Probability and Discrete
Random Variable
Probability
What is Probability?
• When we talk about probability, we are
talking about a (mathematical) measure of
how likely it is for some particular thing to
happen
• Probability deals with chance behavior
– We study outcomes, or results of experiments
– Each time we conduct an experiment, we may
get a different result
Why study Probability?
• Descriptive statistics - describing and
summarizing sample data, deals with data
as it is
• Probability - Modeling the probabilities for
sample results to occur in situations in which the
population is known
• The combination of the two will let us do
our inferential statistics techniques.
Objectives
1. Learn the basic concepts of probability
2. Understand the rules of probabilities
3. Compute and interpret probabilities using
the empirical method
4. Compute and interpret probabilities using
the classical method
5. Compute the probabilities for the
compound events.
Sample Space & Outcomes
• Some definitions
– An experiment is a repeatable process where the results are
uncertain
– An outcome is one specific possible result
– The set of all possible outcomes is the sample space denoted by
a capital letter S
• Example
– Experiment … roll a fair 6 sided die
– One of the outcomes … roll a “4”
– The sample space … roll a “1” or “2” or “3” or “4” or “5” or “6”.
So, S = {1, 2, 3, 4, 5, 6} (Include all outcomes in braces {…}.)
Event
• More definitions
– An event is a collection of possible outcomes … we
will use capital letters such as E for events
– Outcomes are also sometimes called simple events
… we will use lower case letters such as e for
outcomes / simple events
• Example (continued)
– One of the events … E = {roll an even number}
– E consists of the outcomes e2 = “roll a 2”, e4 = “roll a
4”, and e6 = “roll a 6” … we’ll write that as {2, 4, 6}
Example
Consider an experiment of rolling a die again.
– There are 6 possible outcomes, e1 = “rolling a
1” which we’ll write as just {1}, e2 = “rolling a
2” or {2}, …
– The sample space is the collection of those 6
outcomes. We write S = {1, 2, 3, 4, 5, 6}
– One event of interest is E = “rolling an even
number”. The event is indicated by
E = {2, 4, 6}
Probability of an Event
• If E is an event, then we write P(E) as the
probability of the event E happening
• These probabilities must obey certain
mathematical rules
Probability Rule # 1
• Rule # 1 – the probability of any event must be greater
than or equal to 0 and less than or equal to 1, i.e.,
0  P( E )  1
– It does not make sense to say that there is a -30%
chance of rain
– It does not make sense to say that there is a 140%
chance of rain
Note – probabilities can be written as decimals (0, 0.3, 1.0),
or as percents (0%, 30%, 100%), or as fractions (3/10)
Probability Rule # 2
• Rule #2 – the sum of the probabilities of all the
outcomes must equal 1.
 P (e )  1
i
all outcomes
– If we examine all possible outcomes, one of them
must happen
– It does not make sense to say that there are two
possibilities, one occurring with probability 20% and
the other with probability 50% (where did the other
30% go?)
Example
On the way to work Bob’s personal judgment is that he
is four times more likely to get caught in a traffic jam
(TJ) than have an easy commute (EC). What values
should be assigned to P(TJ) and P(EC)?
Solution: Given
P (TJ )  4  P( EC)
Since
P(TJ )  P( EC)  1
Which means
4  P ( EC)  P ( EC)  1
5  P ( EC)  1
1
P ( EC) 
5
1
4
P ( TJ )  4  P ( EC)  4  
 5 5
Probability Rule (continued)
• Probability models must satisfy both of
these rules
• There are some special types of events
– If an event is impossible, then its probability
must be equal to 0 (i.e. it can never happen)
– If an event is a certainty, then its probability
must be equal to 1 (i.e. it always happens)
Unusual Events
• A more sophisticated concept
– An unusual event is one that has a low
probability of occurring
– This is not precise … how low is “low?
• Typically, probabilities of 5% or less are
considered low … events with probabilities
of 5% or lower are considered unusual
• However, this cutoff point can vary by the
context of the problem
How To Compute the Probability?
The probability of an event may be
obtained in three different ways:
– Theoretically (a classical approach)
– Empirically (an experimental approach)
– Subjectively
Compute Probability
theoretically
Equally Likely Outcomes
• The classical method of calculating the probability applies
to situations (or by assuming the situations) where all
possible outcomes have the same probability
which is called equally likely outcomes
• Examples
– Flipping a fair coin … two outcomes (heads and tails)
… both equally likely
– Rolling a fair die … six outcomes (1, 2, 3, 4, 5, and 6)
… all equally likely
– Choosing one student out of 250 in a simple random
sample … 250 outcomes … all equally likely
Equally Likely Outcomes
• Because all the outcomes are equally likely, then
each outcome occurs with probability 1/n where
n is the number of outcomes
• Examples
– Flipping a fair coin … two outcomes (heads and tails)
… each occurs with probability 1/2
– Rolling a fair die … six outcomes (1, 2, 3, 4, 5, and 6)
… each occurs with probability 1/6
– Choosing one student out of 250 in a simple random
sample … 250 outcomes … each occurs with
probability 1/250
Theoretical Probability
• The general formula is
P (E )  Number of ways E can occur
Number of possible outcomes
• If we have an experiment where
– There are n equally likely outcomes (i.e. N(S) = n)
– The event E consists of m of them (i.e. N(E) = m)
then
m
N (E )
P (E ) 

