Counting Principles (10.1)
Counting Principles
The subject of this section is counting. Specifically, we will be learning to count things
using mathematical methods instead of listing things out.
The Multiplication Principle
When counting the number of ways two events can both happen, use a very simple
counting idea called the Multiplication Principle.
Definition - Multiplication Principle: When there are two or more choices to make, the
total number of possible ways that they can both occur is found by multiplying the
numbers of options for each choice.
In other words, if event A can occur m ways and event B can occur n ways, then A and B
occur n∙ m ways.
Why does the Multiplication Principle work? It is nothing fancy, just counting like
you’ve been doing your whole life, formalized. Look at this example:
Example Someone has 5 shirts and 3 jeans, how many “outfits” (shirt and jeans) can
they make?
Each shirt can be matched with each pair of jeans. So each shirt can make 1∙3 = 3 outfits.
Since there are 5 shirts, each of which can make 3 outfits, there are 5∙3 = 15 outfits total.
Visually, a tree diagram can be used to convey the same idea:
Shirt 1
Shirt 2
Shirt 3
Shirt 4
Shirt 5
Jeans 1
Jeans 2
Jeans 3
Jeans 1
Jeans 2
Jeans 3
Jeans 1
Jeans 2
Jeans 3
Jeans 1
Jeans 2
Jeans 3
Jeans 1
Jeans 2
Jeans 3
A diagram like this is called a tree diagram. Each
branch of the tree represents a choice about either a
shirt or a pair of jeans. If you count each branch
that has both a shirt and a pair of jeans, you’ll find
15 such branches.
This confirms what the Multiplication Principle
suggests, that the total number of outfits must be
5∙3 = 15.
Example From a list of 10 songs, you are asked to list your 1st, 2nd, and 3rd favorite
songs. How many different lists could you make?
You have 10 choices for the 1st song, 9 choices for the 2nd song (because you can’t
choose the same song again), and 8 choices for the 3rd song. So, the total number of lists
is 10 · 9 · 8 = 720.
1
Counting Principles (10.1)
The Addition Principle
When counting the number of ways either one of two events can happen, use another
simple counting idea called the Addition Principle.
Definition - Addition Principle: If event A can occur n ways and event B can occur m
ways, then A or B occur n + m – (number of ways both A and B occur together).
Example The Simpson family consists of five characters: Bart, Lisa, Marge, Homer, and
Abe (Grandpa) Simpson. The Flanders family consists of four characters: Ned, Maude,
Rod, and Todd. How many characters are in either the Simpson or Flanders families?
Since there is no overlap between the
two families, by the Addition Principle
there are 5 + 4 – 0 = 9 characters.
Simpsons
Flanders
Homer
This is easily seen by basic counting
with a Venn Diagram.
Ned
Maude
Rod
Todd
Marge
Bart
Abe
Lisa
Example The Simpson family consists of five characters: Bart, Lisa, Marge, Homer, and
Abe (Grandpa) Simpson. There are five characters whose name begins with the letter B:
Bart, Brandine, Barney, Bob, and Bumblebee Man. How many characters either belong
to the Simpson family or have a name that starts with the letter B.
There is overlap between these two sets,
so one must be careful. By the Addition
Principle, the number should be:
(Simpsons) + (B-Characters) – (Both
Simpson and B-Character)
5 + 5 – 1 = 9.
Again, this is easily verified by counting
on a Venn Diagram.
Simpsons
B-Characters
Homer
Marge
Abe
Lisa
Brandine
Bart
Barney
Bob
(sideshow)
Bumblebee man
2
Counting Principles (10.1)
Exercises
1. The Class Council is selling T-shirts that come in 3 colors (yellow, blue, white) and
4 sizes (small, medium, large, extra large).
a. Using the multiplication principle, how many different kinds of T-shirts are
available?
b. List all of the different kinds of T-shirts that are available. (example: “yellow
small” or “YS” for short)
2. A card game uses a special deck where each card has a number and a color.
The numbers range from 1 to 12. The colors are red, orange, yellow, green, and blue.
For example, there is a “5 blue” card. If the deck has one card with each of the
possible combinations of number and color, how many cards are there in the deck?
3. The table d’hôte (complete meal) at a French restaurant consists of a salad, an entree,
and a dessert. If there are 2 salad choices, 5 entree choices, and 3 dessert choices, how
many different meals are there?
4. At a bank, every ATM card has a number password called a PIN.
Here are some examples of PIN’s: 7486, 0122, 9999.
