Chapter 5
Discrete Random Variables
Probability Distributions
Discrete Random Variables
Probability Distributions
 Overview
 Random Variables
 Mean and Standard Deviation for Random
Variables
 Binomial Probability Distributions
 Mean, Standard Deviation for the Binomial
Distribution
Overview
 This chapter will deal with the construction of
discrete probability distributions by combining
the methods of descriptive statistics presented
in Chapter 2 and 3 and those of probability
presented in Chapter 4.
Combining Descriptive Methods
and Probabilities
In this chapter we will construct probability distributions
by presenting possible outcomes along with the relative
frequencies we expect.
Random Variables
 A random variable is a variable whose value is a


numerical outcome of a probability experiment.
We usually denote random variables by capital
letters near the end of the alphabet, such as X or Y
Mathematically speaking, a random variable X
is a function that assigns a number to every
outcome of the sample space S
Random Variables
 There are two kinds of random variables
 Discrete Random Variables
 Continuous Random Variables
Random Variables
 A random variable is discrete if it has a finite or
countable number of possible outcomes that can be
listed.
x
0
2
4
6
8
10
 A random variable is continuous if it has an infinite
number of outcomes, represented by the intervals on a
number line.
x
0
2
4
6
8
10
Random Variables
 Example: Decide if the random variable X is discrete or
continuous.
a.) The distance your car travels on a tank of gas
The distance your car travels is a continuous
random variable because it is a measurement that
cannot be counted. (All measurements are
continuous random variables.)
b.) The number of students in a statistics class
The number of students is a discrete random
variable because it can be counted.
Discrete Random Variables
More Discrete Random
Variable Examples
Experiment
Random
Variable
Possible Values
Make 100 Sales Calls # Sales
0, 1, 2, ..., 100
Inspect 70 Radios
# Defective
0, 1, 2, ..., 70
Answer 33 Questions
# Correct
0, 1, 2, ..., 33
Count Cars at Toll
# Cars
Between 11:00 & 1:00 Arriving
0, 1, 2, ..., 
Notation
 If X a random variable, then its numerical values


are denoted by the corresponding lower case
letters x.
The notation P (X = x) will be used to denote
“probability of the event that makes X = x”
This probability is denoted by p(x) that is,
p(x) = P (X = x) is the probability distribution
function (pdf in the TI-83 notation)
Discrete Probability Distributions
 A discrete probability distribution for the
discrete random variable X lists each possible
value the variable X can assume, together with
its probability.
 This probability distribution is often expressed
in the format of a graph, table, or formula.
 In this course a discrete random variable X
will always have a finite number of possible
values.
Discrete Probability Distributions
 The probability distribution of X lists the values
and their probabilities as shown in the table
 The probabilities pi must satisfy two requirements:
1. Every probability pi is a number between 0 and 1
2. p1 + p2 +・ ・ ・+pk = 1
Probability Distributions and
Histograms
Example
The spinner below is divided into two sections. The probability
of landing on the 1 is 0.25. The probability of landing on the 2
is 0.75. Let X be the number the spinner lands on. Construct a
probability distribution for the random variable X.
1
2
x
P (X = x)
1
2
0.25
0.75
Each probability is
between 0 and 1.
Sum of probabilities is 1.
Example Continued
The spinner below is spun two times. The probability of
landing on the 1 is 0.25. The probability of landing on the 2
is 0.75. Let X be the sum of the two spins. Construct a
probability distribution for the random variable X.
The possible sums are 2, 3, and 4.
P (sum of 2) = 0.25  0.25 = 0.0625
1
2
Spin a 1 on the
first spin.
“and”
Spin a 1 on the
second spin.
Example Continued
P (sum of 3) = 0.25  0.75 = 0.1875
1
2
Spin a 1 on the
first spin.
“and”
Spin a 2 on the
second spin.
“or”
P (sum of 3) = 0.75  0.25 = 0.1875
X = Sum
P (X =x)
of spins
Spin a 2 on the
2
3
4
0.0625
0.375
first spin.
“and”
0.1875 + 0.1875
Spin a 1 on the
second spin.
Example Continued
1
P (sum of 4) = 0.75  0.75 = 0.5625
2
X = Sum
P (X =x)
of spins
2
3
4
0.0625
0.375
0.5625
Spin a 2 on the
first spin.
“and”
Spin a 2 on the
second spin.
Each probability is between
0 and 1, and the sum of the
probabilities is 1.
Example Continued
Graph the probability distribution using a histogram.
X = Sum
P (X =x)
of spins
0.0625
0.375
0.5625
Sum of Two Spins
0.6
0.5
Probability
2
3
4
p(x)
0.4
0.3
0.2
0.1
x
0
2
3
Sum
4
Another Quick Example
Experiment: Toss 2 Coins. X= Count # of Tails
Probability Distribution
Values of X
Probabilities, p (x )
0
p(0) = P(X = 0) = 1/4 = .25
1
p(1) = P(X = 1) = 2/4 = .50
2
p(2) = P(X = 2) = 1/4 = .25
Visualizing The Discrete
Probability Distribution
Table
X = # Tails
0
1
2
p(x) = P(X=x)
.25
.50
.25
p(x)
.50
.25
.00
Graph
x
0
1
2
Function
n
!
x
n x
P (X  x ) 
p (1  p ) .
(n  x )! x !
More Coins
 What is the probability distribution of the
discrete random variable X that counts the
number of heads in four tosses of a coin?
 We can derive this distribution if we make two


