HONR 399
October 1, 2010
Chapter 16 Answers
1. Consider a Poisson random variable X with parameter λ = 5. What is the exact
probability that X > 2?
We could compute P (X > 2) = P (X = 3) + P (X = 4) + P (X = 5) + · · · , but there
is an easier way to calculate the answer than computing an infinite sum. Instead, we could
compute the probability of the complementary event:
P (X > 2) = 1 − P (X ≤ 2) = 1 − P (X = 0) − P (X = 1) − P (X = 2)
e−5 50 e−5 51 e−5 52
=1−
−
−
0!
1!
2!
25
= 1 − e−5 1 + 5 +
2
37 −5
=1− e
2
Now use your calculator to give this answer numerically (i.e., type the exact answer into
your calculator, and give the result).
We compute
P (X > 2) = 1 −
37 −5
e ≈ .8753.
2
So X > 2 about 87.53 percent of the time.
2. Suppose that the number of Roseate Spoonbills (a very rare bird in Indiana) that fly
overhead in 1 hour has a Poisson distribution with mean 2. Also suppose that the number
of Roseate Spoonbills is independent from hour to hour (e.g., the number of birds between
noon and 1 PM does not affect the number of birds between 1 PM and 2 PM, etc.).
A bird watcher sits and looks for the birds for 3 hours. How many of these birds does
she expect to see?
The number of birds each hour is Poisson, and the number of birds in different hours
are independent, so the number of birds during the entire 3-hour time period is Poisson too.
The expected number of birds in each hour is 2, so the expected number of birds in a 3-hour
time period is 6.
What is the probability that she sees exactly 5 of these birds during a 3 hour time period?
The number of birds X in a 3-hour time period (as discussed above) is Poisson with
expected value 6, and thus λ = 6. So the probability that she sees exactly 5 birds during a
−6 5
e−6 ≈ .1606.
3 hour time period is P (X = 5) = e 5!6 = 324
5
1
3. In a person’s lifetime, he plays a lottery game 10,000 times, but his chances of winning
each time are only 1 in 5,000. Approximate the probability that he wins at least one time
during his lifetime.
The number of times he wins in his life is a binomial random variable X with parameters
n = 10,000 and p = 1/5,000. So the probabilty that he wins at least once in his life is
actually
0 10,000
4999
10,000
1
≈ .8646918
P (X ≥ 1) = 1 − P (X = 0) = 1 −
5000
5000
0
(Just FYI: I did this approximate value on the computer.)
For the approximate value of P (X ≥ 1), we can use a Poisson random variable Y that
has parameter λ = np = (10,000)(1/5,000) = 2. So the approximate value is
e−2 20
= 1 − e−2 ≈ .8646647.
0!
So this approximation agrees with the actual value for the first four decimal places.
P (Y ≥ 1) = 1 − P (Y = 0) = 1 −
4. There is a certain disease in America which is very rare. Each person has a probability
of 1 in 100,000,000 of having the disease. Assume that there are 310,000,000 people in
America.
Approximate the probability that nobody in America actually has the disease.
The number of people who have the disease is a binomial random variable X with parameters n = 310,000,000 and p = 100,000,000. So the probabilty that nobody in America
has the disease is actually
0 310,000,000
99,999,999
310,000,000
1
P (X = 0) =
100,000,000
100,000,000
0
For the approximate value of P (X = 0), let Y be a Poisson random variable with parameter λ = np = (310,000,000)(1/100,000,000) = 3.1. So the approximate value is
P (Y = 0) =
e−3.1 3.10
= e−3.1 ≈ 0.0450492.
0!
Approximate the probability that at most 3 people (i.e., 3 or less) in America have the disease.
The approximate probability is
P (Y ≤ 3) = P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3)
e−3.1 3.10 e−3.1 3.11 e−3.1 3.12 e−3.1 3.13
+
+
+
=
0!
1!
2!
3!
≈ 0.6248399.
2