Chapter 6: Discrete Probability
Distributions
6.1 Discrete Random Variables
6.2 The Binomial Probability Distribution
6.3 The Poisson Probability Distribution
November 12, 2008
1
Discrete Random Variables
Consider the probability experiment of flipping a coin two times. The
possible outcomes of this experiment, the sample space, is:
S = {HH,HT,TH,TT}
Where H is a “head” and T is a “tail.” We call the outcome of such an
experiment a random event since we are never sure of its value.
Mathematicians and statisticians like to work with numbers or things that can
be represented by numbers. For this reason, we try to assign the random
event a number or a set of numbers.
Section 6.1
2
Random Variable
Definition: A random variable is a numerical measure or
representation of a random event in a probability experiment.
Remark: Since a random variable comes from something that is
random (a random event), it may take several numerical values.
We will denote the random variable by using capital letters, e.g.,
X, its values by small letter variables, e.g., x.
Example: Flipping a coin twice. S = {HH,HT,TH,TT}. Let X be the
random variable that represents the number of tails. Then the number
of values assume by X are: x = 0,1,2.
3
Examples
Example: Consider the probability experiment of measuring your systolic
blood pressure. This is a random variable X and its range of values is 50
to 230 mm Hg. For example, it is normal if x = 120. We call X a
continuous random variable since it can take any value in the interval
[50,230].
Example: Consider the probability experiment of rolling once a single die.
The random variable X in this case is that a particular number that comes
up. The possible values for this random variable are: x = 1, 2, 3, 4, 5 or 6.
We call X a discrete random variable since it can take only discrete values.
4
Random Variable Types
• A discrete random variable has a finite or countable number of
values. That is, the values can be put into a 1-1 correspondence with
the integers or a subset of the integers.
• A continuous random variable has an infinite number of possible
values that have a 1-1 corresponds with number in an interval [a,b].
5
Examples
• Let X be the random variable of the number of students in Math 127A that will
receive A’s in the course. Here, x = 0,1,2,….,132. This is a discrete random
variable.
• Let X be the random variable for the weight of students in Math 127A. Here, x > 0
and this is a continuous random variable.
6
Discrete Probability Distributions
Definition: Let X be a discrete random variable. The discrete probability
distribution of X, which we denote by P, is a function that maps the values
of the discrete random X into the interval [0,1]. P is often represented by
a graph or table that gives all of the possible values of X (i.e., x) and the
corresponding probabilities of x, P(x). Sometimes it is given as a
mathematical formula.
x
P(x)
---
---
P(x)  n C x p x 1  p 
---
---
---
---
where n and p are parameters
and x  0,1, 2,..., n
---
---
n x
7
Example
Suppose that we perform a probability experiment of flipping a coin 3
times in a row. Let X be the random variable of the number of times that
head appears. The possible values for X are x = 0,1,2,3. Using a
simulation of flipping the coin 3 times, we find the following:
x
P(x)
0
0.10
1
0.01
This is a discrete probability distribution for X.
Notice that
2
0.51
• 0 ≤ P(x) ≤ 1
3
0.38
• P(0) + P(1) + P(2) + P(3) = 1
8
Rules for Discrete Probability Distributions
Let X be a discrete random variable with values x1 , x2 ,K , xN  and probabilites
P(x j ), j  1, 2,K , n. Then the following conditions must be satisfied:
(a) 0  P(x j )  1, j  1, 2,K , N
N
 
