MATH 264

Statistics for
Social Sciences

Chapter 6

Discrete Random Variables and

Probability Distributions
Definition

A Random variable is a variable that takes on

numerical values determined by the outcome of a
random experiment
 Introduction to
Probability Distributions

 Random Variable
 Represents a possible numerical value from
a random experiment
Random
Variables

Ch. 6 Discrete Continuous Ch. 7

Random Variable Random Variable
 Discrete Random Variables
 Can only take on a countable number of values
Examples:

 Roll a die twice

Let X be the number of times 4 comes up
(then X could be 0, 1, or 2 times)

 Toss a coin 5 times.

Let X be the number of heads
(then X = 0, 1, 2, 3, 4, or 5)
Definition

A Probability Distribution Function P(X), of a

discrete random variable X expresses that the
random variable X takes the value x, as a function of
X.

That is;

P(X) = P(X=x) for all values of x

 Discrete Probability Distribution
Experiment: Toss 2 Coins. Let X = # heads.
Show P(x) , i.e., P(X = x) , for all values of x:

4 possible outcomes
Probability Distribution
T T x Value Probability
0 1/4 = .25
T H 1 2/4 = .50
2 1/4 = .25
H T
Probability

.50

.25
H H
0 1 2 x
 Probability Distribution
Required Properties

 P(x)  0 for any value of x

 The individual probabilities sum to 1;

 P(x)  1
x

(The notation indicates summation over all possible x values)

 Cumulative Probability Function

 The cumulative probability function, denoted

F(x0), shows the probability that X is less than or
equal to x0

F(x 0 )  P(X  x 0 )

 In other words,

F(x 0 )   P(x)
xx0
Properties of F(X0)

 0 ≤ F(X0) ≤ 1 for every number x0

 İf x0 and x1 are two numbers with x0 < x1

 F(X0) < F(X1)
Descriptive Measures
 Functions of Random Variables

 If P(x) is the probability function of a discrete

random variable X , and g(X) is some function of
X , then the expected value of function g is

E[g(X)]   g(x)P(x)
x

 Note that expected value takes the average of g(x)

 Expected Value
 Expected Value (or mean) of a discrete
distribution (Weighted Average)
μ  E(x)   xP(x)
x

x P(x)
 Example: Toss 2 coins, 0 .25
x = # of heads, 1 .50

compute expected value of x: 2 .25

E(x) = (0 x .25) + (1 x .50) + (2 x .25)

= 1.0
 Variance and Standard
Deviation
 Variance of a discrete random variable X

σ  E(X  μ)   (x  μ) P(x)
2 2 2

 Standard Deviation of a discrete random variable X

σ  σ2  
x
(x  μ)2
P(x)
Shortcut for variance formula

Var(X) = s²
= E(X²) – m²
= S X² P(X) – m²
 Standard Deviation Example

 Example: Toss 2 coins, X = # heads,

compute standard deviation (recall E(x) = 1)

σ x
(x  μ)2
P(x)

σ  (0  1)2 (.25)  (1 1)2 (.50)  (2  1)2 (.25)  .50  .707

Possible number of heads

= 0, 1, or 2
Rules for Expectation & Variance

 E(c) = c  V(c) = 0
 E(cX) = cE(X)  V(cX) = c²V(X)
 E(cX+Y)= cE(X)+E(Y)  V(cX+Y)= c²V(X)+V(Y)
 E(X-cY)= E(X)-cE(Y)  V(X-cY)= V(X)+c²V(Y)

Where X and Y random

variable & c constant
Solve CE4,1 and CE4,2

CE4,1: Consider the following probability distribution function (pdf). Let X denotes number of cars sold in a
day.

X 0 1 2 3 4 5 6
P(X) 0.07 0.19 0.23 0.17 0.16 0.14 0.04

a) What is P(X>3) =?
b) P(2<X<5) = ?
c) P(X≥2) = ?
d) P(X<6) = ?