n
N(S )
A More Complex Example
Here we consider an example of select two subjects at random
instead of just one subject:
Three students (Katherine (K), Michael (M), and Dana (D)) want to
go to a concert but there are only two tickets available. Two of the
three students are selected at random.
Question 1: What is the sample space of who goes?
Solution: S = {(K,M),(K,D),(M,D)}
Question 2: What is the probability that Katherine goes?
Solution: Because 2 students are selected at random,
each outcome in the sample space has equal
chance to occur.
Therefore, P( Katherine goes) = 2/3.
Another Example
A local automobile dealer classifies purchases by number of doors and
transmission type. The table below gives the number of each
classification.
Manual
Automatic
Transmission Transmission
2-door
75
155
4-door
85
170
If one customer is selected at random, find the probability that:
1)
The selected individual purchased a car with automatic
transmission
2)
The selected individual purchased a 2-door car
Solutions
Apply the formula
P (E ) 
m
N (E )

n
N(S )
1) P(Automatic Transmission)
155  170
325 65



75  85  155  170 485 97
2) P(2 - door )
75  155
230 46



75  85  155  170 485 97
Compute Probability
empirically
Empirical Probability
• If we do not know the probability of a certain event E, we
can conduct a series of experiments to approximate it by
P (E ) 
frequency of E
number of trials of the experiment
• This is called the empirical probability or experimental
probability. It becomes a good approximation for P(E) if
we have a large number of trials (the law of large
numbers)
Example
We wish to determine what proportion of
students at a certain school have type A
blood
– We perform an experiment (a simple random
sample!) with 100 students
– If 29 of those students have type A blood,
then we would estimate that the proportion of
students at this school with type A blood is
29%
Example (continued)
We wish to determine what proportion of
students at a certain school have type AB blood
– We perform an experiment (a simple random
sample!) with 100 students
– If 3 of those students have type AB blood,
then we would estimate that the proportion of
students at this school with type AB blood is
3%
– This would be an unusual event
Another Example
Consider an experiment in which we roll two six-sided fair
dice and record the number of 3s face up. The only
possible outcomes are zero 3s, one 3, or two 3s. Here are
the results after 100 rolls of these two dice, and also after
1000 rolls:
100 Rolls
Outcome Frequency
0
80
1
19
2
1
1000 Rolls
Outcome Frequency
0
690
1
282
2
28
Using a Histogram
• We can express these results (from the 1000 rolls) in
terms of relative frequencies and display the results
using a histogram:
0.7
0.6
Relative
Frequency
0.5
0.4
0.3
0.2
0.1
0.0
0
1
Three’s Face Up
2
Continuing the Experiment
If we continue this experiment for several thousand
more rolls, the relative frequency for each possible
outcome will settle down and approach to a
constant. This is so called the law of large
numbers.
Coin-Tossing Experiment
Consider tossing a fair coin. Define the event H as the occurrence of a head. What
is the probability of the event H, P(H)?
•
Theoretical approach –
If we assume that the coin is fair, then there are two equally likely outcomes in a
single toss of the coin. Intuitively, P(H) = 50%.
•
Empirical approach –
If we do not know if the coin is fair or not. We then estimate the probability by
tossing the coin many times and calculating the proportion of heads occurring.
To show you the effect of applying large number of tosses on the accuracy of the
estimation. What we actually do here is to toss the coin 10 times each time and
repeated it 20 times. The results are shown in the next slide. We cumulate the total
number of tosses over trials to compute the proportion of heads. We plot the
proportions over trials in a graph as shown in the following slide. We observe that the
proportion of heads tends to stabilize or settle down near 0.5 (50%). So, the
proportion of heads over larger number of tosses is a better estimate of the true
probability P(H).
Experimental results of tossing a coin 10 times on
each trial
Trial
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Number of
Relative
Cumulative
Heads Observed Frequency Relative Frequency
5
5/10
5/10 = 0.5000
4
4/10
9/20 = 0.4500
4
4/10
13/30 = 0.4333
5
5/10
18/40 = 0.4500
6
6/10
24/50 = 0.4800
7
7/10
28/60 = 0.4667
6
6/10
34/70 = 0.4857
4
4/10
38/80 = 0.4750
7
7/10
45/90 = 0.5000
3
3/10
48/100 = 0.4800
4
4/10
52/110 = 0.4727
6
6/10
58/120 = 0.4838
7
7/10
65/130 = 0.5000
4
4/10
69/140 = 0.4929
3
3/10
72/150 = 0.4800
7
7/10
79/160 = 0.4938
6
6/10
85/170 = 0.5000
3
3/10
88/180 = 0.4889
6
6/10
94/190 = 0.4947
4
4/10
98/200 = 0.4900
Cumulative Relative Frequency
0.6
Expected value = 1/2
0.55
0.5
0.45
0.4
0
5
10
15
20
25
Trial