reasonable assumptions.
The coin is balanced, so each toss is equally likely to
give H or T.
The coin has no memory, so tosses are independent.
That is, the outcome of a toss does not depend on the
outcome of the previous ones.
 The outcome of four tosses is a sequence of heads and

tails such as HTTH. There are 16 possible outcomes.
The picture below gives the sample space along with
the value of X for each outcome.
The Table for this
Probability Distribution
Probability Histogram for
this Distribution
Some Probability Calculations
 The probability of tossing at least two heads is
P( X  2)  0.375  0.25  0.0625  0.6875
 The probability of tossing at least one head is
found by use of the complement rule
P( X  1)  1  P( X  1)  1  P( X  0)
 1  0.0625  0.9375
Mean and Standard Deviation
of a Discrete Random Variable
Introducing the Mean of a Discrete
Random Variable - Example
Ages of eight students
Probability distribution of X, the
age of a randomly selected student
Express the mean age of the eight students in terms of the
probability distribution of the random variable X.
Express the mean age of the eight students in terms of the
probability distribution of the random variable X.
Express the mean age of the eight students in terms of the
probability distribution of the random variable X.
Express the mean age of the eight students in terms of the
probability distribution of the random variable X.
Express the mean age of the eight students in terms of the
probability distribution of the random variable X.
Express the mean age of the eight students in terms of the
probability distribution of the random variable X.
Mean of a Discrete Random Variable
Mean of a Discrete Random Variable
The mean of a discrete random variable is given by
μ = Σ xP(X=x)
Each value of x is multiplied by its corresponding
probability and the products are added.
Example: Find the mean of the probability distribution for
the sum of the two spins.
x
2
3
4
xP (x)
P (x)
0.0625 2(0.0625) = 0.125
0.375 3(0.375) = 1.125
0.5625 4(0.5625) = 2.25
Σ xP(X=x) = 3.5
The mean for the
two spins is 3.5.
Interpretation
 The following interpretation is commonly known as the
law of averages and in mathematical circles as the law
of large numbers.
Variance of a Discrete Random
Variable
The variance of a discrete random variable is given by
 2 = Σ(x – μ)2P (X = x)
Example: Find the variance of the probability distribution
for the sum of the two spins. The mean is 3.5.
x–μ
(x – μ)2
p(x)(x – μ)2
0.0625
–1.5
2.25
 0.141
3
0.375
–0.5
0.25
 0.094
4
0.5625
0.5
0.25
 0.141
x
p (x)
2
ΣP(X=x)(x – 2)2
 0.376
Variance of a Discrete Random
Variable
The standard deviation of a discrete random variable is
σ = σ2 
 (x
  )2P (X  x ).
Example: Find the variance of the probability distribution
for the sum of the two spins. The mean is 3.5.
x–μ
(x – μ)2
p(x)(x – μ)2
0.0625
–1.5
2.25
 0.141
3
0.375
–0.5
0.25
 0.094
4
0.5625
0.5
0.25
 0.141
x
p (x)
2
σ  σ 2  0.376
 0.613
Expected Value
The expected value of a discrete random variable is equal to
the mean of the random variable.
Expected Value = E(x) = μ = ΣxP(X=x)
Example: At a raffle, 500 tickets are sold for $1 each for a
prize of $100. What is the expected value of your gain?
Your gain for the $100 prize is $100 – $1 = $99.
Write a probability distribution for the possible gains.
Expected Value
At a raffle, 500 tickets are sold for $1 each for a prize of
$100. What is the expected value of your gain?
Gain, x
P (x)
$99
1
500
$-1
499
500
Winning
no prize
E(x) = ΣxP(x)
= $99 ×
1
499
+ (-$1) ×
500
500
= -$0.80
Because the expected value is
negative, you can expect to lose
$0.80 for each ticket you buy.
Binomial Distributions
Factorial Notation
Binomial Experiments
The interpretation for this number is the following:
The binomial coefficient represents the number of different ways
in which x objects can be selected from a list of n distinct objects
when the order of the selection is not important
Binomial Experiments
A binomial experiment is a probability experiment
that satisfies the following conditions.
1. The experiment is repeated for a fixed number of trials, where
each trial is independent of other trials.
2. There are only two possible outcomes of interest for each trial.
The outcomes can be classified as a success s, or as a failure f .
3. The probability of a success p = P (s ) is the same for each trial.
p is called the success probability.
4. The random variable X counts the number of successful trials.
Notation for Binomial Experiments
Symbol
n
Description
The number of times a trial is repeated.
p = P (s )
The probability of success in a single trial.
q = P (f )
The probability of failure in a single trial.
q = 1– p.
X
The random variable represents a count of the
number of successes in n trials: The possible
values of X are: x = 0, 1, 2, 3, … , n.
Example: Decide whether the experiment is a binomial
experiment. If it is, specify the values of n, p, and q, and list
the possible values of the random variable X. If it is not a
binomial experiment, explain why.
You randomly select a card from a deck of cards, and note if
the card is an Ace. You then put the card back and repeat
this process 8 times.
This is a binomial experiment. Each of the 8 selections
represent an independent trial because the card is
replaced before the next one is drawn. There are only
two possible outcomes: either the card is an ace or not.
q  1  1  12
n 8 p  4  1
x  0,1,2,3,4,5,6,7,8
52 13
13 13
Example: Decide whether the experiment is a binomial
experiment. If it is, specify the values of n, p, and q, and list
the possible values of the random variable X. If it is not a
binomial experiment, explain why.
You roll a die 10 times and note the number the die lands
on.
This is not a binomial experiment. While each trial (roll)
is independent, there are more than two possible
outcomes: 1, 2, 3, 4, 5, and 6.
More Binomial Experiments
# of reds in 15 spins of roulette wheel
# of defective items in a batch of 25 items
# of correct on a 33 question exam
# of customers who purchase out of 100
customers who enter a store
Binomial Probability Formula
In a binomial experiment, the probability of exactly x successes in
n trials is
n  x n x
n!
P (X  x )    p q