(b)  P x j  1.
j 1
(c) The values, P(x1 ), P(x2 ),..., P(xN ) are called the probability distribution function
for the discrete random variable X.
Remark: If any of the above conditions is violated, then P is not a discrete
probability distribution.
9
Examples
Which of the following are discrete probability distributions?
x
P(x)
x
P(x)
x
P(x)
1
0.40
1
0.40
1
0.40
2
0.35
2
0.35
2
0.35
3
0.12
3
0.12
3
0.12
4
-0.07
4
0.01
4
0.01
5
0.20
5
0.20
5
0.12
No
No
Yes
10
Example
Consider the following discrete probability distribution for the number
of homeruns in a single game by a Boston Red Sox team. Let X be
the random variable which is the number of HR hit in a single game.
x (# of HR)
P(x)
0
0.23
1
0.38
2
0.22
3
0.13
4
0.03
5
0.01
6 or more
0.00
11
Question: What is the probability that the team will hit three or more
homeruns in a single game?
Let A = event that they will hit 3 HR in a game.
Let B = event that they will hit 4 HR in a game.
Let C = event that they will hit 5 HR in a game..
Let D = event that they will hit 6 or more HR in a game.
Then P(A or B or C or D) = P(A) + P(B) + P(C) + P(D)
= 0.13 + 0.03 + 0.01 + 0.00 = 0.17 (17%)
Note: A, B, C, D are disjoint events.
12
Remark
Consider a discrete probability distribution: x1  x2  L  x N with probabilities
P(x1 ), P(x2 ),K , P(x N ). Consider one of the values of the discrete random variable,
say xk , where 1  k  N.
Question : What is the probability that the value of the random variable X is less than or equal to xk
i.e., x  xk ?
Answer :
k
P(x  xk )  P(x  x1 or x  x2 or K or x  xk )  P(x1 )  P(x2 )  K  P(xk )   P(x j )
j 1
k
Remark : The quantity, P(x  xk )   P(x j ), is called the cummulative distribution function.
j 1
13
Graphical Representations of Discrete
Probability Distributions
If a discrete probability distribution is given as a probability table:
x
P(x)
---
---
---
---
---
---
---
---
then we construct a histogram (relative frequencies) from this table. This
is called a probability histogram.
14
Example
In the 2004 baseball season, Ichiro Suzuki of the Settle Mariners set the record for the most hits
in a season with a total of 262 hits. Let X be the discrete random variable for the number of hits
per game. The following table gives the probabilities of the number of hits per game by Suzuki.
Here the number of hits per game is the random variable and the sum of the probabilities is 1.
Construct the probability histogram.
x
P(x)
0
0.1677
1
0.3354
2
0.2857
3
0.1491
4
0.0373
5 or more
0.0248
15
Parameters of a Discrete Probability Distribution
Recall: A parameter of a population is a numerical characteristic of the
population (e.g., mean, standard deviation, etc.).
Definition: A parameter of a discrete probability distribution is a number
that summarizes a characteristic of the distribution.
Parameters:
• Mean
• Standard Deviation
16
The Mean of a Discrete Probability
Distribution
Defintion: Consider a discrete probability distribution: D  x1 , P(x1 ), x2 , P(x2 ),K , x N , P(x N ) for a random variable X.
N
 
