Multinomial Experiments
What if there are more than 2 possible outcomes?
(e.g., acceptable, scrap, rework)
That is, suppose we have:
 n independent trials
 k outcomes that are
 mutually exclusive (e.g., ♠, ♣, ♥, ♦)
 exhaustive (i.e., ∑all kpi = 1)
Then
n

 x1 x2
 p1 p2 ... pkxk
f(x1, x2, …, xk; p1, p2, …, pk, n) = 
 x1, x2 ,..., xk 
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 1
Multinomial Examples
 Example 5.7 refer to page 150
 Problem 22, page 152
 Convert ratio 8:4:4 to probabilities (8/16, 4/16)
x1 = _______
p1 = 0.50
x2 = _______
p2 = 0.25
x3 = _______
p3 = 0.25
 f( __, __, __; ___, ___, ___, __) =___.25, 8)
 = (8 choose 5,2,1)(0.5)5(0.25)2(0.25)1
 = 8!/(5!2!1!)* )(0.5)5(0.25)2(0.25)1
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
 = 21/256 or 0.082031
Slide 2
Binomial vs.Hypergeometric Distribution
 Replacement and Independence
 Binomial (assumes sampling “with replacement”)
and hypergeometric (sampling “without
replacement”)
 Binomial assumes independence, while
hypergeometric does not.
 Hypergeometric: The probability associated with
getting x successes in the sample (given k successes
in the lot.)
 k  N  k 
 

x nx 
P ( X  x )  h( x; N, n, k )   
N 
 
n 
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 3
Hypergeometric Example
 Example from Complete Business Statistics, 4th ed (McGraw-Hill)
 Automobiles arrive in a dealership in lots of 10. Five out of
each 10 are inspected. For one lot, it is known that 2 out of 10
do not meet prescribed safety standards. What is probability
that at least 1 out of the 5 tested from that lot will be found not
meeting safety standards?
 This example follows a hypergeometric distribution:
 A random sample of size n is selected without replacement
from N items.
 k of the N items may be classified as “successes” and N-k are
“failures.”
 The probability associated with getting x successes in the sample
 k  N  k 
(given k successes in the lot.)
 

JMB Ch. 5 Part 2 rev 2012b
 x  n  x 

P( X  x)  h( x; N , n, k )   
N
 
n 
EGR 252.001 2012 9th ed.
Slide 4
Solution: Hypergeometric Example
 In our example,
k = number of “successes” = 2
N = the lot size = 10
n = number in sample = 5
x = number found = 1 or 2
P ( X  x )  P ( X  1)  P ( X  2)
 2 10  2 
 

1 5 1 
h(1;10,5,2)  h( 2;10,5,2)   

10 
 
5 
 2 10  2 
 

 2  5  2 
10 
 
5 
P(X > 1) = 0.556 + 0.222 = 0.778
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 5
Expectations: Hypergeometric Distribution
 The mean and variance of the hypergeometric distribution
are given by
nk