This stabilizing effect, or long-term average value, is often referred to
as the law of large numbers.
Law of Large Numbers
If the number of times an experiment is repeated is increased,
the ratio of the number of successful occurrences to the number
of trials will tend to approach the theoretical probability of the
outcome for an individual trial
– Interpretation: The law of large numbers says: the
larger the number of experimental trials, the closer the
empirical probability is expected to be to the true
probability P(A)
Subjective Probability
1. Suppose the sample space elements are not equally
likely and empirical probabilities cannot be used
2. Only method available for assigning probabilities may be
personal judgment
3. These probability assignments are called subjective
probabilities
Summary
• Probabilities describe the chances of events occurring …
events consisting of outcomes in a sample space
• Probabilities must obey certain rules such as always
being greater than or equal to 0 and less then or equal to
1.
• There are various ways to compute probabilities,
including empirical method and classical method for
experiments with equally likely outcomes.
Compute Probabilities for
Compound Events
Venn Diagram
• Venn Diagrams provide a useful way to visualize
probabilities
– The entire rectangle represents the sample space S
– The circle represents an event E
S
E
Example
• In the Venn diagram below
– The sample space is {0, 1, 2, 3, …, 9}
– The event E is {0, 1, 2}
– The event F is {8, 9}
– The outcomes {3}, {4}, {5}, {6}, {7} are in neither event
E nor event F
Mutually Exclusive Events
• Two events are disjoint if they do not have any outcomes in common
• Another name for this is mutually exclusive
• Two events are disjoint if it is impossible for both to happen at the
same time
• E and F below are disjoint
Example
The following table summarizes visitors to a local amusement park:
Male
Female
Total
All-Day
Pass
1200
900
2100
Half-Day
Pass
800
700
1500
Total
2000
1600
3600
One visitor from this group is selected at random:
1) Define the event A as “the visitor purchased an all-day pass”
2) Define the event B as “the visitor selected purchased a half-day pass”
3) Define the event C as “the visitor selected is female”
Solutions
1) The events A and B are mutually exclusive
2) The events A and C are not mutually exclusive. The intersection of A
and C can be seen in the table above or in the Venn diagram below:
3)
A
C
1200
900
700
800
Addition Rule for Disjoint Events
• For disjoint events, the outcomes of (E or F) can be
listed as the outcomes of E followed by the outcomes of
F
• There are no duplicates in this list
• The Addition Rule for disjoint events is
P(E or F) = P(E) + P(F)
• Thus we can find P(E or F) if we know both P(E) and
P(F)
Addition Rule for More than Two
Disjoint Events
• This is also true for more than two disjoint events
• If E, F, G, … are all disjoint (none of them have any
outcomes in common), then
P(E or F or G or …) = P(E) + P(F) + P(G) + …
• The Venn diagram below is an example of this
Example
• In rolling a fair die, what is the chance of
rolling a {2 or lower} or a {6}
– The probability of {2 or lower} is 2/6
– The probability of {6} is 1/6
– The two events {1, 2} and {6} are disjoint
• The total probability is 2/6 + 1/6 = 3/6 =
1/2
Note
•
•
•
•
The addition rule only applies to events that are disjoint
If the two events are not disjoint, then this rule must be modified
Some outcomes will be double counted
The Venn diagram below illustrates how the outcomes {1} and {3} are
counted both in event E and event F
Example
In rolling a fair die, what is the chance of rolling a {2 or
lower} or an even number?
– The probability of {2 or lower} is 2/6
– The probability of {2, 4, 6} is 3/6
– The two events {1, 2} and {2, 4, 6} are not disjoint
– The total probability is not 2/6 + 3/6 = 5/6
– The total probability is 4/6 because the event is {1, 2,
4, 6}
Note: When we say A or B, we include outcomes either
in A or in B or both.
General Addition Rule
• For the formula P(E) + P(F), all the outcomes that are in both events
are counted twice
• Thus, to compute P(E or F), these outcomes must be subtracted
(once)
• The General Addition Rule is
P(E or F) = P(E) + P(F) – P(E and F)
• This rule is true both for disjoint events and for not disjoint events.
when E and F are disjoint, P(E and F) = 0 which leads to
P(E or F) = P(E) + P(F)
Example
• When choosing a card at random out of a
deck of 52 cards, what is the probability of
choosing a queen or a heart?
– E = “choosing a queen”
– F = “choosing a heart”
• E and F are not disjoint (it is possible to
choose the queen of hearts), so we must
use the General Addition Rule
Solution
• P(E) = P(queen) = 4/52
• P(F) = P(heart) = 13/52
• P(E and F) = P(queen of hearts) = 1/52, so
P (queen or heart)
 P (queen)  P (heart)
 P (queen and heart)
4 13 1