CE4,2: Suppose that the PDF for the number of errors, X, on pages from business textbook is

P(0) = 0.81 P(1) = 0.17 P(2) = 0.02

Find the mean standard deviation of number of errors per page

Solution: CE4,3

CE4,3: Consider the following probability distribution function (pdf). Let X denotes number of cars
sold in a day.

X 0 1 2 3 4 5 6
P(X) 0.07 0.19 0.23 0.17 0.16 0.14 0.04

Find mean and standard deviation.

Mean= m = E(X)=SX*P(X) =(0)(0.07)+(1)(0.19)+………+ (6)(0.04)=2.74 3 cars will be sold in any

random day

Variance=s2 = SX-m)2P(X) = (0-2.74)2(0.07)+(1-2.74)2(0.19)+ ……+(6-2.74)2(0.04)=2.63

s= 2.63 = 1.62  2 cars is the expected deviation from center
 Probability Distributions
Probability
Distributions

Ch. 6 Discrete Continuous Ch. 7

Probability Probability
Distributions Distributions

Bernoulli Normal

Binomial

Hypergeometric
Bernoulli Distribution
 Bernoulli Distribution

 Consider only two outcomes: “success” or “failure”

 Let P denote the probability of success
 Let 1 – P be the probability of failure
 Define random variable X:
x = 1 if success, x = 0 if failure
 Then the Bernoulli probability function is

P(0)  (1 P) and P(1)  P

 Bernoulli Distribution
Mean and Variance

 The mean is µ = P

μ  E(X)   xP(x)  (0)(1 P)  (1)P  P

X

 The variance is σ2 = P(1 – P)

σ 2  E[(X  μ)2 ]   (x  μ)2 P(x)

X

 (0  P) (1 P)  (1 P) P  P(1 P)

2 2
Binomial Distribution
 Binomial Distribution
 A fixed number of observations, n
 e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse
 Two mutually exclusive and collectively exhaustive
categories
 e.g., head or tail in each toss of a coin; defective or not defective
light bulb
 Generally called “success” and “failure”
 Probability of success is P , probability of failure is 1 – P
 Constant probability for each observation
 e.g., Probability of getting a tail is the same each time we toss
the coin
 Observations are independent
 The outcome of one observation does not affect the outcome of
the other
 Possible Binomial Distribution
Settings

 A manufacturing plant labels items as

either defective or acceptable
 A firm bidding for contracts will either get a
contract or not
 A marketing research firm receives survey
responses of “yes I will buy” or “no I will
not”
 New job applicants either accept the offer
or reject it
 Developing Binomial Distribution
Lets repeat success and failure type experiment n independent times.
Assume that we obtain below sequence.

S S S……S S F F F……F F
Probability of observing this sequence is
p*p* ….p(1-p)*(1-p)*……*(1-p)

Since X defined as number of success we can write

px(1-p)n-x

Is this the only possible sequence that we can observe?

How many different sequences can be observed?

 Sequences of x Successes
in n Trials

 The number of sequences with x successes in n

independent trials is:

n!
C 
n

x! (n  x)!
x

Where n! = n·(n – 1)·(n – 2)· . . . ·1 and 0! = 1

 These sequences are mutually exclusive, since no two

can occur at the same time
 Developing Binomial Distribution
Lets repeat success and failure type experiment n independent times.
Assume that we obtain below sequence.

S S S……S S F F F……F F
Probability of observing this sequence is
p*p* ….p(1-p)*(1-p)*……*(1-p)

Since X defined as number of success we can write

px(1-p)n-x
We can observe 𝑋𝑛 different sequences

𝑛
P(X)= 𝑋
px(1-p)n-x
Example
The random variable X represents the number of students
who prefer news from the Internet among a random
sample of students from a large university. If the
population proportion of students who prefer Internet
news is 0.6. And, if we relabel the outcome “Internet” as
a success ( S ) and “not Internet” as a failure (F). List the
associated probabilities, and the value of X for the
elementary outcomes of sampling 4 students.
 Example: Answer

Sample points FFFF SFFF SSFF SSSF SSSS

(basic FSFF SFSF SSFS
outcomes) FFSF SFFS SFSS
FFFS FSSF FSSS
FSFS
FFSS
 Example: Answer

