Answer:
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.
This mathematical problem falls under the subject of Mathematical Probabilities. There are various kinds of probabilities in mathematics. This one is related to Poisson Distribution.
The Poisson Distribution is a discrete probability distribution in the theory of probability and statistics that examines the probability of occurrence of a number of events in a determined interval of time or space in relation to a constant mean rate and exclusive of the time of the previous event.
Solution (a1) - The probability that there is at least one occurrence of a hazard in a single year is given as:
Given that the mean is λ=0.000001, the function is given as Pr (X ≥ 1).
Therefore,
Pr (X ≥ 1) = 1 − Pr (X < 1) = 1 − Pr (X ≤ 0)
= 1 - 1 = 0
Hence the probability of a hazard occurring in a single year is Zero.
Solution (2) - The probability that there is an occurrence of a hazard sometime in the next 100 years, from each of the following potentially damaging events, all of which arise from independent Poisson processes is given as:
Where Pr (probability of occurrence) is at least once in 100 years, this means that
λ(Mean Rate) = 100 * 0.00001 = 0.0001
Hence Pr (X ≥ 1) = 1 − Pr (X < 1) = 1 − Pr (X ≤ 0)
= 1 - 0.9999, Hence
The probability of occurrence is at least once in 100 years (Pr) = 0.0001.
Solution (b1) Pr (at least 1 in a year)
The mean is given as 1/500
Hence, λ = 0.002.
Therefore, Pr(X ≥ 1) = 1 − Pr (X < 1)
= 1 − Pr(X ≤ 0)
= 1 -0.998
Pr (at least 1 in a year) = 0.002
Solution (b2) Pr (at least 1 in 100 years)
This means that we are looking at a mean = 100 x 0.002.
That is λ = 0.2
Pr(X ≥ 1 = 1 − Pr(X < 1)
= 1 − Pr( X ≤ 0)
= 1−0.8187
= 0.1813
Solution (c1) Pr (at least 1 in 1 year)
Recall that the mean is λ = 0.00001. To compute for Pr(X ≥ 1), we say:
Pr(X ≥ 1) = 1 − Pr( X < 1)
= 1 − Pr(X ≤ 0)
= 1-1
=0
Solution (c2) Pr (at least 1 in 100 years)
λ = 100 X 0.00001 = 0.001; Hence, Pr (X ≥ 1)will be computed as;
Pr(X ≥ 1) = 1 − Pr(X < 1) = 1 − Pr( X ≤ 0)
= 1 - 0.999
= 0.001
Learn more about probabilities at:
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