p xq n  x
(n  x )! x !
x 
Example: A bag contains 10 chips; 3 of the chips are red, 5 are
white, and 2 are blue. Three chips are selected, with replacement.
Find the probability that you select exactly one red chip.
p = the probability of selecting a red chip  3  0.3
10
q = 1 – p = 0.7
3
1
2
P
(
X

1)

(0.3)
(0.7)
 0.441
n=3
 
1 
x=1
Example: A bag contains 10 chips. 3 of the chips are red, 5
are white, and 2 are blue. Four chips are selected, with
replacement. Create a probability distribution for the number
of red chips selected.
p = the probability of selecting a red chip  3  0.3
q = 1 – p = 0.7
n=4
x = 0, 1, 2, 3, 4
10
x
P (X=x)
0
1
2
3
4
0.240
0.412
0.265
0.076
0.008
The binomial
probability
formula is used
to find each
probability.
Finding Probabilities
Example: The following probability distribution represents
the probability of selecting 0, 1, 2, 3, or 4 red chips when 4
chips are selected.
x
0
1
2
3
4
P (X=x)
0.24
0.412
0.265
0.076
0.008
a.) Find the probability of selecting no
more than 3 red chips.
b.) Find the probability of selecting at
least 1 red chip.
a.) P (no more than 3) = P (x  3) = P (0) + P (1) + P (2) + P (3)
= 0.24 + 0.412 + 0.265 + 0.076 = 0.993
b.) P (at least 1) = P (x  1) = 1 – P (0) = 1 – 0.24 = 0.76
Graphing the distribution
Example: The following probability distribution represents
the probability of selecting 0, 1, 2, 3, or 4 red chips when 4
chips are selected. Graph the distribution using a histogram.
P (x)
P (X=x)
0.24
0.412
0.265
0.076
0.008
Selecting Red Chips
0.5
Probability
x
0
1
2
3
4
0.4
0.3
0.2
0.1
x
0
0
1
2
3
Number of red chips
4
Mean and Standard Deviation
of a Binomial Distribution
Population Parameters of a Binomial Distribution
Mean: μ  np
Variance: σ 2  npq
Standard deviation: σ  npq
Example: One out of 5 students at a local college say that
they skip breakfast in the morning. Find the mean, variance
and standard deviation if 10 students are randomly selected.
n  10
p  1  0.2
5
q  0.8
μ  np
σ 2  npq
σ  npq
 10(0.2)
 (10)(0.2)(0.8)
 1.6
2
 1.6
 1.3
Thinking Challenge
 You’re taking a 33 question
multiple choice test. Each
question has 4 choices.
Clueless on 1 question, you
decide to guess. What’s the
chance you’ll get it right?
 If you guessed on all 33
questions, what would your
grade most likely be? pass?
X
Answer
Guessing 33 Answers on a Test
Data
Sample size
Probability of success
33
0.25
Statistics
Mean
Variance
Standard deviation
8.25
6.1875
2.487469
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
P(X)
0.000075339
0.000828733
0.004419907
0.015224123
0.038060308
0.073583262
0.114462852
0.147166524
0.159430401
0.147620742
0.118096593
0.082309747
0.050300401
0.027084831
0.012897539
0.005445627
0.002042110
0.000680703
0.000201690
0.000053076
0.000012384
0.000002556
0.000000465
