Discrete Distributions

1
Discrete Distributions
2
Discrete Distributions

• What is the binomial distribution?
• The binomial distribution is a discrete
  probability distribution. It is a distribution
  that governs the random variable, X, which is the
  number of successes that occur in "n" trials.
• The binomial probability distribution gives us
  the probability that a success will occur x times
  in the n trials, for x 0, 1, 2, , n.
• Thus, there are only two possible outcomes. It is
  conventional to apply the generic labels
  "success" and "failure" to the two possible
  outcomes.

3
Discrete Distributions

• A success can be defined as anything! "the axle
  failed" could be the definition of success in an
  experiment testing the strength of truck axles.
• Examples
• 1. A coin flip can be either heads or tails
• 2. A product is either good or defective
•  
• Binomial experiments of interest usually involve
  several repetitions or trials of the same basic
  experiments. These trials must satisfy the
  conditions outlined below

4
Discrete Distributions

• When can we use it?
• Condition for use
• Each repetition of the experiment (trial) can
  result in only one of two possible outcomes, a
  success or failure. See example BD1.
• The probability of a success, p, and failure
  (1-p) is constant from trial to trial.
• All trials are statistically independent i.e. No
  trial outcome has any effect on any other trial
  outcome.
• The number of trials, n, is specified constant
  (stated before the experiment begins).

5
Discrete Distributions

• Example 1 binomial distribution
• A coin flip results in a heads or tails
• A product is defective or not
• A customer is male or female
• Example 4 binomial distribution
• Say we perform an experiment flip a coin 10
  times and observe the result. A successful flip
  is designated as heads.
• Assuming the coin is fair, the probability of
  success is .5 for each of the 10 trials, thus
  each trial is independent.
• We want to know the number of successes (heads)
  in 10 trials.
• The random variable that records the number of
  successes is called the binomial random variable.
• Random variable, x, the number of successes that
  occur in the n 10 trials.

6
Binomial Random Variable

• Do the 4 conditions of use hold?
• We are not concerned with sequence with the
  binomial. We could have several successes or
  failures in a row. Since each experiment is
  independent, sequence is not important.
• The binomial random variable counts the number of
  successes in n trials of the binomial experiment.
• By definition, this is a discrete random variable.

7
Calculating the Binomial Probability

• Rather than working out the binomial
  probabilities from scratch each time, we can use
  a general formula.
• Say random variable "X" is the number of
  successes that occur in "n" trials.
• Say p probability of success in any trial
• Say q probability failure in any trial where q
  (1 p)
• In general, The binomial probability is
  calculated by

Where x 0, 1, 2, , n
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Calculating the Binomial Probability
9
Discrete Distributions
Each pair of values (n, p) determines a distinct
binomial distribution. Two parameters n and p
where a parameter is Any symbol defined in the
functions basic mathematical form such that the
user of that function may specify the value of
the parameter.
10

• Developing the Binomial Probability Distribution

P(SSS)p3
S3
P(S3S2,S1)
P(S3)p
S2
P(S2)p
P(S2S1)
S1
P(F3)1-p
F3
P(SSF)p2(1-p)
P(F3S2,S1)
P(S3F2,S1)
S3
P(S3)p
P(SFS)p(1-p)p
P(F2S1)
P(S1)p
P(F2)1-p
P(F3)1-p
Since the outcome of each trial is independent
of the previous outcomes, we can replace the
conditional probabilities with the marginal
probabilities.
F2
P(F3F2,S1)
F3
P(SFF)p(1-p)2
S3
P(S3S2,F1)
P(FSS)(1-p)p2
P(S3)p
S2
P(S2)p
P(F1)1-p
P(S2F1)
P(F3)1-p
P(F3S2,F1)
F3
P(FSF)(1-p)P(1-p)
S3
P(S3F2,F1)
P(FFS)(1-p)2p
F1
P(S3)p
P(F2F1)
P(F2)1-p
F2
P(F3)1-p
P(F3F2,F1)
F3
P(FFF)(1-p)3
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Let X be the number of successes in three
trials. Then,
P(SSS)p3
SSS
P(SSF)p2(1-p)
SS
X 3 X 2 X 1 X 0
P(X 3) p3
P(SFS)p(1-p)p
S S
P(X 2) 3p2(1-p)
P(SFF)p(1-p)2
P(X 1) 3p(1-p)2
P(FSS)(1-p)p2
SS
P(X 0) (1- p)3
P(FSF)(1-p)P(1-p)
P(FFS)(1-p)2p
This multiplier is calculated in the following
formula
P(FFF)(1-p)3
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Binomial Example

• 5 of a catalytic converter production run is
  defective.
• A sample of 3 converter s is drawn. Find the
  probability distribution of the number of
  defectives.
• Solution
• A converter can be either defective or good.
• There is a fixed finite number of trials (n3)
• We assume the converter state is independent on
  one another.
• The probability of a converter being defective
  does not change from converter to converter
  (p.05).

