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!