```Chapter 4 Discrete Random Variables Part II
Ex. Suppose we have 20 computers. Unknown to you, 4 are
defective and will explode when you turn them on. You are
to test a random sample of 3 computers.
What is the probability that exactly 1 of the computers
explodes?
What is the probability that exactly 2 of the computers
explodes?
What is the probability that exactly 3 of the computers
explodes?
What is the probability that exactly 0 of the computers
explodes?
You first thought may be to use the binomial distribution, with
n = 3, p = 4/20 = 1/5 = .2. But this is not correct because the
population (20) is not large.
Sure the probability of selecting a defective computer on the
first selection is 4/20, but what about the second selection?
If you selected the a defective computer on the first try, then
the probability of selecting a defective computer on the
second selection is 3/19 = .158, but if you did not select a
defective computer on the first try, then the probability of
selecting a defective computer on the second selection is
4/19 = .210.
This violates the assumption of independence and a
constant p of the binomial distribution. So how to do this
problem:
Use the Hyper-Geometric Distribution.
X is the random variable which counts the number of
successes in the sample of size n from a population of size
N and r is the total number of successes in the population. It
is much like the Binomial, but used when
n > .05 N.
The pdf for the Hyper-Geometric is:
∗
( = ) =

Where X is the DRV counting the number of successes in
the sample, r is the number of successes in the population.
The other number can actually be calculated from the
previous, y = n – x which is the number of failures in the
sample and b = N – r = number of failures in the population.
max[0, n – b] ≤ x ≤ min [r, n]
In the computer example above:
N = 20, n = 3, r = 4 where X counts the number of defective
computers.
P(X = 1) = 4 C 1 * 16 C 2 / 20 C 3 = 480 / 1140 = .421
x
0
1
2
3
P(X = x)
0.491
0.421
0.084
0.004
The above probabilities were found using the
HYPGEOMDIST function in Excel. Sad to say that your
calculators do not have a Hyper-Geometric function.
Note that as N increases the Hyper-Geometric distribution
approaches the Binomial distribution. In fact most problems
that we use the Binomial for, are theoretically HyperGeometric. However, we rarely know the population size, N,
and the approximation is very good.
Note also that we did problems like this already in chapter 3
when we did combination problems.
Ex. There are 15 students in a class. There are 8 women
and 7 men. 4 are randomly selected to be on a committee.
Let X be the number of women on the committee. Find the
probability distribution of X.
N = 15, n = 4, r = 8. Possible values for X = 0, 1, 2, 3, 4
x
0
1
2
3
4
P(X = x)
0.026
0.205
0.431
0.287
0.051
Also note that if X is Hyper-Geometric with N, n and r then
rn
E( X ) 
N
r ( N  r ) n( N  n)
Var ( X ) 
N 2 ( N  1)
so for the previous example:
E(X) = 8*4/15 = 2.133
Var(X) = 8(7)(4)(11)/225*14 = .783
σ = .884
In class examples:
The Geometric Distribution:
A Geometric RV: counts the number of “trials” necessary to
realize the first “success”. Trials mean the number of times
the experiment is run.
Success is not necessarily a good thing, just what you are
counting.
If X has a geometric distribution, then the pmf is
p(X= x) = p(1 – p)x – 1
for x = 1, 2, 3, …
p(X=x) = 0
otherwise
1
() =
() =
1−
2
p is the probability of a success at each trial.
So if p = .75 then the P(X = 2) = p(2) = .75 (.25) = .1875
Characteristics of a Geometric Random Variable:
1. The experiment consists of identical trials.
2. Each trial results in either a Success or Failure
3. The trials are independent
4. The probability of success, p, is constant from trial to
trial
Note how this differs from Binomial:
1. Binomial has a fixed number of trials. Geometric does
not.
2. Binomial counts number of successes. Geometric
counts number of trial until first success.
On the TI83/84
If p = .75 and you want the P(X = 2) [2nd][DISTR] go down to
geometpdf(p, x)  geometpdf(.75, 2)[ENTER] = .1875
On the TI83/84
If p = .75 and you want the P(X ≤ 2)
[2nd][DISTR] go down to geometcdf(p, x)  geometcdf(.75,
2)[ENTER]
.9375
P(X ≤ 2) = P(X = 1) + P(X = 2)
Example:
fracture occurs. 20% of the fractures occur in the beam and
not the weld. If X = the number of trials up to and including
the 1st beam fracture? Find the probability that the first test
with a beam fracture is the 3rd test. Find the probability that
the first test with a beam fracture is at most on the 3rd test.
What are the mean and standard deviation of X
X ~ Geom(p = .20)
P(X = 3) = .128
P(X ≤ 3) = .488
E(X) = 1/.2 = 5
Var(X) = .8/.04 = 20
The Poisson Distribution
Characteristics:
1.
The experiment consists of counting the number of
times a certain event occurs during a specific time or in an
area.
2.
The Probability that the event occurs in a given time
or area is the same for all units.
3.
The number of events that occur in one unit is
independent of the other units of time or area.
4.
The mean or expected number of events in each unit
is denoted by the Greek letter lambda = μ. Other books use
λ
If Y has a Poisson Distribution with mean μ then the pdf for Y
is:
( = ) =
−
!
for k = 0, 1, 2, …
E(Y) = μ and Var(Y) = μ
Ex.
The average number of occurrences of a certain
disease in a certain population is 2 per 1000 people. In a
random sample of 1000 people from this population, what is
the probability that exactly 1 person in the sample has the
disease? What is the probability that at least 1 person in the
sample has the disease?
Y = counts the number of people in the sample with the
disease.
Y has a Poisson distribution with μ = 2.
P(Y = 1) = 21 * e -2 / 1! = .270
P(Y ≥ 1) = 1 – P(Y = 0) = 1 - .135 = .865
On the TI83/84 page 214
[2nd] [DIST] (VARS key)
C: poissonpdf(μ, k)
poissonpdf(μ, k) gives P(Y = k)
for the last example:
poissonpdf(2, 1) = .270
poissonpdf(2, 0) = .135
On the TI83/84
[2nd] [DIST] (VARS key)
poissoncdf(μ, k)
poissoncdf(μ, k) gives P(Y ≤ k)
In Class examples:
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