52 52 52
16

52
Another Example
A manufacturer is testing the production of a new product on two assembly
lines. A random sample of parts is selected and each part is inspected for
defects. The results are summarized in the table below:
Line 1 (1)
Line 2 (2)
Total
Good (G)
70
80
150
Defective (D)
40
25
65
Total
110
105
215
Suppose a part is selected at random:
1) Find the probability the part is defective
2) Find the probability the part is produced on Line 1
3) Find the probability the part is good or produced on Line 2
Solutions
65
1) P( D) 
215
(total defective divided by total number of parts)
110
2) P(1) 
215
(total produced by Line 1 divided by total number of parts)
3) P(G or 2)  n(G or 2)  175
n( S )
215
(total good or produced on Line 2 divided by total parts)
175
 P(G )  P(2)  P (G and 2)  150  105  80 
215 215 215 215
Complement Events
•
The set of all sample points in the sample space that do not belong to
event E. The complement of event E is denoted by E 
or E c (read “E complement”). For example,
 The complement of the event “success” is “failure”
 The complement of the event “rain” is “no rain”
 The complement of the event “at least 3 patients recover” out of
5 patients is “2 or fewer recover”
•
The probability of the complement of an event E is 1 minus the probability
of E
P( E )  1  P( E )
•
This can be shown in one of two ways
 It’s obvious … if there is a 30% chance of rain, then there is a 70%
chance of no rain. E = Rain, E  = No rain
 E and E are two disjoint events that add up to the entire sample space
Illustration
The Complement Rule can also be illustrated using a Venn
diagram
Entire region
The area of the
region outside the
circle represents Ec
Notes:
Complementary events are also mutually exclusive/disjoint.
Mutually exclusive events are not necessarily complementary
Example
All employees at a certain company are classified as only
one of the following: manager (A), service (B), sales (C), or
staff(D). It is known that P(A) = 0.15, P(B) = 0.40,
P(C) = 0.25, and P(D) = 0.20
Solution:
P(A)  1  P(A)  1  0.15  0.85
P( A and B )  0 (A and B are mutually exclusive )
P ( B or C )  P ( B)  P ( C )  0.40  0.25  0.65
P ( A or B or C )  P ( A )  P ( B)  P ( C )  0.15 + 0.40 + 0.25 = 0.80
Summary
• Probabilities obey Additional Rules
• For disjoint events, the Addition Rule is used for
calculating “or” probabilities
• For events that are not disjoint, the Addition Rule
is not valid … instead the General Addition Rule
is used for calculating “or” probabilities
• The Complement Rule is used for calculating
“not” probabilities
Independence
• Definition of independence:
Events E and F are independent if the occurrence of E in
a probability experiment does not affect the probability of
event F
• Other ways of saying the same thing
– Knowing E does not give any additional information
about F
– Knowing F does not give any additional information
about E
– E and F are totally unrelated
Examples of Independence
– Flipping a coin and getting a “tail” (event E)
and choosing a card and getting the “seven of
clubs” (event F)
– Choosing one student at random from
University A (event E) and choosing another
student at random from University B (event F)
– Choosing a card and having it be a heart
(event E) and having it be a jack (event F)
Dependent Events
• If the two events are not independent, then they
are said to be dependent
• Dependent does not mean that they completely
rely on each other … it just means that they are
not independent of each other
• Dependent means that there is some kind of
relationship between E and F – even if it is just a
very small relationship
Examples of Dependence
– Whether Jack has brought an umbrella (event
E) and whether his roommate Joe has
brought an umbrella (event F)
– Choosing a card and having it be a red card
(event E) and having it be a heart (event F)
– The number of people at a party (event E)
and the noise level at the party (event F)
Multiplication Rule
Let A and B be two events defined in sample space S. If A and B are
independent events, then:
P(A and B)  P(A)  P( B)
This formula can be expanded. If A, B, C, …, G are independent events,
then P(A and B and C and ... and G)  P(A)  P( B)  P(C) P(G)
Example: Suppose the event A is “Allen gets a cold this winter,” B is “Bob
gets a cold this winter,” and C is “Chris gets a cold this winter.” P(A) =
0.15, P(B) = 0.25, P(C) = 0.3, and all three events are independent.
Find the probability that:
1. All three get colds this winter
2. Allen and Bob get a cold but Chris does not
3. None of the three gets a cold this winter
Solutions
1) P(All three get colds this winter )
 P(A and B and C)  P(A )  P( B)  P(C)
= (0.15)(0.25)(0.30) = 0.0113
2) P ( Allen and Bob get a cold, but Chris does not)
 P ( A and B and C )  P ( A )  P ( B)  P ( C )
= (0.15)(0.25)(0.70) = 0.0263
3) P ( None of the three gets a cold this winter)
 P ( A and B and C )  P ( A )  P ( B)  P ( C)
= (0.85)(0.75)(0.70) = 0.4463
Example
A fair coin is tossed 5 times, and a head(H) or a tail (T) is recorded
each time. What is the probability A = {at least one head in 5
tosses}
Solution: Apply the complement rule:
P(A)  1  P(A)
 1  P(0 heads in 5 tosses )
1 31
 1

32 32
Note: P(0 heads in 5 tosses) = P( all tails in 5 tosses )
= P(1st toss is a head)  P(2nd toss is a head)…  p(5th toss is a head)
1
= 1 1 1 1 1
=
(due to independence of tosses)
   
2 2 2 2 2
32
Example
In a sample of 1200 residents, each person was asked if he or she
favored building a new town playground. The responses are
summarized in the table below:
Age
Less than 30 (Y)
30 to 50 (M)
More than 50 (O)
Total
Favor (F)
250
600
100
950
Oppose
50
75
125
250
Total
300
675
225
1200
If one resident is selected at random, what is the probability the resident will:
1) Favor the new playground?
2) Be between 30 to 50 years old?
3) Are the events F and M independent?
Solutions
1. P(F)  950  19
1200
2.
3.
24
675
9
P(M ) 

1200 16
600 1
P(F & M ) 

