥1). Therefore, the following is obtained:Pr(X≥1)=1−Pr(X<1)=1−Pr(X≤0)=1−0.8187 =0.1813which completes the calculation.(c) (1)  P(at least 1 in a year) impliesThe provided mean is λ=0.00001.We need to compute Pr(X≥1). Therefore, the following is obtained:Pr(X≥1)=1−Pr(X<1)=1−Pr(X≤0)=1−1   =0(2) P(at least 1 in a 100 years) impliesMean = 100 X 0.00001=0.001Then, The provided mean is λ=0.001.We need to compute Pr(X≥1). Therefore, the following is obtained:Pr(X≥1)=1−Pr(X<1)=1−Pr(X≤0)=1−0.999  =0.001which completes the calculation.