Sample points FFFF SFFF SSFF SSSF SSSS

(basic FSFF SFSF SSFS
outcomes) FFSF SFFS SFSS
FFFS FSSF FSSS
FSFS
FFSS
Value of X 0 1 2 3 4
 Example: Answer

Sample points FFFF SFFF SSFF SSSF SSSS

(basic FSFF SFSF SSFS
outcomes) FFSF SFFS SFSS
FFFS FSSF FSSS
FSFS
FFSS
Value of X 0 1 2 3 4
Probability of 0404 0.601 0.403 0.602 0.402 3 1
0.60 0.40 0.604
each outcome
 Example: Answer

Sample points FFFF SFFF SSFF SSSF SSSS

(basic FSFF SFSF SSFS
outcomes) FFSF SFFS SFSS
FFFS FSSF FSSS
FSFS
FFSS
Value of X 0 1 2 3 4
Probability of 0404 0.601 0.403 0.602 0.402 3
0.60 0.40 1
0.604
each outcome
Number of 4 4 4 4 4
0
=1 1
=4 2
=6 3
=4 4
=1
outcomes
 Example: Answer
Sample points FFFF SFFF SSFF SSSF SSSS
(basic FSFF SFSF SSFS
outcomes) FFSF SFFS SFSS
FFFS FSSF FSSS
FSFS
FFSS
Value of X 0 1 2 3 4
Probability of 0404 0.601 0.403 0.602 0.402 3
0.60 0.40 1
0.604
each outcome
Number of 4 4 4 4 4
0
=1 1
=4 2
=6 3
=4 4
=1
outcomes
P(X) 0404 4*0.601 0.403 6*0.602 0.402 4*0.603 0.401 0.604

𝑛
P(X)= 𝑋
px(1-p)n-x
 Binomial Distribution Formula

n! X nX
P(x)  P (1- P)
x ! (n  x )!
P(x) = probability of x successes in n trials,
with probability of success P on each trial
Example: Flip a coin four
times, let x = # heads:
x = number of ‘successes’ in sample,
n=4
(x = 0, 1, 2, ..., n)
P = 0.5
n = sample size (number of trials
or observations) 1 - P = (1 - 0.5) = 0.5
P = probability of “success” x = 0, 1, 2, 3, 4
 Example:
Calculating a Binomial Probability
What is the probability of one success in five
observations if the probability of success is 0.1?
x = 1, n = 5, and P = 0.1

n!
P(x  1)  P X (1 P)n X
x! (n  x)!
5!
 (0.1)1(1 0.1)51
1!(5  1)!
 (5)(0.1)(0.9)4
 .32805
 Binomial Distribution
 The shape of the binomial distribution depends on the
values of P and n
Mean P(x) n = 5 P = 0.1
.6
 Here, n = 5 and P = 0.1 .4
.2
0 x
0 1 2 3 4 5

P(x) n = 5 P = 0.5
 Here, n = 5 and P = 0.5 .6
.4
.2
0 x
0 1 2 3 4 5
 Binomial Distribution
Mean and Variance

 Mean
μ  E(x)  nP
 Variance and Standard Deviation

σ  nP(1- P)
2

σ  nP(1- P)
Where n = sample size
P = probability of success
(1 – P) = probability of failure
 Binomial Characteristics
Examples
μ  nP  (5)(0.1)  0.5
Mean P(x) n = 5 P = 0.1
.6
.4
σ  nP(1- P)  (5)(0.1)(1 0.1) .2
 0.6708 0 x
0 1 2 3 4 5

μ  nP  (5)(0.5)  2.5 P(x) n = 5 P = 0.5

.6
.4
σ  nP(1- P)  (5)(0.5)(1 0.5) .2
 1.118 0 x
0 1 2 3 4 5
CE4,4
The cubs are to play a series of 5 games in St. Louis
against the Cardinals. For any one game it is estimated
that the probability of a Cubs win is 0.4. The
outcomes of the five games are independent of one
other.
Solution: CE4,4
a) What is the probability that the Cubs will win all five games?