The mean of the discrete probability distribution is the number:  X   x j P x j .
j 1
x
P(x)
0
0.10
1
0.01
2
0.51
3
0.38
Example : Here N  4.
 X  0  0.1  1 0.01  2  0.51  3  0.38  2.17
17
Mean - Mean?
Recall that we had defined the mean of a set of numbers (population) in
a different way.
 1 N
  z1, z2 ,K , zN       z j
 N  j 1
Question: Is the mean of a population and the mean of a discrete
probability distribution really the same thing? We consider the
following example to show that they are the same.
18
Example
Suppose that we consider a group of people and we ask them how many
hours of TV have they watched during the past 24 hours. Suppose that there
responses are summarized in the list: {2,4,6,6,4,4,2,3,5,5}. The mean of this
set (population) is (2+4+6+6+4+4+2+3+5+5)/10 = 41/10 = 4.1.
We can set up a probability table from this data consider the random variable
X to be the number of hours of TV watching. Hence, x = 2,3,4,5,6.
x
P(x)
2
2/10
The mean of this discrete probability distribution is
3
1/10
4
3/10
X = 2(0.2) + 3(0.1) + 4(0.3) + 5(0.2) + 6(0.2) = 4.1
5
2/10
6
2/10
and hence, we compute the same value! This is not
by “chance.”
19
Expected Value of X
The mean of a discrete probability distribution for X is a weight average of
its probabilities P(xi) where xi are the weights. In a way, we can think of it
as the value of P(X). This mean is the the average of all of the possible
outcomes of X. For this reason it is also called the expected value of X.
We can think of X as the mean outcome of the all events in X. That is, if
we were to repeat the probability experiment many, many times, then the
average of the outcomes would approach the expected value of X.
20
Example
x (# of HR)
P(x)
0
0.23
1
0.38
2
0.22
3
0.13
4
0.03
5
0.01
6 or more
0.00
Find the expected value of X where X is
the random variable for the number of
homeruns in a single game by the
Boston Red Sox baseball team.
X = 0(0.23) + 1(0.38) + 2(0.22) + 3(0.13) + 4(0.03) + 5(0.01) + 6(0.00) = 1.38
21
Example
Suppose that you want to invest $100 in the stock market. Let X be the
random variable for the results of your $100 investment. For simplicity,
suppose that there are two possible outcomes: x = $0 or $1000 such that
P(0) = 0.50 and P(1000) = 0.50. What is the expected value of your
investment?
X = 0(0.5) + 1000(0.5) = 500 i.e., we have an expected return of $500.
22
Standard Deviation of a Discrete
Probability Distribution
Defintion: Consider a discrete probability distribution: D  x1 , P(x1 ), x2 , P(x2 ),K , x N , P(xN ) for a random variable X.
N
N
The variance of the probability distribution is the number:    xi   X  P xi    xi  P xi    X2
2
X
j 1
2
2
j 1
where  X is the mean of the random variable X. The standard deviation of the probability distribution
is defined as  X 
N
 x
i
  X  P xi .
2
j 1
The standard deviation of a probability distribution is a measure of its spread. In other
words, it is the variation in xi from its mean, weighted by the probabilities.
23
Example
Find the mean and standard deviation for the following discrete
probability distribution.
x
P(x)
0
0.25
1
0.25
x = 0(0.25) + 1(0.25) + 2(0.1) +4(0.40) = 2.05
2
0.10
X = [02(0.25)+12(0.25)+22(0.10)+42(0.4) - (2.05)2]1/2 = 1.69
4
0.40
24
Comparing Discrete Probability
Distributions
 = 2.18
 = 0.64
 = 1.50
 = 1.12
25
The Binomial Probability Distribution
The binomial probability distribution is a discrete probability distribution that is used
to determine probabilities of events in a probability experiment in which there are two
mutual exclusive (disjoint) outcomes are possible. For example, the probability
experiment might be flipping a coin. The two mutually exclusive outcomes are “head”
or “tails.” Another example, the probability experiment might be one of asking an
individual if he has a driver’s license. He or she “does” or “doesn’t.”
Section 6.2
26
Binomial Probability Experiment
Binomial Probability Experiment:
• The experiment is performed a n times with each repetition of the
experiment being a trial.
• The trials are independent.
• Each trial results in one of two mutually exclusive outcomes (success
or failure).
• The probability of a success is p and the probability of failure is 1 - p,
e.g., p = 0.5.
• A random variable for this type of probability experiment is the
number of successes in the n trials. .
27
The Binomial Random Variable
• We consider a binomial probability experiment.
• For each of the n trials of a binomial probability experiment, the probability
of “success” is p.
• For each of the n trials of a binomial probability experiment, the probability
of “failure” is 1 - p.
• Each of the n trials are independent i.e., the outcome of one trial does not
dependent on any other trial.
Let X be the number of successes in the n trials. We call X a binomial
random variable.
28
Binomial Probability Distribution Function
Definition : Consider a binomial probability experiment of n trials. Suppose that the
probability of success for an individual trial is p. let X be the number of successes in the n trials.
Then the probability of x successes in the n trials is given by the discrete probability distribution:
ΚΚΚΚΚΚΚΚΚΚΚΚΚΚΚΚP(x) 
n!
p x (1  p)n  x  n C x p x (1  p)n  x
x!(n  x)!
for x  0,1, 2,K , n. The discrete probability distribution for X is called the Binomial Distribution.
29
Example
According to the Uniform Crime Report, 2003, 66.9% of murders in the the U.S. were committed with a
firearm.
(a) If 100 murders are randomly selected, how many would you expect be committed with a firearm?
(b) What is the probability that you would observe 75 murders out of the 100 randomly selected
murders?
(a) 66.9  67 murders out of the randomly selected murders.
(b) Here p  0.669 and n  100
 P(x  75) 
100 C 75 0.669  1  0.669 
75
100  75