The conditions required for the binomial
experiment are met
13

• Let X be the binomial random variable indicating
  the number of defectives.
• Define a success as a converter is found to be
  defective.

X P(X) 0 .8574 1 .1354 2 .0071 3
.0001
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Discrete Distributions
Example The quality control department of a
manufacturer tested the most recent batch of 1000
catalytic converters produced and found that 50
of them to be defective. Subsequently, an
employee unwittingly mixed the defective
converters in with the nondefective ones. Of a
sample of 3 converters is randomly selected from
the mixed batch, what is the probability
distribution of the number of defective
converters in the sample? Does this situation
satisfy the requirements of a binomial
experiment? n 3 trials with 2 possible outcomes
(defective or nondefective). Does the probability
remain the same for each trial? Why or why
not? The probability p of selecting a defective
converter does not remain constant for each trial
because the probability depends on the results of
the previous trial. Thus the trials are not
independent.
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Discrete Distributions
The probability of selecting a defective
converter on the first trial is 50/1000 .05.
If a defective converter is selected on the
first trial, then the probability changes to
49/999 .049. In practical terms, this violation
of the conditions of a binomial experiment is
often considered negligible. The difference would
be more noticeable if we considered 5 defectives
out of a batch of 100.
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Discrete Distributions
If we assume the conditions for a binomial
experiment hold, then consider p .5 for each
trial. Let X be the binomial random variable
indicating the number o defective converters in
the sample of 3. P(X 0) p(0)
3!/0!3!(.05)0(.95)3 .8574 P(X 1) p(1)
3!/1!2!(.05)1(.95)2 .1354 P(X 2) p(2)
3!/2!1!(.05)2(.95)1 .0071 P(X 3) p(3)
3!/3!0!(.05)3(.95)0 .0001 The resulting
probability distribution of the number of
defective converters in the sample of 3, is as
follows
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Discrete Distributions
x p(x) 0 .8574 1 .1354 2 .0071 3 .0001
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Cumulative Binomial Distribution
F(x) S from k0 to x nCx p k q (n-k) Another
way to look at things cummulative
probabilities Say we have a binomial with n 3
and p .05 x p(x) 0 .8574 1 .1354 2 .0071 3
.0001
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Cumulative Binomial Distribution
this could be written in cumulative form from x
0 to x k x p(x) 0 .8574 1 .9928 2 .9999
3 1.000
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Cumulative Binomial Distribution
What is the advantage of cummulative? It allows
us to find the probability that X will assume
some value within a range of values. Example 1
Cumulative p(2) p(xlt2) p(xlt1) .9999 -
.9928 .0071 Example 2 Cumulative Find the
probability of at most 3 successes in n5 trials
of a binomial experiment with p .2. We locate
the entry corresponding to k 3 and p .2 P(X lt
3) SUM p(x) p(0) p(1) p(2) p(3) .993
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Mean and Variance of Binomial Random Variable

• E(X) µ np
• V(X) s2 np(1-p)
• Example 6.10
• Records show that 30 of the customers in a shoe
  store make their payments using a credit card.
• This morning 20 customers purchased shoes.
• Use the Cummulative Binomial Distribution Table
  (A.1 of Appendix) to answer some questions stated
  in the next slide.

22

• What is the probability that at least 12
  customers used a credit card?
• This is a binomial experiment with n20 and
  p.30.

.01.. 30
0 . . 11
P(At least 12 used credit card)
P(Xgt12)1-P(Xlt11) 1-.995 .005
.995
23

• What is the probability that at least 3 but not
  more than 6 customers used a credit card?

