Originally Posted by
jamesmartinn Hi! I would greatly appreciate if someone could look these over for me:
Suppose Oils spills off the coast of BC in any given year has a Poisson Distribution with mean 3.
A) what is the chance that there will be exactly 2 spills in 2010?
Answer:
P(x=2) = (e^-3)(3^2)(1/2!) = 0.22404
B) what is the chance that there will be at least one spill over the next three years?
Answer:
our mean rate of 3 is for one year. For three years it is 9.
P(X>=1) = 1 - P(x=0)
= 1 - (9^0)*(e^-9)*(1/0!)
= 1 - e^-9
= 0.999876
C) if there are 3 oil spills over the next two years, what is the chance that exactly two will occur in 2010?
Answer:
Let X1 = # of spills in 2010, X2 = # of spills in 2011.
Then X1 ~ Poisson(3), X2~Poisson(3)
P(X1 = 2 | X1 + X2 = 3) ~ Binomial (3,p) where p = mu1 / mu1 + mu2
= (3 choose 2) (1/2)^2 (1/2)^1
= 0.375
Let X be the number of defects on a silicon wafer for computer chips manufactured by a high-tech company. Suppose defects occur randomly and uniformly over the surface of the wafer at a mean rate of 2 defects per wafer so that X follows a Poisson Distribution with mean 2. X ~ Poisson (2)
A) Find the probability that a wafer has at least two defects
Answer:
P(X >= 2) = 1 - P(X=0) - P(X=1)
= 0.59399415
B) Find the probability that 3 wafers have each 1 defect.
Answer:
I found the probability that one wafer has one defect, and cubed it due to independence.
I got 0.0198300174
C) Find the probability that if we examine 3 wafers, the TOTAL number of defects is exactly one.
Im thinking the following:
find the probability of exactly one defect using poisson distribution
then use this probability as "p" in
3 choose 1 * p^1 * (1-p)^2
Thanks!