1200 2
Since P(F and M) is not equal to P(F)  P(M), the two
events F and M are not independent.
Mutually Exclusive Events Are Not
Independent
• What’s the difference between disjoint events
and independent events?
• Disjoint events can never be independent
–
–
–
–
Consider two events E and F that are disjoint
Let’s say that event E has occurred
Then we know that event F cannot have occurred
Knowing information about event E has told us much
information about event F
– Thus E and F are not independent
Summary
• Compound Events are formed by combining several
simple events:
 The probability that either event A or event B will
occur: P(A or B)
 The probability that both events A and B will
occur: P(A and B)
• The “disjoint” concept corresponds to “or” and the
Addition Rule … disjoint events and adding probabilities
• The concept of independence corresponds to “and” and
the Multiplication Rule … independent events and
multiplying probabilities
Discrete Probability
Distributions
Discrete Random Variables,
Discrete Probability Distribution
The Binomial Probability Distribution
Learning Objectives
1. Distinguish between discrete and continuous
random variables
2. Identify discrete probability distributions
3. Construct probability histograms
4. Compute and interpret the mean of a discrete
random variable
5. Interpret the mean of a discrete random variable as
an expected value
6. Compute the variance and standard deviation of a
discrete random variable
Discrete Random Variable
Random Variables
If the outcomes from an experiment are quantitative type
(i.e. numbers), we denote the outcomes of this type with
a variable using a capital letter such as X, Y , Z … That
is, the variable contains all possible values (i.e.
outcomes) from the experiment. Each possible value is
denoted with lower case letter such as x, y, z… Since
there is a probability/chance for each value of the
variable to occur, we call the variable as a random
variable.
Examples of Random Variables
• Tossing four coins and counting the number of heads
– The number could be 0, 1, 2, 3, or 4
– The number could change when we toss another four
coins
• Measuring the heights of students
- The heights could vary from student to student
• Recording the number of computers sold per day by a
local merchant with a random variable. Integer values
ranging from zero to about 50 are possible values.
Discrete Random Variables
• A discrete random variable is a random
variable that has either a finite or a
countable number of values
– A finite number of values such as {0, 1, 2, 3,
and 4}
– A countable number of values such as {1, 2,
3, …}
• Discrete random variables are often
“counts of …”
Examples of Discrete Random
Variables
• The number of heads in tossing 3 coins There
are four possible values – 0 heads, 1 head, 2
heads, and 3 heads
– A finite number of possible values – a discrete
random variable
– This fits our general concept that discrete random
variables are often “counts of …”
• The number of pages in statistics textbooks
– A countable number of possible values
• The number of visitors to the White House in a
day
– A countable number of possible values
Continuous Random Variables
• A continuous random variable is a random
variable that has an infinite, and more than
countable, number of values
– The values are any number in an interval
• Continuous random variables are often
“measurements of …”
Examples of Continuous Random
Variables
• The possible temperature in Chicago at noon tomorrow,
measured in degrees Fahrenheit
– The possible values (assuming that we can measure
temperature to great accuracy) are in an interval. So, 20 degrees
can be recorded as 20.4 degrees or 20.41 degrees … etc.
– The interval may be something like from -20 to 110 degrees.
– This fits our general concept that continuous random variables
are often “measurements of …”
• The height of a college student
– A value in an interval between 3 and 8 feet
• The number of bytes of storage used on a 80 GB (80
billion bytes) hard drive
– Although this is discrete, it is more reasonable to model it as a
continuous random variable between 0 and 80 GB
Discrete Probability
Distribution
Probability Distribution
• Probability Distribution is a distribution of the
probabilities associated with each of the
values of a random variable.
• The probability distribution is a theoretical
distribution because the probabilities are
theoretical probabilities; it is used to represent
populations.
Discrete Probability Distribution
• The probability distribution of a discrete
random variable X relates the values of X
with their corresponding probabilities
• A distribution could be
– In the form of a table
– In the form of a graph
– In the form of a mathematical formula
Probability Function
• If X is a discrete random variable and x is
a possible value for X, then we write P(x)
as the probability that X is equal to x
• Examples
– In tossing one coin, if X is the number of
heads, then P(0) = 0.5 and P(1) = 0.5
– In rolling one die, if X is the number rolled,
then
P(1) = 1/6
Properties of P(x)
• Since P(x) form a probability distribution,
they must satisfy the rules of probability
– 0 ≤ P(x) ≤ 1
– Σ P(x) = 1
• In the second rule, the Σ sign means to
add up the P(x)’s for all the possible x’s
Probability Distribution Table
• An example of a discrete probability distribution. All of the possible
values x are listed in one column of a table, the corresponding
probability P(x) for each value is listed on the next column as shown
below:
x
1
2
5
6
P(x)
.2
.6
.1
.1
• All of the P(x) values need to be not only positive but also add up to
1
Not a Probability Distribution
• An example that is not a probability distribution
x
1
2
5
6
• Two things are wrong
P(x)
.2
.6
-.3
.1
– P(5) is negative
– The P(x)’s do not add up to 1
Example
The number of people staying in a randomly selected
room at a local hotel is a random variable ranging in
value from 0 to 4. The probability distribution is known
and is given in various forms below:
x
P (x )
0
2/15
1
4/15
2
5/15
3
3/15
4
1/15
Notes:
 This chart implies the only values x takes on are 0, 1, 2,
3, and 4
5

P( the randomvariablex equals 2)  P(2) 
15
Probability Histogram
• A probability histogram is a histogram
where
– The horizontal axis corresponds to the
possible values of X (i.e. the x’s)
– The vertical axis corresponds to the
probabilities for those values (i.e. the P(x)’s)
• A probability histogram is very similar to a
relative frequency histogram
Probability Histogram
• An example of a probability histogram
• The histogram is drawn so that the height
of the bar is the probability of that value
Notes
• The histogram of a probability distribution uses
the area of each bar to represent its assigned
probability
• The width of each bar is 1 and the height of each
bar is the assigned probability, so the area of each
bar is also equal to the corresponding probability
• The idea of area representing probability is
important in the study of continuous random
variables later
Probability Function
Sometimes the probability distribution for a random variable x is given
by a functional expression. For example,
8 x
P( x ) 
15
for
x  3, 4, 5, 6, 7
Find the probability associated with each value by using the probability
function.
Solution:
83 5
P(3) 

15 15
86 2
P(6) 