.01.. 30
0 2 . 6
P(3ltXlt6) P(X3 or 4 or 5 or 6) P(Xlt6)
-P(Xlt2) .608 - .035 .573
.035
.608
24

• What is the expected number of customers who used
  a credit card?
• E(X) np 20(.30) 6
• Find the probability that exactly 14 customers
  did not use a credit card.
• Let Y be the number of customers who did not use
  a credit card.P(Y14) P(X6) P(Xlt6) -
  P(xlt5) .608 - .416 .192
• Find the probability that at least 9 customers
  did not use a credit card.
• Let Y be the number of customers who did not use
  a credit card.P(Ygt9) P(Xlt11) .995

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Poisson Distribution

• What if we want to know the number of events
  during a specific time interval or a specified
  region?
• Use the Poisson Distribution.
• Examples of Poisson
• Counting the number of phone calls received in a
  specific period of time
• Counting the number of arrivals at a service
  location in a specific period of time how many
  people arrive at a bank
• The number of errors a typist makes per page
• The number of customers entering a service
  station per hour

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Poisson Distribution

• Conditions for use
• The number of successes that occur in any
  interval is independent of the number of
  successes that occur in any other interval.
• The probability that a success will occur in an
  interval is the same for all intervals of equal
  size and is proportional to the size of the
  interval.
• The probability that two or more successes will
  occur in an interval approaches zero as the
  interval becomes smaller.
• Example
• The arrival of individual dinners to a restaurant
  would not fit the Poisson model because dinners
  usually arrive with companions, violating the
  independence condition.

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Poisson Random Variable

• The Poisson variable indicates the number of
  successes that occur during a given time interval
  or in a specific region in a Poisson experiment
• Probability Distribution of the Poisson Random
  Variable

• ? average number of successes occurring in a
  specific interval
• Must determine an estimate for ? from historical
  data (or other source)
• No limit to the number of values a Poisson random
  Variable can assume

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Poisson Probability Distribution With m 1
The X axis in Excel Starts with x1!!
0 1 2 3 4 5
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Poisson probability distribution with m 2
0 1 2 3 4 5 6
Poisson probability distribution with m 5
0 1 2 3 4 5 6 7 8 9
10
Poisson probability distribution with m 7
0 1 2 3 4 5 6 7 8 9
10 11 12 13 14 15
30
Poisson Example

• Cars arrive at a tollbooth at a rate of 360 cars
  per hour.
• What is the probability that only two cars will
  arrive during a specified one-minute period? (Use
  the formula)
• The probability distribution of arriving cars for
  any one-minute period is Poisson with µ 360/60
  6 cars per minute. Let X denote the number of
  arrivals during a one-minute period.

31

• What is the probability that only two cars will
  arrive during a specified one-minute period? (Use
  table 2, Appendix B.)
• P(X 2) P(Xlt2) - P(Xlt1) .062 - .017 .045

32

• What is the probability that at least four cars
  will arrive during a one-minute period? Use
  Cummulative Poisson Table (Table A.2 , Appendix)
• P(Xgt4) 1 - P(Xlt3) 1 - .151 .849

33
Poisson Approximation of the Binomial

• When n is very large, binomial probability table
  may not be available.
• If p is very small (plt .05), we can approximate
  the binomial probabilities using the Poisson
  distribution.
• Use ? np and make the following approximation

With parameters n and p
With m np
34
Example of Poisson
Example Poisson Approximation of the Binomial A
warehouse engages in acceptance sampling to
determine if it will accept or reject incoming
lots of designer sunglasses, some of which
invariably are defective. Specifically, the
warehouse has a policy of examining a sample of
50 sunglasses from each lot and accepting the lot
only if the sample contains no more than 2
defective pairs. What is the probability of a
lot's being accepted if, in fact, 2 of the
sunglasses in the lot are defective? This is a
binomial experiment with n 50 and p .02. Our
binomial tables include n values up to 25, but
since p lt .05 and the expected number of
defective sunglasses in the sample is np
50(.02) 1, the required probability can be
approximated by using the Poisson distribution
with µ 1. From Table A.1, we find that the
probability that a sample contains at most 2
defective pairs o sunglasses is .920.
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Poisson Example

• What is the probability of a lot being accepted
  if, in fact, 2 of the sunglasses are defective?
• Solution
• This is a binomial experiment with n 50, p
  .02.
• Tables for n 50 are not available plt.05
  thus, a Poisson approximation is appropriate ?
  (50)(.02) 1
• P(Xpoissonlt2) .920 (true binomial probability
  .922)