N
N n
k 
k
2
 
* n * 1  
N 1
N
N
 What are the expected number of cars that fail inspection
in our example? What is the standard deviation?
μ
= nk/N = 5*2/10 = 1
σ2
σ
= (5/9)(5*2/10)(1-2/10) = 0.444
= 0.667
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 6
Additional problems …
A worn machine tool produced defective parts for a period
of time before the problem was discovered. Normal
sampling of each lot of 20 parts involves testing 6 parts and
rejecting the lot if 2 or more are defective. If a lot from the
worn tool contains 3 defective parts:
1. What is the expected number of defective parts in a
sample of six from the lot?
N = 20 n = 6 k = 3
μ = nk/N = 6*3/20 =18/20=0.9
2. What is the expected variance?
σ2 = (14/19)(6*3/20)(1-3/20) = 0.5637
3. What is the probability that the lot will be rejected?
P(X>2) = 1 – [P(0)+P(1)]
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 7
Binomial Approximation
 Note, if N >> n, then we can approximate the
hypergeometric with the binomial distribution.
 Example:
Automobiles arrive in a dealership in lots of 100. 5 out
of each 100 are inspected. 2 /10 (p=0.2) are indeed below safety
standards. What is probability that at least 1 out of 5 inspected will
not meet safety standards?
 Recall: P(X ≥ 1) = 1 – P(X < 1) = 1 – P(X = 0)
Hypergeometric distribution
1 - h(0;100,5,20)
= 0.6807
Binomial distribution
1 - b(0;5,0.2)
1 - 0.3277 = 0.6723
(See also example 5.12, pg. 155-6)
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 8
Negative Binomial Distribution
b*
 A binomial experiment in which trials are
repeated until a fixed number of successes
occur.
 Example:
Historical data indicates that 30% of all bits
transmitted through a digital transmission channel
are received in error. An engineer is running an
experiment to try to classify these errors, and will
start by gathering data on the first 10 errors
encountered.
What is the probability that the 10th error will occur
on the 25th trial?
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 9
Negative Binomial Equation
 This example follows a negative binomial distribution:
 Repeated independent trials.
 Probability of success = p and probability of failure = q = 1-p.
 Random variable, X, is the number of the trial on which the kth
success occurs.
 The probability associated with the kth success occurring
on trial x is given by,
 x  1 k x k
 p q , x  k, k  1, k  2,...
b * ( x; k, p )  
 k  1
Where,
k = “success number”
x = trial number on which k occurs
p = probability of success (error)
q=1–p
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 10
Example: Negative Binomial Distribution
What is the probability that the 10th error will
occur on the 25th trial?
k = “success number” = 10
x = trial number on which k occurs = 25
p = probability of success (error) = 0.3
q = 1 – p = 0.7
 25  1
(0.3)10 (0.7) 2510
b * (25;10,0.3)  
10  1 
 24 
b * (25;10,0.3)   (0.3)10 (0.7)15  0.037
9
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 11
Geometric Distribution
Continuing with our example in which p =
probability of success (error) = 0.3
What is the probability that the 1st bit received in
error will occur on the 5th trial?
This is an example of the geometric distribution,
which is a special case of the negative binomial
in which k = 1.
 The probability associated with the 1st success
occurring on trial x is
g ( x; p)  pq x 1
x  1 , 2 , 3, ...
 P = (0.3)(0.7)4 = 0.072
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 12
Additional problems …
A worn machine tool produces 1% defective parts. If we
assume that parts produced are independent:
1.What is the probability that the 2nd defective part will be the
6th one produced?
2.What is the probability that the 1st defective part will be seen
before 3 are produced?
3.How many parts can we expect to produce before we see the
1st defective part?
Negative binomial or geometric? Expected value = ?
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 13
Poisson Process
The number of occurrences in a given interval or
region with the following properties:
 “memoryless” ie number in one interval is independent of
the number in a different interval
 P(occurrence) during a very short interval or small region
is proportional to the size of the interval and doesn’t
depend on number occurring outside the region or
interval.
 P(X>1) in a very short interval is negligible
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 14
Poisson Process Situations
Number of bits transmitted per minute.
Number of calls to customer service in an hour.
Number of bacteria present in a given sample.
Number of hurricanes per year in a given region.
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 15
Poisson Distribution Probabilities
The probability associated with the number of
occurrences in a given period of time is given by,
e   t ( t ) x
p( x ;  t ) 
, x  0, 1, 2, ...
x!
Where,
λ = average number of outcomes per unit time or region
t = time interval or region
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 16
Service Call Example - Poisson Process
Example
An average of 2.7 service calls per minute are
received at a particular maintenance center.
The calls correspond to a Poisson process. To
determine personnel and equipment needs to
maintain a desired level of service, the plant
manager needs to be able to determine the
probabilities associated with numbers of
service calls.
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 17
Our Example: λ = 2.7 and t = 1 minute
 What is the probability that fewer than 2 calls will be
received in any given minute?
 The probability that fewer than 2 calls will be received in
any given minute is
P(X < 2) = P(X = 0) + P(X = 1)
e 2.7 (2.7) 0 e 2.7 (2.7)1
P( X  2) 

0!
1!
 The mean and variance are both λt, so
μ = λt =________________
 Note: Table A.2, pp. 732-734, gives Σt p(x;μ)
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 18
Service Call Example
(Part 2)
 If more than 6 calls are received in a 3-minute period,
an extra service technician will be needed to maintain
the desired level of service. What is the probability of
that happening?
 μ = λt = (2.7) (3)= 8.1
 8.1 is not in the table; we must use basic equation
 e 8.1 (8.1) 0 e 8.1 (8.1) 1
e 8.1 (8.1) 6 
p( X  6)  1  p( X  6)  1  



0!
1!
6!


 Suppose λt = 8; see table with μ = 8 and r = 6
P(X > 6) = 1 – P(X < 6) = 1 - 0.3134 = 0.6866
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 19
Frequency
Poisson Distribution
50
45
40
35
30
25
20
15
10
5
0
Calls per minute
JMB Ch. 5 Part 2 rev 2012b
EGR 252.001 2012 9th ed.
Slide 20