1. 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
f(x1, x2, …, xk; p1, p2, …, pk, n) =
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

2. Multinomial Examples Example 5.7 refer to page 150
Problem 22, page 152
Convert ratio 8:4:4 to probabilities (8/16, 4/16)
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
= 21/256 or
x1 =
_______
p1 =
0.50
x2 =
p2 =
0.25
x3 =
p3 =
f( 5,2,1; 0.5, 0.25, 0.25, 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 = 21/256 or
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

3. 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.)
Note the difference between binomial (assumes sampling “with replacement”) and hypergeometric (sampling “without replacement”) Also, binomial assumes independence, while hypergeometric does not.
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

4. 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 (given k successes in the lot.)
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

5. Solution: Hypergeometric Example
In our example,
k = number of “successes” = n = number in sample = 5
N = the lot size = x = number found = 1 or 2
P(X > 1) = = 0.778
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

6. Expectations: Hypergeometric Distribution
The mean and variance of the hypergeometric distribution are given by
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 = (5/9)(5*2/10)(1-2/10) = 0.444
σ =
μ = nk/N = 5*2/10 = 1
σ2 =(5/9)(5*2/10)(1-2/10) = 0.444
σ = 0.667
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

7. 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:
What is the expected number of defective parts in a sample of six from the lot? N = 20 n = 6 k = μ = nk/N = 6*3/20 =18/20=0.9
What is the expected variance? σ2 = (14/19)(6*3/20)(1-3/20) =
What is the probability that the lot will be rejected?
P(X>2) = 1 – [P(0)+P(1)]
N = 20 n = 6 k = 3
μ = nk/N = 6*3/20 =18/20=0.9
σ2 = (14/19)(6*3/20)(1-3/20) =
P(X>2) = 1 – [P(0)+P(1)] = =0.2018
= P(2)+P(3) = [3 choose 2]*[17 choose 4] / [20 choose 6] + [3 choose 3]*[17 choose 3] / [20 choose 6]
= =
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

8. 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 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
Binomial distribution
1 - h(0;100,5,20) =
1 - b(0;5,0.2) =
JMB 2012
h(0;100,5,20) = (20 choose 0)(80 choose 5)/(100 choose 5) =
1-P(0) = =
From Table A1, n=5, p=0.2
b(0;5,0.2) =
1-P(0) = =
NOTE: If N = 200, p=0.2 and n=5, then the hypergeometric distribution yields P(X > 1) = 0.676
Using Excel, = 1 - HYPGEOMDIST(0,5,40,200)
We can see that the binomial approximation gets very close to the hypergeometric value as N gets very large relative to n.
(See also example 5.12, pg )
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

9. 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 2016
EGR JMB th ed.

10. 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,
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 2016
EGR JMB th ed.

11. 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
b*(15;10,0.3) = (24 choose 9)(.3)10(.7)15 = 0.037
JMB 2012
=NEGBINOMDIST(15,10,0.3)
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

12. 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
P = (0.3)(0.7)4 = 0.072
JMB 2012
(0.3)(0.7)4 = 0.072
=NEGBINOMDIST(4,1,0.3)
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

13. Additional problems …
A worn machine tool produces 1% defective parts. If we assume that parts produced are independent:
What is the probability that the 2nd defective part will be the 6th one produced?
What is the probability that the 1st defective part will be seen before 3 are produced?
How many parts can we expect to produce before we see the 1st defective part?
Negative binomial or geometric? Expected value = ?
JMB 2012
Problem x=6 k=2 x-1=5 k-1= b*(6:2,0.01) = (5 choose 1)(.01)2(.99)4 =
Problem P(X<3) = P(x=1)+P(x=2) = (0.01)*(0.99)1-1 + (0.01)*(0.99)2-1 =
0.01 =NEGBINOMDIST(0,1,0.01) =NEGBINOMDIST(1,1,0.01)
Problem 3
(Based on Theorem 5.3, pg. 160 μ = 1/p for geometric)
Problem μ = 1/p = 1/0.01 = 100
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

14. 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
memoryless  number in one interval is independent of the number in a different interval
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

15. 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.
memoryless  number in one interval is independent of the number in a different interval
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

16. Poisson Distribution Probabilities
The probability associated with the number of occurrences in a given period of time is given by,
Where,
λ = average number of outcomes per unit time or region
t = time interval or region
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

17. Service Call Example - Poisson Process
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 2016
EGR JMB th ed.

18. 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)
The mean and variance are both λt, so
μ = λt =________________
Note: Table A.2, pp , gives Σt p(x;μ)
P(x=0) = e /0! + e /1! =
μ = 2.7 (1) = 2.7
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

19. 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
Suppose λt = 8; see table with μ = 8 and r = 6
P(X > 6) = 1 – P(X < 6) = =
μ = 2.7*3 = 8.1 ≈ 8
see page 668 , with μ = 8 and r = 6, P(X < 6) =
P = =
9th ed: page 733
JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.

20. Poisson Distribution JMB Ch. 5 Part 2 rev 2016
EGR JMB th ed.