36
Example of Poisson
So how well does the Poisson approximate the
Binomial? Consider the following
table x Binomial (n 50, p .02) Poisson (µ
np 1) 0 .364
.368 1 .372
.368 2 .186
.184 3
.061
.061 4 .014
.015 5 .003
.003 6 .000
.001
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Example of Poisson

• A tollbooth operator has observed that cars
  arrive randomly at an average rate of 360 cars
  per hour.
• Using the formula, calculate the probability that
  only 2 cars will arrive during a specified 1
  minute period.
• Using Table A.2 on page 360, find the probability
  that only 2 cars will arrive during a specified 1
  minute period.
• Using Table A.2 on page 360, find the probability
  that at least 4 cars will arrive during a
  specified 1 minute period.
• P(X2) (e-6)(62) / 2! (.00248)(36) / 2
  .0446
• P(X2) P(X lt 2) - P(X lt 1) .062 .017 .045
• P(X gt 4) 1 - P(X lt 3) 1 - .151 .849

38
Example of Poisson
What if we wanted to know the probability of a
small number of occurrences in a large number of
trials and a very small probability of
success? We use Poisson as a good approximation
of the answer. When trying to decide between the
binomial and the Poisson, use the following
guidelines n gt 20 n gt 100 or p lt
.05 or np lt 10
39
Hypergeometric Distribution
What about sampling without replacement? What is
likely to happen to the probability of
success? Probability of success is not constant
from trial to trial. We have a finite set on N
things, and "a" of them possess a property of
interest. Thus, there are "a" successes in N
things. Let X be the number of successes that
occur in a sample, without replacement of "n"
things from a total of N things. This is the
hypergeometric distribution P(x) (aCx)(N-a C
n-x) x 0,1,2. NCn
40
Binomial Approximation of Hypergeometric
Distribution
If N is large, then the probability of success
will remain approximately constant from one trial
to another. When can we use the binomial
distribution as an approximation of the
hypergeometric distribution when N/10 gt n
41
Bernoulli Distribution

• What if we want to perform a single experiment
  and there are only 2 possible outcomes?
• We use a special case of the binomial
  distribution where n1
• P(x) 1Cx px (1-p) 1-x x 0,1 which yields
  p(0) 1-p
• px (1-p) 1-x x0,1 p(1) p
• In this form, the binomial is referred to as the
  Bernoulli distribution.

42
Geometric Distribution

• Now, instead of being concerned with the number
  of successes in n Bernoulli trials, lets
  consider the number of Bernoulli trial failures
  that would have to be performed prior to
  achieving the 1st success.
• In this case, we use the geometric distribution,
  where X is the random variable representing the
  number of failures before the 1st success.
• Mathematical form of geometric distribution
• P(x)p (1-p)x x 0,1,2

43
Negative Binomial Distribution

• What if we wanted to know the number of failures,
  x, that occur before the rth success (r 1,2.)?
• In this case, we use the negative binomial
  distribution.
• The number of statistically independent trials
  that will be performed before the r success X
  r
• The previous r 1 successes and the X failures
  can occur in any order during the X r 1
  trials.
• Negative binomial distribution mathematical
  form
• P(x) rx-1 Cx pr (1-p)x x 0,1,2..

44
Problem Solving

• A Suggestion for Solving Problems Involving
  Discrete Random Variables
• An approach
• Understand the random variable under
  consideration and determine if the random
  variable fits the description and satisfies the
  assumptions associated with any of the 6 random
  variables presented in Table 3.3.
• If you find a match in Table 3.3 use the software
  for that distribution.
• If none of the 6 random variables in table 3.3
  match the random variable associated with your
  problem, use the sample space method

45
Discrete Bivariate Probability Distribution
Functions

• The first part of the chapter considered only
  univariate probability distribution functions.
  One variable is allowed to change.
• What if two or more variables change?
• When considering situations where two or more
  variables change, the definitions of
• sample space
• numerically valued functions
• random variable
• still apply.

