Chapter 5 Discrete Probability Distribution
I. Basic Definitions
II. Summary Measures for Discrete Random Variable
• Expected Value (Mean)
• Variance and Standard Deviation
III. Two Popular Discrete Probability Distributions
• Binomial Distribution
• Poisson Distribution
I. Basic Definitions
• Random Variable:
(p.195)
a numerical description of the outcomes of an
experiment.
Assign values to outcomes of an experiment so the
experiment can be represented as a random variable.
Example:
Test scores: 0  X  100
Toss coin: X = {0, 1}
Roll a die: X = {1, 2, 3, 4, 5, 6}
I. Basic Definitions
• Discrete Random Variable: It takes a set of discrete
values. Between two possible values some values are
impossible.
• Continuous Random Variable: Between any two
possible values for this variable, another value
always exists.
Example:
Roll a die: X = {1, 2, 3, 4, 5, 6}. X is a discrete random
variable.
Weight of a person selected at random: X is a
continuous random variable.
I. Basic Definitions
Two ways to present a discrete random variable
• Probability Distribution: (p.198 Table 5.3)
a list of all possible values (x) for a random variable
and probabilities (f(x)) associated with individual
values.
• Probability Function: probability may be
represented as a function of values of the random
variable.
Example: p.199
Consider the experiment of rolling a die and define
the random variable X to be the number coming up.
1. Probability distribution
2. Probability function: f(x) = 1/6.
I. Basic Definition
• Valid Discrete Probability Function
0  f(x)  1 for all f(x) AND
 f(x) = 1
Example: p.200 #7
a. The probability distribution is proper because all
f(x) meet requirements 0  f(x)  1 and  f(x) = 1.
b. P(X=30) = f(30) = .25
c. P(X25) = f(25) + f(20) = .15 + .20 = .35
d. P(X>30) = f(35) = .40
Homework: p.201 #10, p.201 #14
I. Basic Definitions
Example: p.202 #14
a. Find valid f(200):
.1+.2+.3+.25+.1+f(200) = 1,
so f(200) = .05
b. P(?)
P(X>0) = f(50)+f(100)+f(150)+f(200) = .7
c. P(?)
P(X100) = f(100)+f(150)+f(200) = .4
(“at least”)
II. Summary Measures for Discrete Random Variable
• Expected Value (mean): E(X) or 
E(X) = xf(x)
(p.203)
• Variance: Var(X) or 2
(p.203)
Var(X) = (x-  )2f(x)
Homework: p.204 #16
Example: Consider the experiment of tossing coin and
define X to be 0 if head and 1 if tail. Find
the expected value and variance.
E(X) = xf(x)
= (0)(.5)+(1)(.5) = .5
Var(X) = (x-  )2f(x)
= (0 - .5)2(.5)+(1 - .5)2(.5) = .25
III. Two Popular Discrete Probability Distributions
— Binomial and Poisson
Outlines:
1. Probability Distribution:
Binomial Distribution: Table 5 (p.989 - p.997).
Given n, p, x  f(x).
Poisson Distribution: Table 7 (p.999 - p.1004).
Given , x  f(x).
2. Applications: Difference between Binomial and
Poisson.
3. Applications: criterion to define “success”.
III. Two Popular Discrete Probability Distributions
1. Binomial Distribution
p.207
• Random variable X: x “successes” out of n trials (the
number of “successes” in n trials).
• Three conditions for Binomial distribution:
— n independent trials
— Two outcomes for each trial: “success” and
“failure”.
— p: probability of a “success” is a constant from
trial to trial.
III. Two Popular Discrete Probability Distributions
Example: Toss a coin 10 times. We define the
“success” as a head and X is the number of heads
from 10 trials. Does X have a binomial distribution?
Answer: Yes.
Follow-up: Why? n? p?
Example: Roll a die 10 times. We define the “success”
as 5 or more points coming up and X is the
number of successes from 10 trials. Does X
have a binomial distribution?
Answer: Yes.
Why? n? p?
III. Two Popular Discrete Probability Distributions
• Summary Measures for Binomial Distribution
E(X) = np
p.214
Var(X) = np(1-p)
p.214
• Binomial Probability Distribution
p.212
 n x
n x
— f(x) = P(X=x) = 