100!
0.669 75 1  0.669 25  0.0193823
75!(100  75)!
30
Example
Probability Experiment: Does a person have ESP?
• A person in one room picks one the integers (1-5) at random and thinks about
this particular number for 1 minute.
• In another room, the person who claims to have ESP tries to identify the
number that was chosen by the person in the first room.
• The is done 3 times (3 trials). We assume that each trial is independent.
• The ESP claimant has the correct answer twice i.e., he or she is correct two out
the three trials.
Question: Does the person have ESP?
Answer: If the person has ESP, then their success rate would be greater than
guessing the number each trial. Hence, we must compute the probability that
one could guess 2 out of the 3 correct answers?
31
Possible Outcomes of Guessing Three Times (a compound event)
S = success and F = failure to guess correctly. The probability of guessing the
correct number is 1/5 = 0.20 . Hence, P(S) = 0.2 and P(F) = 1 - P(S) = 0.8 .
Different Possible Outcomes
Outcome
Probability
SSS
(0.2)(0.2)(0.2) = 0.008
The probability of having any two
successes is 0.032 from the following:
SSF
(0.2)(0.2)(0.8) = 0.032
If A = SSF, B = SFS, C = FSS, then
SFS
(0.2)(0.8)(0.2) = 0.032
P(A or B or C) = P(A) + P(B) + P(C)
FSS
(0.8)(0.2)(0.2) = 0.032
SFF
(0.2)(0.8)(0.8) = 0.128
FSF
(0.8)(0.2)(0.8) = 0.128
FFS
(0.8)(0.8)(0.2) = 0.128
FFF
(0.8)(0.8)(0.8) = 0.512
= 3(0.032) = 0.096
since A, B and C are disjoint events.
Hence, this will happen approximately
10% of the time. Hence, the ESP
claimant has a success rate that is much
higher than one would expect from
guessing.
32
Calculation using
Binomial Distribution Function
The calculation of the probability can also be done with the Binomial Distribution. Let x
denote the number of successes. In the calculation of the probability of having two out of
three successes, we find for x  2 : P(x)  3 C2 p x 1  p 
n x