46
Bivariate Distributions

• To consider the relationship between two random
  variables, the bivariate (or joint) distribution
  is needed.
• Bivariate probability distribution
• The probability that X assumes the value x, and Y
  assumes the value y is denoted
• p(x,y) P(Xx, Y y)

47
Discrete Bivariate Probability Distribution
Functions
Example Consider the following real estate
data We want to know how the size of the house
varies with the cost.
48
Discrete Bivariate Probability Distribution
Functions
49
Discrete Bivariate Probability Distribution
Functions
What is the next step? Construct frequency
distributions Frequency distribution of house
size
50
Discrete Bivariate Probability Distribution
Functions
Frequency distribution of selling price
51
Discrete Bivariate Probability Distribution
Functions
What next? Construct a bivariate frequency
distribution of X and Y
52
Discrete Bivariate Probability Distribution
Functions
A Bivariate (or joint) probability distribution
of X and Y is a table that gives the joint
probabilities p(x,y) for all pairs of values
(x,y). Cumulative bivariate probability
distribution function is F(x1, x2) P(X1 lt x1
? X2 lt x2) Marginal probability distribution
function of X , p (x ) Is a univariable
probability distribution function Px1 (X1) Sum
p(x1, x2) Px2 (X2) Sum p(x1, x2)
53
Discrete Bivariate Probability Distribution
Functions

• Example Xavier and Yvette are two real estate
  agents. Let X denote the number of houses that
  Xavier will sell in a week, and let Y denote the
  number of houses that Yvette will sell in a week.
  Suppose that the joint probability distribution
  of X and Y is as shown in the table below.

54
Discrete Bivariate Probability Distribution
Functions

• Notice that a bivariate probability distribution
  is similar to a bivariate frequency distribution,
  with the probabilities having replaced the
  frequencies.
• The none probabilities in the interior of the
  table are the joint probabilities p(x,y).
•  
• p(0,0) P(X 0 and Y 0) .12
• p(0,1) P(X 0 and Y 1) .21
• p(0,2) P(X 0 and Y 2) .07

55
Discrete Bivariate Probability Distribution
Functions

• Summing these three probabilities, we obtain the
  marginal probability P(X 0) .40 (named
  marginal because it appears in the margin of the
  table)
• Summing the probabilities in each of the other
  columns and rows, we obtain the other marginal
  probabilities.
• Thus the marginal probability distributions of X
  and Y are

56
Example

• The bivariate (joint) probability distribution

p(0,0)
P(Y1), the marginal probability.
p(0,1)
p(0,2)
P(X0) The marginal probability
57
0.42
x p(x) y p(y) 0
.4 0 .6 1 .5
1 .3 2 .1 2
.1 E(X) .7 E(Y) .5 V(X) .41
V(Y) .45
p(x,y)
0.21
0.12
0.06
X
y0
0.06
0.03
0.07
0.02
y1
0.01
Y
y2
X0
X2
X1
58
Calculating Conditional Probability
Example - continued
59
Conditions For Independence

• Two random variables are said to be independent
  when
• This leads to the following relationship for
  independent variables
• Example 6.7 - continued
• Since P(X0Y1).7 but P(X0).4, The variables
  X and Y are not independent.

P(XxYy)P(Xx) or P(YyXx)P(Yy).
P(Xx and Yy) P(Xx)P(Yy)
60

• Additional example
• The table below represent the joint probability
  distribution of the variable X and Y. Are the
  variables X and Y independent

Find the marginal probabilities of X and Y. Then
apply the multiplication rule.
p(y) .7 .3
P(y) .40 .60
Compare the other two pairs. Yes, the two
variables are independent
61
The sum of Two Variables

• To calculate the probability distribution for a
  sum of two variables X and Y observe the example
  below.
• Example 6.7 - continued
• Find the probability distribution of the total
  number of houses sold per week by Xavier and
  Yvette.
• Solution
• XY is the total number of houses sold. XY can
  have the values 0, 1, 2, 3, 4.
• We find the distribution of XY as demonstrated
  next.

62
P(XY0) P(X0 and Y0) .12
P(XY1) P(X0 and Y1) P(X1 and Y0) .21
.42 .63
The probabilities P(XY)3 and P(XY) 4 are
calculated the same way. The distribution
follows
P(XY2) P(X0 and Y2) P(X1 and Y1) P(X2
and Y0) .07 .06 .06 .19

• ..
• ..

63
Expected value and variance of XY

• When the distribution of XY is known (see the
  previous example) we can calculate E(XY) and
  V(XY) directly using their definitions.
• An alternative is to use the relationships
• E(aXbY) aE(X) bE(Y)
• V(aXbY) a2V(X) b2V(Y) if X and Y are
  independent.
• When X and Y are not independent, (see the
  previous example) we need to incorporate the
  covariance in the calculations of the variance
  V(aXbY).