15
15
P(4) 
84 4

15 15
P(7) 
P(5) 
85 3

15 15
87 1

15
15
Note: P( x) follows the probability rules; each number is between 0 and 1 and
sum of the probabilities is 1.
Mean of Probability Distribution
• Probability distribution is a population distribution, because the
probability is regarded as an idealized relative frequency for an
outcome to occur if the experiment is repeated large number of
times (mostly infinite times if we can).
• Since the mean of a population is denoted by a parameter m, the
mean of a probability distribution is denoted by m as well
• The mean of a probability distribution can be thought of in this way:
– There are various possible values of a discrete random variable
– The values that have the higher probabilities are the ones that
occur more often
– The values that occur more often should have a larger role in
calculating the mean
– The mean of the probability distribution is the weighted average
of the values, weighted by the probabilities
Mean of Discrete Probability
Distribution
• The mean of a discrete random variable is
μ = Σ [ x • P(x) ]
• In this formula
– x are the possible values of X
– P(x) is the probability that x occurs
– Σ means to add up all of the products of these
terms for all the possible values x
Mean of Discrete Probability
Distribution
• Example of a calculation for the mean
Multiply
x
P(x)
x • P(x)
1
0.2
0.2
2
0.6
1.2
Multiply again
5
0.1
0.5
Multiply again
6
0.1
0.6
Multiply again
• Add: 0.2 + 1.2 + 0.5 + 0.6 = 2.5
• The mean of this discrete random variable
is 2.5
Mean of Discrete Probability
Distribution
• The calculation for this problem written out
μ
= Σ [ x • P(x) ]
= [1• 0.2] + [2• 0.6] + [5• 0.1] + [6• 0.1]
= 0.2 + 1.2 + 0.5 + 0.6
= 2.5
• The mean of this discrete random variable is 2.5
• The mean is an average value, so it does not
have to be one of the possible values for X or an
integer.
Interpret the Mean of a Probability
Distribution
• The mean can also be thought of this way
(as in the Law of Large Numbers)
– If we repeat the experiment many times
– If we record the result each time
– If we calculate the mean of the results (this is
just a mean of a group of numbers)
– Then this mean of the results gets closer and
closer to the mean of the random variable
Expected Value
• The expected value of a random variable is another term
for its mean. That is, we often interpret the mean of a
discrete random variable as an expected value
• The term “expected value” illustrates the long term
nature of the experiments as described on the previous
slide – as we perform more and more experiments, the
mean of the results of those experiments gets closer to
the “expected value” of the random variable
Variance and Standard Deviation of Discrete
Probability Distribution
Variance of a discrete random variable denoted by s2, is found by
multiplying each possible value of the squared deviation from the
mean, (x  m)2, by its own probability and then adding all the products
together:
s 2   [( x  m ) 2 P( x )]
  [ x 2 P( x )]  { [ xP( x )]}
2
  [ x 2 P( x )]  m 2
Standard deviation of a discrete random variable is the positive square
root of the variance:
s  s2
Short-cut Formula for Variance of Discrete
Probability Distribution
• The variance formula
σ2 = Σ [ (x – μ)2 • P(x) ]
can involve calculations with many decimals or
fractions
• An short-cut formula is
σ2 = [ Σ x2 • P(x) ] – μ2
This formula is often easier to compute
Example
• To find the mean, variance and standard deviation of a probability
distribution, you can extended the probability table as below:
x2
P( x )
xP ( x )
x 2 P( x )
x
3
4
5
6
7
5/15
4/15
3/15
2/15
1/15
15/15
Totals
15/15
16/15
15/15
12/15
7/15
65/15
[ xP( x)]
m   [ xP ( x )] 
s 
2


65
 4.33
15
2
305  65 
x P(x)  m 
    1.56
15  15 
2
2
s  s 2  1.56  1.25
9
16
25
36
49
45/15
64/15
75/15
72/15
49/15
305/15
2
[
x
 P( x)]
Calculate Mean and Variance of Discrete
Probability Distribution
• You can get the mean and variance of a discrete probability
distribution from a TI graphing calculator. The procedure is the same
as the one for getting the sample statistics from a frequency
distribution (grouped data).
• Enter all possible values of a variable in one list, say L1, and their
corresponding probabilities in another list, say L2. Then,
STAT CALC 1 ENTER  L1, L2  ENTER
• The sample mean x shown on the screen is actually the mean of
the discrete probability distribution. The population standard
deviation s x is the standard deviation s of the discrete probability
distribution. ( Ignore other calculated statistics)
m
Summary
• Discrete random variables are measures of outcomes that have
discrete values
• Discrete random variables are specified by their probability
distributions which are regarded as population distributions
• The mean of a discrete random variable is a parameter, can be
interpreted as the long term average of repeated independent
experiments
• The variance of a discrete random variable is a parameter,
measures its dispersion from its mean
The Binomial
Probability Distribution
Learning Objectives
1. Determine whether a probability experiment
is a binomial experiment
2. Compute probabilities of binomial
experiments
3. Compute the mean and standard deviation
of a binomial random variable
4. Construct binomial probability histograms
Binomial Experiment
• A binomial experiment has the following structure
– The first trial of the experiment is performed … the result is either
a success or a failure (Outcomes are classified into two
categories, so it is term binomial experiment.)
– The second trial is performed … the result is either a success or
a failure. This result is independent of the first and the chance of
success is the same as the first trial.
– A third trial is performed … the result is either a success or a
failure. The result is independent of the first two and the chance
of success is the same
– The process can go on and on.
Example
– A card is drawn from a deck. A “success” is for
that card to be a heart … a “failure” is for any
other suit
– The card is then put back into the deck
– A second card is drawn from the deck with the
same definition of success.
– The second card is put back into the deck
– We continue for drawing 10 cards
Binomial Experiment
A binomial experiment is an experiment with the following
characteristics
– The experiment is performed a fixed number of times, each time
called a trial
– The trials are independently performed
– Each trial has two possible outcomes, usually called a success
(desired outcomes) and a failure (the rest of other outcomes)
– The probability of success is the same for every trial
Note: If an experiment contains more than 2 outcomes. We can
always classify the outcome into two categories, success and
failure. For instance, tossing a die, success is having a number less
than 3, then failure is having a number not less than 3.
Binomial Probability Distribution
• Notation used for binomial distributions
– The number of trials is represented by n
– The probability of a success is represented by p
– The total number of successes in n trials is the
ransom variable X, the outcome observed
• Because there cannot be a negative number of
successes, and because there cannot be more than n
successes (out of n attempts)
0≤X≤n
Example
• In our card drawing example
– Each trial is the experiment of drawing one card
– The experiment is performed 10 times, so n = 10
– The trials are independent because the drawn card is
put back into the deck so that a card drawn before will
not affect a card drawn next.
– Each trial has two possible outcomes, a “success” of
drawing a heart and a “failure” of drawing anything
else
– The probability of success is 0.25 (Since there are
four suits in a deck where heart is one of the suits),
the same for every trial, so p = 0.25
– X, the number of successes, is between 0 and 10
Notes
• The word “success” does not mean that
this is a good outcome or that we want this
to be the outcome
• A “success” in our card drawing
experiment is to draw a heart
• If we are counting hearts, then this is the
outcome that we are measuring
• There is no good or bad meaning to
“success”
Calculate Binomial Probability
• We would like to calculate the probabilities
of X, i.e. P(0), P(1), P(2), …, P(n)
• Do a simpler example first
– For n = 3 trials (e.g. toss a coin 3 time)
– With p = .4 probability of success (e.g. Head
is a success. The coin is loaded such that
P(head) = 0.4)
– Calculate P(2), the probability of 2 successes
(i.e. 2 heads)
Calculate Binomial Probability
• For 3 trials, the possible ways of getting exactly
2 successes are
–S S F
–S F S
–F S S
• The probabilities for each (using the
multiplication rule of independent events) are
– 0.4 • 0.4 • 0.6 = 0.096
– 0.4 • 0.6 • 0.4 = 0.096
– 0.6 • 0.4 • 0.4 = 0.096
Calculate Binomial Probability
• The total probability is
P(2) = 0.096 + 0.096 + 0.096 = 0.288
• But there is a pattern
– Each way had the same probability, … the probability
of 2 success (0.4 times 0.4) times the probability of 1
failure (0.6 times 0.6), because each way contained 2
successes and 1 failure regardless of the order of the
success and failure.
• The probability for each case is
(0.4)2 • (0.6)1
Counting formula
• There are 3 possible sequences of success and failure
– S S F could represent choosing a combination of
2 out of 3 … choosing the first and the second
– S F S could represent choosing a second
combination of 2 out of 3 … choosing the first and the
third
– F S S could represent choosing a third
combination of 2 out of 3
• These are the 3 ways to choose 2 out of 3, which can be
computed by a counting formula 3C2 or  3  ( read as 3
 2
choose 2)
Calculate Binomial Probability
• Thus the total probability
P(2) = 0.096 + 0.096 + 0.096 = 0.288
can also be written as
P(2) = 3C2 • (0.4)2 • (0.6)1
• In other words, the probability is
– The number of ways of choosing 2 out of 3,
times
– The probability of 2 successes, times
– The probability of 1 failure
General Formula for Binomial
Probabilities
• The general formula for the binomial
probabilities is just this:
• For P(x), the probability of x successes, the
probability is
– The number of ways of choosing x out of n, times
– The probability of x successes, times
– The probability of n-x failures
• This formula is
P(x) = nCx px (1 – p)n-x
Calculate nCx
•
The formula nCx (called the binomial coefficient) for counting the
number of ways to have x number of successes out of n trials can
be computed as below:
n!
nC x =
x!(n  x )!
Where n! is an abbreviation for n factorial:
n! n (n  1)( n  2)  (3)( 2)(1)
•
It is easier to compute it from the calculator
For instance, to compute 3C2:
1. Enter 3
2. Click MATH  PRB  3:nCr
3. Enter 2
Solution: 3C2 = 3
Example
• A student guesses at random on a multiple
choice quiz
– There are n = 10 questions in total
– There are 5 choices per question so that the
probability of success (guess correctly) p =
1/5 = .2 (due to random guessing, each
choice has equal chance)
• What is the probability that the student
gets 6 questions correct?
Example Continued
First, check if this is a binomial experiment
– There are a finite number n = 10 of trials (10
questions, answer one question each trial)
– Each trial has two outcomes (a correct guess and an
incorrect guess)
– The probability of success is independent from trial to
trial (every question is done by a random guess)
– The probability of success p = .2 is the same for each
trial
Example Continued
• The probability of 6 correct guesses is
P(x) = nCx px (1 – p)n-x
= 10C6 (0.2)6 (0.8)4
= 210 • 0.000064 • 0.4096
= 0.005505
• This is less than a 1% chance
• In fact, the chance of getting 6 or more correct
(i.e. a passing score) is also less than 1%
Example
According to a recent study, 65% of all homes in a certain county
have high levels of radon gas leaking into their basements. Four
homes are selected at random and tested for radon. The random
variable x is the number of homes with high levels of radon (out of
the four).
Solution: First check if it is a binomial experiment:
1. There are 4 repeated trials: n = 4. The trials are independent.
2. Each test for radon is a trial, and each test has two outcomes:
radon or no radon
3. p = P(radon) = 0.65, q = P(no radon) = 0.35
4. x is the number of homes with high levels of radon out of 4
homes selected, possible values: 0, 1, 2, 3, 4
Yes! So apply the binomial probability distribution with n= 4, p=0.65 to
compute the probabilities
Example continued
 4


P ( x )   (0.65 ) x (0.35 ) 4 x , for x  0, 1, 2, 3, 4
 x
 4
0
4
P ( 0 )    (0.65 ) (0. 35 )  (1)( 1)( 0 .0150 )  0 .0150
 0
 4
1
3