3!
0.2 2 0.8   0.096
2!(3  2)!
where p  0.2, x  2, n  3.
33
Binomial Distributions for n = 10
and Different p
34
Mean and Standard Deviation of
Binomial Distributions
For the Binomial Distribution, P(x) 
n
n
i0
i0
is  X   iP(i)   i
X 
n
 i  np 
i0
2
n!
p x (1  p)n  x , x  0,1, 2,K , n, the mean
x!(n  x)!
n!
p i (1  p)ni  np and the standard deviation by
i!(n  i)!
P(i) 
n
2
 i  np 
i0
n!
p i (1  p)n i  np(1  p).
i!(n  i)!
35
Example
Example: Suppose the a random variable is distributed by the
binomial distribution with n = 12 and p = 0.65. Calculate the
probability that x = 10. Find the mean and standard deviation.
P(10) 
12!
(12)(11)
(0.65)10 (1  0.65)12 10 
(0.65)10 (0.35)2  0.109
10!(12  10)!
(1)(2)
 X  np  (12)(0.65)  7.8
 X  np(1  p)  2.73  1.65227
36
Calculation on TI-83
•2nd VARS (DISTR) key
• Select binompdf( [ENTER]
• Complete entry e.g., binompdf(n,p,x) [ENTER]
Remark:
• In binompdf, if x is omitted, then it calculates the Binomial
Distribution for all of the possible values of x.
• The CDF of the Binomial Distribution can be calculated with
binomcdf(n,p,{xk,xk+1,…,xm})
37
Example
Question: Are women fairly treated in the selection for managerial training?
Situation:
• A pool of 1,000 employees from which 10 will be selected.
• Of the 1,000 employees, 50% are women and 50% are men.
Result: None of the 10 selected employees for management training were
women.
Question: Does this show bias against women?
Analysis: Assuming that the 1,000 employees are equally qualified for training
and there is equal chance of selecting woman or a man, what is the probability
of selecting 10 male trainees?
In the long-run, we would expect that for a sample of 10, 5 would be women
and 5 would be men. We will assume that the 10 selected employees are
distributed according to a Binomial Distribution (n = 10, p = 0.5). What does the
probability distribution look like? Let “success” be the outcome of picking a
female employee.
38
x
P(x)
0
0.00097
1
0.00975
2
0.04394
3
0.11718
4
0.20507
5
0.24609
6
0.20507
7
0.11718
8
0.04394
9
0.00976
10
0.00097
P(0) = 0.00097 which is very small and hence, picking no
women is very unlikely.
Note that the highest probability is when x = 5 i.e, P(5) =
0.24609. The expect value is (n)(p) = (10)(0.5) = 5.
39
Example
Question: Can we check racial profiling by the police?
Situation:
• Statistics from Philadelphia Police Department in 1997
• 262 car stops
Result: 207 of 262 involve African Americans i.e., 79% of total number of stops.
Analysis: Assume that the car stops are binomially distributed with n = 262. In
Philadelphia, African Americans (AA) make up 42.2% of the population. Hence,
we would expect that approximately 4 out 10 car stops would involve AA and we
choose p = 0.422 . Let X be the binomial random variable be the event of
stopping an AA. For the binary outcome, “success” is stopping of a car driven by
an AA.
Mean: X = (p)(n) = (0.422)(262) = 110.564
Standard Deviation: X = [(n)(p)(1-p)]1/2 = [(110.564)(0.5780]1/2 = 9.275
40
P(207) = 3.9 x 10-34
This is probability of stopping 207 AA out
of the 262 stops. We would expect
approximately 111 stops using the
expected value. Hence, 207 out of 262
is very, very unlikely.
41
Bell-shaped Binomial Distributions
Remark: If a probability distribution is bell-shaped, then the Empirical Rule
applies.
Let  denote the standard deviation of a probability distribution and  the mean of the distribution.
 68% of the observations fall within the interval    ,    .
 95% of the observations fall within the interval   2 ,   2 .
 Greater than 99.7% of the observations fall within the interval   3 ,   3 .
Observation: If a binomial distribution is bell-shaped, then the Empirical Rule applies.
42
When is a Binomial Distribution
Bell-shaped?
Rule of Thumb : Consider a Binomial Distribution with the parameter p and n.
The distribution is approximately bell-shape if np(1  p)  10.
43
Example
According to the Higher Education Research Institute, 55% of college freshmen in 4-year colleges
and universities during the 2003-2004 academic year were female. Suppose 12 freshmen are
randomly selected and the number of female recorded. Find the following probabilities:
(a) Exactly 7 of the 12 are female.
(b) Five or more are female.
(c) Fewer than five are female.
(d) Between 7 and 10, inclusively, are female.
Answers: P(x)  n C x p x 1  p 
n x
(a) P(x  7) 