p
(
1

p
)
 x
 
— Table 5 (p.989-p.997):
Given n, p and x, find f(x).
Homework: p.216 #26, #27, #29, #30 c, d.
Example: p.216 #25
Given:
Binomial, n = 2, p = .4
“Success”
X: # of successes in 2 trials
Answer:
b. f(1) = ? (Table 5)
f(1) = .48
c. f(0) = ?
f(0) = .36
d. f(2) = ?
e. P(X1) = ?
f(2) = .16)
P(X1) = f(1) + f(2) = .64
e. E(X) = np = .8 Var(X) = np(1-p) = .48  = ?
Example: p.217 #35
(Application)
Binomial distribution?
1. Is there a criterion for “success” and “failure”?
2. n trials? n=?
3. p=?
Given:
“Success”: withdraw; n = 20; p = .20
Answer:
a. P(X  2) = f(0) + f(1) + f(2)
= .0115 + .0576 + .1369 = .2060
b. P(X=4) = f(4) = .2182
c. P(X>3) = 1 - P(X  3) = 1 – [f(0) + f(1) + f(2) + f(3)]
= 1 - .0115 - .0576 - .1369 - .2054 = .5886
d. E(X) = np = (20)(.2) = 4.
III. Two Popular Discrete Probability Distributions
2. Poisson Distribution p.218
• Random variable X: x “occurrences” per unit (the
number of “occurrences” in a unit - time, size, ...).
Example: X = the number of calls per hour. Does X
have a Poisson distribution?
Answer: Yes.
Because
“Occurrence”: a call.
Unit: an hour.
X: the number of “occurrences” (calls) per unit (hour).
Reading: p.216 #29, #30, and p.220 #40, p.221 #42
Binomial or Poisson? If Binomial, “success”? p? n?
If Poisson, “occurrence”? ? unit?
III. Two Popular Discrete Probability Distributions
• Summary Measures for Poisson Distribution
E(X) = 
( is given)
Var(X) = 
• Poisson Probability Distribution
— f(x) = P(X=x) =
 xe
p.219
x!
— Table 7 (p.999 through p.1004): Given 
and x, find f(x).
Note: Keep the unit of  consistent with the question.
Homework: p.220 #38, p.221 #42, #43
Example: p.220 #39
Given:
•Poisson distribution, because
X = the number of occurrences per time period.
•  = 2 and unit is “a time period”.
Note: Unit in questions may be different.
Answer:
a. f(x) =
 xe
x!
2 x e 2
=
x!
b. What is the average number of occurrences in
three time periods? (Different unit!)
3 = (3)( ) = 6.
d. f(2) = P(X=2) = .2707
e. f(6) = P(X=6) = .1606
6 x e 6
c. f(x) =
x!
(Table 7.  =2, x = 2)
(Table 7.  =6, x = 6)
Example: p.221 #43
Given:
• Poisson distribution, because
X = the number of arrivals per time period.
•  = 10 and unit is “per minute”.
Note: Unit in questions may be different.
Answer:
a. f(0) = 0
(Table 7.  =10, x = 0)
b. P(X3) = f(0)+f(1)+f(2)+f(3) (Table 7.  =10, x = ?)
= 0 + .0005 + .0023 + .0076 = .0104
c. P(X=0) = f(0) = .0821
(Table 7.  =(10)(15/60)=2.5 for unit of 15 seconds, x = 0)
d. P(X1) = ?
(Table 7.  =2.5, x = ?)
P(X1) = f(1) + f(2) + f(3) + … ???
= 1 - P(X<1) = 1 - f(0) = 1 - .0821 = .9179.
Chapter 5 Summary
• Binomial distribution: X = # of “successes” out of n trials
• Poisson distribution: X = # of “occurrences” per unit
• For Binomial distribution, E(X)=np, Var(X)=np(1-p)
• For Poisson distribution, E(X) =  = Var(X)
• Binomial distribution table: Table 5
• Poisson distribution table: Table 7
• Poisson Approximation of Binomial distribution
If p  .05 AND n  20, Poisson distribution (Table 7) can
be used to find probability for Binomial distribution.
Example: A Binomial distribution with n = 250 and p = .01,
f(3) = ?
Answer:  = np =(250)(.01)=2.5. From Table 7, f(x)=.2138
• Sampling with replacement and without replacement.
Example: A bag of 100 marbles contains 10% red ones.
Assume that samples are drawn randomly with
replacement.
a. If a sample of two is drawn, what is the probability
that both will be red?
b. If a sample of three is drawn, what is the probability
that at least one will be red?
(“without replacement”: Hypergeometric distribution
p.214)
Answer:
a. n = 2, p = .1, f(2) = ?
From Table 5, f(2) = .01
b. n = 3, p = .1, P(X1) = f(1) + f(2) + f(3)
From Table 5, P(X1) = .2430 + .0270 + .0010 = .271