0.000000074
0.000000010
0.000000001
0.000000000
0.000000000
0.000000000
0.000000000
0.000000000
0.000000000
0.000000000
0.000000000
Porbality Distribution
of Total
Number
Right Answers
Probability
Distribution
for
TotalofNumber
of Right Answers
0.180000000
0.160000000
0.140000000
0.120000000
P(X)
0.100000000
0.080000000
0.060000000
0.040000000
0.020000000
0.000000000
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
Number of Successes
Another Thinking Challenge
 You’re a telemarketer selling
service contracts for Macy’s. You
have sold 20 in your last 100 calls
(p = .20). If you call 12 people
tonight,
what’s the probability of
A. No sales?
B. Exactly 2 sales?
C. At most 2 sales?
D. At least 2 sales?
Solution
 Using the Binomial probability formula:
A. p(0) = .0687
B. p(2) = .2835
C. p(at most 2) = p(0) + p(1) + p(2)
= .0687 + .2062 + .2835
= .5584
D. p(at least 2) = p(2) + p(3)...+ p(12)
= 1 - [p(0) + p(1)]
= 1 - .0687 - .2062
= .7251
More Discrete
Probability Distributions
Geometric Distribution
A geometric distribution is a discrete probability
distribution of a random variable X that satisfies the
following conditions.
1. A trial is repeated until a success occurs.
2. The repeated trials are independent of each other.
3. The probability of a success p is constant for each trial.
The probability that the first success will occur on trial x is
p(x)=P (X = x) = p(q)x – 1, where q = 1 – p
Geometric Distribution
Example: A fast food chain puts a winning game piece on
every fifth package of French fries. Find the probability
that you will win a prize,
A. with your third purchase of French fries,
B. with your third or fourth purchase of French fries.
p = 0.20
A. x = 3
q = 0.80
B. x = 3, 4
P (3) = (0.2)(0.8)3 – 1
= 0.128
P (3 or 4) = P (3) + P (4)
 0.230
Poisson Distribution
The Poisson distribution is a discrete probability distribution
of a random variable x that satisfies the following conditions.
1. The experiment consists of counting the number of times an event, x, occurs
in a given interval. The interval can be an interval of time, area, or volume.
2. The probability of the event occurring is the same for each
interval.
3. The number of occurrences in one interval is independent of the
number of occurrences in other intervals.
The probability of exactly x occurrences in an interval is
x μ
μ
P (x )  e
x!
where e  2.71818 and μ is the mean number of occurrences.
Poisson Distribution
Example: The mean number of power outages in the city of
Brunswick is 4 per year. Find the probability that in a given
year,
A. there are exactly 3 outages,
B. there are more than 3 outages.
A.   4, x  3
43(2.71828)-4
P (3) 
3!
 0.195
B. P (more than 3)
 1  P (x  3)
 1  [P (3)  P (2) + P (1) + P (0)]
 1  (0.195  0.147  0.073  0.018)
 0.567