n!
n x
p x 1  p  , n  12, p  0.55, x  0,1, 2,...,12
x!(n  x)!
12!
0.55 7 1  0.55 12  7  0.222498
7!(12  7)!
(b) P(x  5)  P(x  5)  P(x  6)  ....  P(x  12) 
12!
12!
0.55 5 1  0.55 12  5  ... 
0.55 12 1  0.55 12 12  0.88826
5!(12  5)!
12!(12  12)!
Note that P(x  5)  1  P(x  4).
(c) P(x  4)  P(x  0)  P(x  1)  P(x  2)  P(x  3)  P(x  4)  1  P(x  5)  1  0.88826  0.11174
(d) P(7  x  10)  P(x  7)  P(x  8)  P(x  9)  P(x  10)  0.518641
44
Here np(1  p)  2.97.
45
The Poisson Probability Distribution
We now introduce a new discrete random variable that counts the number of times that a
particular event occurs in a particular set (time or space). For example,
• the number of bacterial per unit volume of a fluid
• the number of customers at McDonald’s that order “big Macs” between 4:00-5:00 PM
• the number of machines in a factory that break down per day.
Definition: Let X be a random variable and suppose that the possible values of X
are x = 0,1,2,3,… and these values occur in a fixed interval I (e.g., an interval of
time). We say that the random variable follows a Poisson process if all of the
following conditions hold:
•
The probability of 2 or more occurrence of the event on any sufficiently small
subinterval of I is zero.
•
If we look at the random variable on two subintervals, I1 and I2, the probability of
an event will be same, provide the two subintervals have the same length.
•
For any two non-overlapping subintervals,I1 and I2, the number frequency of
and event in I1 is independent of the frequency of the event on I2.
46
Section 6.3
Poisson Probability Distribution
Function
Definition : Let X be a random variable from a Poisson process on an interval I of
fixed length t.
x
t    t

The probability of x is P(x) 
e , x  0,1, 2, 3,...
x!
where  is the average number of occurences of the event per unit length of I.
Remark : We must know the average number of occurences of the events over some time period.
This is . We must also specify the length of the time interval t  over which we are computing
the probabilities. Note the the variables  and t occur as a product in the definition of P. Hence, we
need only specify their product, t, to completely define the Poisson distribution function. The constant
t   is the average number of events in interval over which one is counting events. When t  1,   .
47
Graphs of the Poisson
Distribution Function
48
Example
The phone calls to a computer software help desk occur at the rate of 2.1 per minute between
11:00 AM and noon. Find the following probabilities for calls between 11:15-11:20 AM:
(a) There will be exactly 8 calls.
(b) There will be fewer than 8 calls.
(c) There will be at least 8 calls.
  2.1
t5
t  10.5
x
t    t

P(x) 
e , x  0,1, 2,...
x!
(a) P(x  8) 
2.1 5 8 e2.15  10.5 8 e10.5  0.100902
8!
8!
(b) P(x  8)  P(x  0 or x  1 or x  2 or ... or x  7)  P(1)  P(2)  P(3)  ...  P(7)
0
1
2
7
10.5  10.5 10.5  10.5 10.5  10.5
10.5  10.5



e

e

e
 ... 
e
0!
1!
2!
7!
 10.5  10.5  10.5 
10.5   e10.5  6582.7  e10.5  0.178511



 .... 
1!
2!
7! 
 0!
0
1
2
7
(c) P(x  8)  1  P(x  8)  1  0.178511  0.821489
49
Example
From 1900 to 2003 (104 years), the state of Florida suffered 24 major hurricanes (category 3 to 5).
What is the probability that in the year 2009, the state will see 3 major hurricanes? What is the
probability that it will see at most 3 major hurricanes?
24
 0.230769 (hurricanes per year)
104
t  1 (one year)

 t  0.230769
x
t    t

P(x) 
e , x  0,1, 2,...
x!
3
0.230769  0.230769

P(x  3) 
e
 0.00162615
3!
P(x  3)  P(x  0)  P(x  1)  P(x  2)  P(x  3)  0.99902
Note : P(x  0)  0.793923
50
Example
The Vanderbilt Printing division duplicates documents for distribution in the university. For
every 500 documents printed, one document is not acceptable for distribution. Documents
are delivered in boxes with each box containing 100 documents. Assuming that the
number of defective documents in a box is distributed according to a Poisson Distribution
and we reject any box with two or more defective documents, what percent of the boxes
are rejected?
1
500
t  100 (one box)
100 1
t 

500 5
x
x
t   t 0.2  0.2

P(x) 
e 
e , x  0,1, 2,...
x!
x!

P(x  2)  1  P(x  2)  1  P(x  0)  P(x  1)  1  0.818731  0.163746   0.0175231
Hence, we reject approximately 1.75% of the boxes.
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