SOLUTION: For the past 111 years, a certain state suffered 32 direct hits from major(category 3 to 5) hurricanes. Assume that this was typical and the number of hits per year follows a Poiss

Question 1086770: For the past 111 years, a certain state suffered 32 direct hits from major(category 3 to 5) hurricanes. Assume that this was typical and the number of hits per year follows a Poisson distribution.
What is the probability that the state will not be hit by any major hurricanes in a single& year?
What is the probability that the state will be hit by at least one major hurricane in a single year?
What is the probability that the state will be hit by at least three major hurricanes in a single& year, as happened several years ago?
What is the probability that the state will be hit by at least two major hurricanes in a single year, as happened last year?

Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!
Question:
For the past 111 years, a certain state suffered 32 direct hits from major(category 3 to 5) hurricanes. Assume that this was typical and the number of hits per year follows a Poisson distribution.
(a) What is the probability that the state will not be hit by any major hurricanes in a single& year?
(b) What is the probability that the state will be hit by at least one major hurricane in a single year?
(c) What is the probability that the state will be hit by at least three major hurricanes in a single& year, as happened several years ago?
(d) What is the probability that the state will be hit by at least two major hurricanes in a single year, as happened last year?

Solution:
Given Poisson distribution, assuming historical trends up to the last century still apply today.
Mean number of direct hits per year, λ = 32/111
PMF(k,λ)=λ^k (e^(-λ)/k!

(a) k=0
P(k=0)=λ^k (e^(-λ)/k!
=32/111^0(e^(-32/111))/0!
=1*(e^(-32/111))/1
=e^(-32/111)
=0.7495

(b) k>=1
P(k)=1-P(k=0)=1-0.7495=0.2505

(c) P(k>=3)=1-(P(k=0)+P(k=1)+P(k=2))
=1-(0.7495+0.2161+0.0311)
=1-0.9967
=0.0033

(d) P(k>=2)=1-(P(k=0)+P(k=1))
=1-(0.7495+0.2161)
=1-0.9656
=0.0344
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