5
P(X=5)= P(X)= 5
(0.4)5(1-0.4)0 =0.0102= 1%

b) What is the probability that the Cubs will win a majority of five games?

P(X≥3)= P(X=3)+P(X=4)+P(X=5)
= 53 (0.4)3(1-0.4)2 + 54 (0.4)4(1-0.4)1 + 5
5
(0.4)5(1-0.4)0 =0.3174= 32%

c) If the Cubs win the first game, what is the probability that they will win a majority of five games?

First game cubs 4 games to play

P(X≥2)=0.5248 = 52%

d) Before the series begins, what is the expected number of Cubs wins in these five games?

Mean=np=5(0.4)=2 games

e) If Cubs win the first game, what is the expected number of Cubs wins in these five games?

First game cubs 4 games to play

Y=1+X
Mean=E(Y)=E(1+X)=1+E(X)=1+(4)(0.4)=2.6  3 games
The Hypergeometric Distribution
 The Hypergeometric Distribution

 “n” trials in a sample taken from a finite

population of size N
 Sample taken without replacement
 Outcomes of trials are dependent
 Concerned with finding the probability of “X”
successes in the sample where there are “S”
successes in the population
 Hypergeometric Distribution
Formula

S! (N  S)!

CSxCNnxS x! (S  x)! (n  x)! (N  S  n  x)!
P(x)  N

Cn N!
n!(N  n)!
Where
N = population size
S = number of successes in the population
N – S = number of failures in the population
n = sample size
x = number of successes in the sample
n – x = number of failures in the sample
Hypergeometric Distribution
Mean and Variance
Mean = E(X)=m  n(S/N)
Variance =Var(X) = s² = n(S/N)(1-S/N)*((N-n)/(N-1))

Where (N-n)/(N-1) is the finite population correction

We assume that we sample from a finite population

without replacement and n/N > 0.05. So, the
probability of a success changes for each trial.
Otherwise you can use Binomial Distribution
 Using the
Hypergeometric Distribution
■ Example: 3 different computers are checked from 10 in
the department. 4 of the 10 computers have illegal
software loaded. What is the probability that 2 of the 3
selected computers have illegal software loaded?
N = 10 n=3
S=4 x=2

CSxCNnxS C24C16 (6)(6)

P(x  2)  N
 10   0.3
Cn C3 120
The probability that 2 of the 3 selected computers have illegal
software loaded is 0.30, or 30%.
CE4,5
A company receives a shipment of 16 items. A random
sample of 4 items is selected, and the shipment is
rejected if any of these items proves to be defective.

N=16 n=4
X=number of defectives in the sample
Decision rule: Reject shipment if X≥1 and accept
shipment if X=0
Solution: CE4,5
N=16 n=4 X=number of defectives in the sample
Decision rule: Reject shipment if X≥1 and accept shipment if X=0

a) What is the probability of accepting a shipment containing 4 defective items?

S=4
4 12
0 4
P(accept)=P(X=0)= 16 = 0.29
4

b) What is the probability of accepting a shipment containing 1 defective item?

S=1
1 15
0 4
P(accept)=P(X=0)= 16 = 0.75
4

c) What is the probability of rejecting a shipment containing 1 defective item?

S=1
P(reject)=1-P(accept) = 1- 0.75 =0.25
Discuss: CE4,6 & CE4,7
CE4,6: An analyst predicted that 3.5% of all small corporations
would file for bankruptcy in the coming year. For a random sample
of 100 small corporations, estimate the probability that at least 3
will file for bankruptcy in the next year.

CE4,7:An auditor reviewing the invoices of a small company finds

that there are errors in 1.5% of them. If auditor looks at 500
invoices, what is the probability that he finds more than 1 invoice
with errors?
 Chapter Summary

 Defined discrete random variables and

probability distributions
 Discussed descriptive measures
 Discussed the Bernoulli distribution
 Discussed the Binomial distribution
 Discussed the Hypergeometric distribution