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 .1

1) 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 site

1) 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 .1 per year.

1) p(at least 1 in a year) =

2) p(at least 1 in a 100 years) =

See the step by step calculations below.

Step-by-step explanation:

(a) (1) P(at least 1 in a year) implies

The 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

=0

which completes the calculation.

(2) P(at least 1 in a 100 years) implies mean =100X0.000001=0.0001

The 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 =0.0001

which completes the calculation.

(b) (1) P(at least 1 in a year)

mean=1/500=0.002

The 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 =0.002

which completes the calculation.

(2) P(at least 1 in a 100 years) implies

mean=100 X 0.002=0.2

The 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 =0.1813

which completes the calculation.

(c) (1)  P(at least 1 in a year) implies

The 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) implies

Mean = 100 X 0.00001=0.001

Then,

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  =0.001

which completes the calculation.

x^5/3y^1/3 the correct answer

hope it

where is the page?

step-by-step explanation: