STA 2023 Elementary Statistics
Lecture Notes
Chapter 3 – Probability
Professor Achenbach
Probability
Probability is a mathematical measure of the likelihood of an event occurring.
Probabilities are always fractions or decimals indicating the portion or percent of the time
that the event occurs.
Examples:



40% chance of rain
Batting Average of .313
The probability of a royal flush is 1:649,740
Interpretations:



A 40% chance of rain means that if we look at all days with similar conditions,
40% of those days had rain.
A batting average of .313 means that the player got a base hit in 31.3% (.313) of
his attempts at bat.
On average, a royal flush will be dealt once for every 649,740 hands of poker.
Probability Experiments
A probability experiment is any process whose result is determined by chance.
Examples:



Roll a die. Flip a Coin.
Draw a card from a shuffled deck
Choose a name from a hat.
1
Outcome – Any possible result of a probability experiment.
Sample Space – The collection of all possible outcomes for an experiment. The sample
space is often denoted by S.
Examples:

The sample space of the experiment Flip a Coin has 2 outcomes heads and tails,
so we could write:
S  {heads, tails}

The experiment of rolling a die has 6 possible outcomes 1-6, so:
S  {1, 2,3, 4,5, 6}
Note: Each outcome for an experiment completely describes the outcome of the
experiment.
Event – Any collection of outcomes from the sample space. We will generally use
capital letters to represent events.
Examples:

Rolling an even number is an event for the experiment of rolling a die. If we call
this event E, we could write: E  {2, 4, 6}

Drawing a king is an event for the experiment of drawing a card from a deck. If
we call this event K, then K contains 4 outcomes: king of hearts, king of spades,
king of diamonds and king of clubs.
An event E for an experiment is said to have occurred if the outcome of the experiment
is one of the outcomes of E.
Example:
If E is the event described above, and if when the die is cast, a 4 is showing, then we
would say that E occurred. If a 1 was rolled, then E did not occur.
2
Calculating the Probability of an Event
The probability of an event E is denoted by P ( E ) and always represents the fraction (or
percentage) of trials of the experiment on which E occurs.
Note: For any event E: 0  P ( E )  1 .
An event E is called impossible if P( E )  0 .
An event E is called certain if P ( E )  1 .
P ( E ) can be calculated in two different ways:
Empirical Probability
Empirical Probability – A probability calculated by collecting data from past trials of
the experiment. The probability obtained is then used to predict the future likelihood of
the event occurring.
Examples:

We could calculate the probability of the event of obtaining a sum of 7 when two
dice are rolled by rolling 2 dice many times (say 1000) and calculating what
percent of the time a 7 was rolled.

We can calculate the probability that a basketball player will make a free throw by
calculating the percentage of the time that the player has made free throws in the
past.
To calculate an empirical probability we use the formula:
P( E ) 
number of trials in which E occurs
total number of trials
Exercises:

A player has made 527 free throws in 698 attempts. Calculate the probability that
the player makes a free throw on the next attempt.

In 2002 there were 62,318 flights departing from Tampa International Airport. Of
these, 8,821 were late in departing. Calculate the probability that a flight
departing from Tampa will be delayed. (Data Source: U.S. Department of
Transportation)
3
Empirical Probability with Frequency Tables
Given a frequency table for a set of data values, we can calculate the probability that a
data value falls into any data class.
Example: Consider the frequency table of exam scores used in our Chapter 2 examples:
Class
Frequency ( f )
90-99
80-89
70-79
60-69
50-59
40-49
4
6
4
3
2
1
If event A is the event that a student scores between 90 and 99 on the exam, then:
P( A) 
number of students scoring 90-99 4

 .20
total number of students
20
Notice that P( A) is just the relative frequency of the 90-99 data class.
Exercises:

Let B be the event that a student scores between 80 and 89 on the exam.
Calculate P( B) .

Let E be the event that a student passes the exam (with a grade of C or higher).
Calculate P ( E )
Classical Probability
Classical Probability (Theoretical Probability) – A probability calculated for an
experiment in which each outcome is assumed to be equally likely.
The probability of an event in such an experiment can be calculated by determining the
fraction (or percentage) of outcomes that are in the event. We use the formula:
P( E ) 
number of outcomes in E
number of outcomes in S
4
Examples:

When flipping a fair coin, the two possible outcomes heads and tails are equally
likely, so:
1
P (heads)  P(tails) 
2

When rolling a fair die each possible outcome from 1 to 6 is equally likely, with
probability

1
.
6
Let K be the event of drawing a king when a card is drawn from a shuffled deck.
Since all of the 52 possible outcomes is equally likely, and since there are 4 kings
in a deck, we have:
P( K ) 
4
1

52 13
Fundamental Counting Principle
Fundamental Counting Principle – The total number of outcomes for any process
which occurs in sequential steps is equal to the product of the number of outcomes for
each step.
Examples:
 If two dice are rolled, there are a total of 6 possible outcomes for the first die and
6 for the second die. The total number of different possible outcomes for rolling 2
dice is: 6  6  36

If two cards are drawn from a deck without replacement, then there are 52
possible outcomes for the first card and 51 possible outcomes for the second card.
The total number of outcomes for the experiment is: 52  51  2652 .

Florida uses three letters followed by 2 numbers followed by another letter for its
license plate numbers. The total number of different license plate numbers
possible is: 26 10  26  45,697,600
3
2
5
Exercises:
1. Consider the experiment of flipping a coin three times.
a. Calculate the number of outcomes in S.
b. Write out the elements of S.
c. Calculate the probability of all tails, one tail, etc.
2. Consider the experiment of rolling two dice.
a. Calculate the probability that the first die is odd.
b. Calculate the probability of rolling a 7, an 11, a 5.
c. Calculate the probability of rolling doubles.
3. Consider the experiment of drawing two cards without replacement.
a. Find the probability that both cards are Kings
b. Find the probability that both cards are Hearts
c. Find the probability that the first card is a King and the second is a Jack.
Complement of an Event
The complement of an event E is denoted by E’, and is the event that contains every
outcome for the experiment except for those in E. It is essentially the event that E does
not occur.
Examples:

If E is the event that a 5 is showing when a die is rolled, then E’ is the event that a
5 is not showing, so E’ contains 5 outcomes: E '  {1, 2,3, 4, 6}

If K is the event of drawing a king from a shuffled deck, then the event K’ is the
event that a king is not drawn, and so K’ contains 48 possible outcomes.
Note: Since every event either occurs or does not occur, the sum of the probabilities of an
event and its complement is always 1.
P( E )  P( E ')  1
This fact can be used to find the probability of the complement of an event as follows:
P( E ')  1  P( E )
Example:

If the probability that a flight will be delayed is .13, then the probability that it
will not be delayed must be 1  .13  .87 .
6
Conditional Probability
A conditional probability for an event is calculated when some additional information
about the outcome of the experiment is known.
Suppose A and B are two events for an experiment, and suppose that it is known that
event B has occurred. We can recalculate the probability of event A based on this
information, and this probability is called the conditional probability of A given B. The
conditional probability of A given B is denoted by P( A | B) .
Example:
Consider the experiment of drawing two cards from a deck without replacement. Let
event K1 be the event that the first card is a king, and event K 2 be the event that the
second card is a king. We can calculate that:
P( K1 ) 
4
52
P( K 2 | K1 ) 
3
4
P( K 2 | K1 ') 
51
51
P( K2 ' | K1 ')  ?
P( K 2 ' | K1 ) 
48
51
Exercise: Titanic Statistics
Survived
Died
Total
Men
332
1360
1692
Women
318
104
422
Boys
29
35
64
Girls
27
18
45
Total
706
1517
2223
Suppose a passenger on the Titanic is chosen at random, and let M be the event that they
are a man, W the event they are a woman, and S that they survived.
Calculate:
P( M ) 
P( S ) 
P( M | S ) 
P( S | M ) 
P (W | D ) 
P( D | W ) 
P( D | M ') 
7
Independence and Dependence
Two events A and B are called independent if
P( A | B)  P( A) or P( B | A)  P( B)
If the above equations do not hold, then the two events are called dependent.
Examples:

Consider the experiment of drawing one card from a deck. The event K of
drawing a king and event H of drawing a heart are independent since:
P( K ) 
4
1

 P( K | H )
52 13

But K and the event F of drawing a face card are dependent since: P( F | K )  1 .

The events M and S from the Titanic experiment are dependent.
Note: Two events are dependent only if one event’s occurrence affects the likelihood of
the other. So events from two different trials of an experiment are always independent.
Combing Events Using AND
Given two events A and B we define the event A and B to be the event that events A and
B both occur at the same time. We find the probability of the event A and B using the
Multiplication Rule.
The Multiplication Rule – If A and B are two events for an experiment, then:
P( A and B)  P( A)  P( B | A)
The formula is often used when the events occur in sequence so that event A is followed
by event B.
8
Example:
If two cards are drawn from a deck without replacement then the event that both cards
are kings is the event K1 that the first card drawn is a king followed by event K 2 that the
second card is a king. The probability of both events occurring is:
P( K1 and K 2 )  P( K1 )  P( K 2 | K1 ) 
4 3
12
 
 .0045
52 51 2652
Exercises:



Find the probability that both cards are face cards.
Find the probability that the first card is a jack and the second is any face card
The multiplication rule can be extended to 3 or more events as well. Find the
probability of drawing 3 kings from a deck of cards.
Finding AND Probabilities – Independent Events
If A and B are independent events, then P( B | A)  P( B) so that the formula for
calculating the probability of A and B becomes:
P( A and B)  P( A)  P( B)
Example:
If two dice are rolled, the probability of rolling double 6 can be found using the above
formula. The probability of getting a 6 on each die is 1 and the events of each roll are
6
independent from each other. Thus the probability of a “double 6” is just:
P(6 on die #1 and 6 on die #2) 
1 1 1
 
6 6 36
This rule can also be extended to 3 or more events. For example the probability of rolling
a 6 on each of 4 rolls would be:
P(four sixes) 
1 1 1 1 1
1
    4 
6 6 6 6 6
1296
Exercise:

The probability of a flight departing from Tampa International Airport being
delayed is 0.14. Find the probability that 3 randomly chosen flights are all
delayed.
9
Calculating Probabilities Using Complement
Often it is easier to calculate the probability of the complement of an event than the
probability of the event itself. We can then use the fact that P ( E )  1  P ( E ') to find
the probability of the original event E.
One common example is when we want to find the probability of at least one occurrence
of an event in some number of trials.
Example:
If a die is rolled 4 times, find the probability that at least one of the rolls is a six. If E is
the event that at least one roll is a six, then E ' is the event that none of the four rolls is a
six. The probability of E ' can be found using the multiplication rule. Since each roll is
5
not a six with probability , and all rolls are independent, the probability of four rolls all
6
being not six is:
4
5 5 5 5 5
625
P( E ')        
 .4823
6 6 6 6  6  1296
And so the probability of at least one six occurring is:
P( E )  1  P( E ')  1 
625
671

 .5177
1296 1296
Exercise:

Using the fact that the probability of a flight from Tampa being delayed is .14,
find the probability that at least one out of 3 flights from Tampa will be delayed.
Combining Events Using OR
Given two events A and B we define the event A or B to be the event that at least one of
the events A or B occurs. We can find the probability of the event A or B using the
Addition Rule.
The Addition Rule – If A and B are two events for an experiment, then:
P( A or B)  P( A)  P( B)  P( A and B)
10
Two events are called mutually exclusive if both events cannot occur at the same time.
In this case P ( A and B )  0 , so the Addition Rule simplifies to:
P( A or B)  P( A)  P( B)
Examples:

If two dice are rolled, then the event A of rolling a 7 and the event B of rolling an
11 are mutually exclusive, so that the probability of rolling 7 or 11 is:
P( A or B)  P( A)  P( B) 

6
2
8 2



36 36 36 9
If a card is drawn from a shuffled deck, then the event K that the card is a king
and the event H that the card is a heart are not mutually exclusive, so the
probability of drawing a king or a heart is:
P( K or H )  P( K )  P( H )  P( K and H ) 
4 13 1 16 4
 


52 52 52 52 13
Exercises:

If two dice are rolled, find the probability that the first die is even or the second
die is 1.

If two flights depart from Tampa, each with a .14 probability of being delayed,
find the probability that the first flight is delayed or that the second flight is not.

If two cards are drawn from a shuffled deck without replacement, find the
probability that the first card is a king or that the second is a face card.

In the Titanic example from above, find P ( M or S ) .
Counting Techniques
Factorials – For any integer n, n factorial is the descending product beginning with n
and ending with 1. n factorial is denoted by n! , and can be written as:
n !  n  (n  1)  (n  2)  ...  2 1
11
Examples:




3!  3 2 1  6
4!  4  3 2 1  24
5!  5  4  3 2 1  120
10!  10  9  8  7  6  5  4  3  2 1  3, 628,800
Note: A special definition is made for the case of 0! namely:
0!  1
Permutations
A permutation of a group of objects is an ordered arrangement of the objects. The
number of different permutations of a group of n objects is n!
Example:

If we want to place a set of 5 names in some order, we have 5 choices for which
name to place first, then 4 choices of which to list second, 3 choices for third, 2
choices for fourth, and only one choice for last. By the Fundamental Counting
Principle, the number of different ways to put 5 names in order is thus:
5  4  3 2 1  5!

The number of ways of dealing the cards of a deck in some order is:
52!  8.066 1067 .
Often, we do not want to place an entire set of objects in order, but only want to calculate
how many ways a few chosen objects can be ordered.
Example:

A horse race has 14 horses, how many different possible ways can the top 3
horses finish? There are 14 possibilities for which horse finishes first, 13 for
second, and 12 for third, so by the Fundamental Counting Principle, there are
14 1312  2184 different possible finishing orders for the top three horses.
Permutations like the one described above are called permutations of n objects taken r
at a time. In the example above, each different finishing order is a permutation of 14
horses taken 3 at a time.
12
Notation – The number of permutations of n objects taken r at a time is denoted by
n
Pr .
Example:

In the horse race example above we would write:
P  2184 .
14 3
Formula for n Pr – Using the Fundamental Counting Principle in the same way as in
the previous example we have:
n
Pr  n  (n  1)  (n  2)  ...  (n  r  1)
r factors
Using factorial notation, we can write this as:
n
Pr 
n!
(n  r )!
Example:

In a qualifying heat for the 100 meter dash, the top 3 finishers out of each group
of 8 moves on to the next round. The number of different permutations for first,
second, and third place in each heat is:
P 
8 3
8!
8! 40320
 
 336
(8  3)! 5!
120
Exercises:

How many different orders are possible for all 8 runners in each heat?

How many different permutations are possible for the runners in the first 5
places?
13
Combinations
When calculating the number of outcomes for experiments involving multiple choices,
often we do not care what order the choices are made:
Examples:


Drawing a 5-card poker hand.
Selecting a 3 person committee from a group of 30.
A choice of r objects from a group of n objects without regard to order, is called a
combination of n objects taken r at a time.
Notation – The number of permutations of n objects taken r at a time is denoted by n Cr .
Examples:

There are

There are 30 C3 ways of selecting a 3 person committee from a group of 30.
52
C5 5-card poker hands.
Formula for n Cr – The number of combinations of n objects taken r at a time can be
calculated by starting with the number of different permutations for the choice of r
objects n Pr . Since each set of the same r objects has r ! different permutations, we have
counted each combination r ! times when calculating n Pr and so the number of
combinations of n objects taken r at a time is:
n
Cr 
Pr
n!

r ! r !(n  r )!
n
Examples:
 The number of different possible 5-card hands of poker is:
52

C5 
52!
52  51 50  49  48

 2,598,960
5!(47)!
5  4  3  2 1
The number of different ways of choosing a committee of 3 employees from a
group of 30 is:
30
C3 
30!
30  29  28

 4060
3!(27)!
3  2 1
14
Exercises:

Florida randomly chooses six balls from a bin containing balls numbered 1 to 53
for its lottery. Find the number of different lottery combinations.

A professor wishes to choose three students from a class of 23 students. How
many different groups of three students are possible? If the order matters, how
many different orders are possible?
Using Several Counting Principles
In some cases, we may need to use more than one combination, permutation or the
Fundamental Counting Principle.
Example:
A 12 person jury is to be selected from a jury pool consisting of 14 men and 16 women.
Suppose the 12 jurors are selected randomly, and let E be the event that the jury chosen
has exactly 6 women and 6 men. Then we can calculate P ( E ) by using our counting
techniques.
The total number of possible outcomes when 12 people are drawn from a group of 30 is
30 C12  86, 493, 225 . Choosing a jury of 6 men and 6 women can be thought of as a
two step process of choosing 6 men followed by choosing 6 women. The number of
ways of choosing the 6 men is
14
C6  3003 , and the number of ways of choosing the 6
women is 16 C6  8008 . By the Fundamental Counting Principle, the total of number
of ways to choose a 6 man and 6 woman jury is:
14
C6  16 C6  3003  8008  24,048,024 , and so:
C  C
24, 048, 024
P( E )  14 6 16 6 
 .278
86, 493, 225
30 C12
15
Exercise:
To choose the winning numbers, the Florida Lottery randomly chooses six balls from a
bin containing balls numbered 1 to 53. If a player matches at least 3 of the numbers, a
prize is won. If all 6 numbers are matched, the player wins the jackpot.
Calculate:




The probability a player matches all 6 winning numbers
The probability a player matches none of the 6 winning numbers
The probability a player matches exactly 3 out of the six winning numbers
The probability a player matches at least 1 of the 6 numbers
Factorials, Permutations & Combinations Using the TI-83
The TI-83 can be used to calculate factorials, permutations, and combinations.
All three functions are found by pressing the [MATH] key and then using the arrow keys
to highlight the PRB menu.
Factorials – to calculate 8!, we would press 8 then [MATH] and highlight PRB and
select #4: ! by pressing [ENTER]. On our screen will read: 8!
Pressing [ENTER] again we have: 8!  40320 .
Permutations – to calculate 10 P3 we would press 10 then [MATH] and highlight PRB
and select #2: nPr by pressing [ENTER]. We would then press 3 so that our screen
would read: 10 nPr 3. Pressing [ENTER] again we have: 10 P3  720
Combinations – to calculate 23 C5 we would press 23 then [MATH] and highlight PRB
and select #3: nCr by pressing [ENTER]. We would then press 5 so that our screen
would read: 23 nCr 5. Pressing [ENTER] again we have:
16
23
C5  33649