P (1)   (0. 65 ) (0. 35 )  ( 4 )( 0 .65 )( 0 .0429 )  0 .1115
 1
 4

P ( 2 )   (0.65 ) 2 (0. 35 ) 2  ( 6 )( 0 .4225 )( 0 .1225 )  0 .3105
 2
 4

P ( 3 )   (0. 65 ) 3 (0. 35 ) 1  ( 4 )( 0 .2746 )( 0 .35 )  0 .3845
 3
 4

P ( 4 )   (0. 65 ) 4 (0. 35 ) 0  ( 1)( 0 .1785 )( 1)  0 .1785
 4
Example
In a certain automobile dealership, 70% of all customers purchase
an extended warranty with their new car. For 15 customers selected
at random:
1) Find the probability that exactly 12 will purchase an extended
warranty
2) Find the probability at most 13 will purchase an extended
warranty
Solutions:
• Let x be the number of customers who purchase an extended warranty.
x is a binomial random variable.
• The probability function associated with this experiment:
 15
P( x )    (0.7) x (0.3)15 x , for x  0, 1, 2, ... ,15
 x
Solutions Continued
1) Probability exactly 12 purchase an extended warranty:
 15
P(12)    (0.7)12 (0.3) 3 0.1700
 12
2) Probability at most 13 purchase an extended warranty:
P( x  13)  P(0)  P(1)  ...  P(13)
 1  P(14)  P (15)
15 
14
1 15 
15
0
 1   (0.7) (0. 3)   (0.7) (0.3) 
15 
14 

 1  [0.0305  0.0047]
 1  0.0352  0.9648
Note: Instead computing P (x  13) directly, It is easier to apply
the complement rule to compute it’s probability by 1 – P (x >13)
Calculate Binomial Probabilities
•
•
•
It is possible to use tables to look up these probabilities
It is best to use a calculator routine or a software program to compute
these probabilities. With TI graphing calculator, follow the steps
below:
Example 1: Compute an individual probability P(6) for a binomial
random variable with n = 10, p = 0.4 using binompdf ( )
1. Distr[2NDVARS]  A:binompdf  ENTER
2. Enter 10, 0.4, 6), then hit ENTER
Solution: P(6) = 0.1115
•
Example 2: Compute a cumulative probability for x = 0 to 5 using
binomcdf( ), i.e., P (x  5) = P(0) + P(1) + P(2) + p(3) + P(4) + p(5)
1. Distr[2NDVARS]  B:binomcdf  ENTER
2. Enter 10, 0.4, 5), then hit ENTER
Solution: P (x  5) = 0.8338
Note: Need to enter n, p, x in order to binompdf(n,p,x) ot binomcdf(n, p, x)
Example
•
For the automobile dealership example considered on the previous slide. Apply a
binomial probability with n = 15 and p = 0.7. Check the answers with a graphing
calculator.
Solution:
P(12) = binompdf(15, 0.7, 12) = 0.1700
P ( x  13) = binomcdf(15, 0.7, 13) = 0.9647
•
Also, find the cumulative probability x = 4 to 8 inclusively (means including 4 and 8).
That is, P(4  x  8) = P(4) + P(5) + P(6) + P(7) + P(8)
Solution:
P(4  x  8) = binomcdf(15, 0.7, 8) – binomcdf(15, 0.7, 3) = 0.1311
Note: binomcdf(15, 0.7, 8) = P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)
binomcdf(15, 0.7, 3) = P(0)+P(1)+P(2)+P(3)
So, the difference of these two cumulative probabilities covers the probabilities
for x from 4 to 8. So, do not subtract binomcdf(15, 0.7, 4) from the
binomcdf(15, 0.7, 8) since P(4) needs to be included.
Mean of Binomial Probability
Distribution
• We would like to find the mean of a binomial distribution
• You may apply the general formula for mean of a discrete probability
distribution using μ = Σ [ x • P(x) ] to find the mean
• It turns out that, for a binomial probability distribution, the mean can
be quickly computed by
μ=np
• Example
– There are 10 questions
– The probability of success (correct guess) is 0.20 on each one
– Then the expected number of correct answers would be
10 • 0.20 = 2
Variance and standard Deviation of a
Binomial Probability Distribution
• We would like to find the variance and standard deviation of a
binomial distribution
• You may also apply the general formula to compute
σ2 = [ Σ x2 • P(x) ] – μ2
• It turns out that we can quickly get the variance and standard
deviation by using the following formula:
The variance is
σ2 = n p (1 – p)
The standard deviation is
s  np(1 - p)
Example
• For our random guessing on a quiz problem
– n = 10
– p = 0.2
– x=6
• Therefore
– The mean is np = 10 • 0.2 = 2
– The variance is np(1-p) = 10 • .2 • .8 = 0.16
– The standard deviation is 0.16 = 0.4
Shape of Binomial Distribution
• With the formula for the binomial probabilities
P(x), we can construct histograms for the
binomial distribution
• There are three different shapes for these
histograms
– When p < .5, the histogram is skewed right
– When p = .5, the histogram is symmetric
– When p > .5, the histogram is skewed left
Right-skewed Binomial Distribution
• For n = 10 and p = .2 (skewed right)
– Mean = 2
– Standard deviation = .4
Symmetrical Binomial Distribution
• For n = 10 and p = .5 (symmetric)
– Mean = 5
– Standard deviation = .5
Left-skewed Binomial Distribution
• For n = 10 and p = .8 (skewed left)
– Mean = 8
– Standard deviation = .4
Notes
• Despite binomial distributions being skewed, the
histograms appear more and more bell shaped
as n gets larger
• This will be important!
Summary
• Binomial random variables model a series
of independent trials, each of which can be
a success or a failure, each of which has
the same probability of success
• The binomial random variable has mean
equal to np and variance equal to np(1-p)