Answer these questions to figure out how many different PIN’s are possible.
o How many possibilities are there for the first digit of the PIN?
Hint: The possible digits are 0 through 9.
____
o How many possibilities are there for the second digit of the PIN? ____
o How many possibilities are there for the third digit of the PIN?
____
o How many possibilities are there for the fourth digit of the PIN? ____
o Now, using the multiplication principle, how many sequences are there?
3
Counting Principles (10.1)
5. A web site requires passwords consisting of exactly 5 capital letters.
Here are some examples of passwords: ABCDE, LEXMA, HELLO.
How many different passwords are possible?
6. A series of Massachusetts license plates has 4 numbers followed by 2 letters.
Here are some examples of license plates: 1234 AB, 5555 XY, 7070 EE.
Answer these questions to figure out how many different license plates are possible.
o How many possibilities are there for the first number?
____
o How many possibilities are there for the second number?
____
o How many possibilities are there for the third number?
____
o How many possibilities are there for the fourth number?
____
o How many possibilities are there for the first letter?
____
o How many possibilities are there for the second letter?
____
o Now, using the multiplication principle, how many license plates are there?
7. A series of New Hampshire license plates has 3 letters followed by 3 numbers.
Here are some examples: ABC 123, SPY 007, ZZZ 789.
a. How many license plates of this kind are possible?
b. If neither numbers nor letters may repeat on a given license plate, how many
plates are possible?
8. There are 15 tracks on a CD. You are asked to choose your 5 favorite tracks, in order
(1st, 2nd, 3rd, 4th, and 5th). How many ways can they be chosen?
4
Counting Principles (10.1)
9. In a standard deck of 52 playing cards, each card shows a different pairing of a suit
(spades, hearts, diamonds, or clubs) and a value (2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen,
King, or Ace).
a. Suppose you draw one card from a standard 52-card deck. In how many ways
could the card you draw be a Heart or an Ace?
b. Suppose you draw one card from a standard 52-card deck. In how many ways
could the card you draw be a Heart and an Ace?
c. Suppose you draw one card from a standard 52-card deck. In how many ways
could the card you draw be a red (hearts, diamonds) or a face card (Jack, Queen,
King, Ace)?
d. Suppose you draw one card from a standard 52-card deck, look at it, replace it in
the deck, totally shuffle the deck, and then draw a second card. In how many
ways could your first draw be a Heart and your second draw be an Ace?
5
Permutations and Word Arrangements (10.2)
Permutations and Word Arrangements
A permutation problem involves counting the number of ways that some objects can be
selected with an order and without repeating elements. Here’s an exercise from a recent
homework assignment:
10.1 #8. There are 15 tracks on a CD. You are asked to choose your 5 favorite tracks, in
order (1st, 2nd, 3rd, 4th, and 5th). How many ways can they be chosen?
Multiply 15 · 14 · 13 · 12 · 11. The answer is 360,360.
Notice that the answer involved multiplying 5 numbers counting down from 15. Any
counting problem that involves selecting 5 out of 15 objects, where the order of the
objects is important, will have this same answer. Here is a general statement of the
Permutation counting method:
Definition - Permutation: To count the number of ways that r out of n objects can be
selected where the order of the objects is important, multiply r whole numbers starting
from n and counting downward. The answer to this problem is called “n permutation r”
and abbreviated nPr . As a formula:
n
Pr  (n)(n  1)(n  2) ... (n  (r  1))
.
r integers starting at n and
decreasing by one each time
The “…” in this expression means there are terms present that have not been written out
explicitly. Also, the “(r – 1)” in the last factor comes from the fact that r – 1 objects will
have already been given out when you get to the last object.
Example I have 10 different candy bars and I am giving one each to my four best friends.
How many ways could I hand out the candy bars?
I have 10 options for what I give to friend 1, then 9 options left for friend 2, and so on.
Thus, the total number of possibilities is 10 P4  10  9  8  7  5040 .
Example The grocery store has 10 varieties of candy bars. If I want to buy four candy
bars, how many possibilities do I have?
I could buy 4 snickers bars if I wanted to, so in this case I can repeat elements. Thus I do
not use permutation counting here, and just apply the multiplication principle:
10 10 10 10  10,000 possibilities.
6
Permutations and Word Arrangements (10.2)
Word Arrangements
An important type of counting problem involves rearrangements with repeated elements.
Here is an example:
Example How many ways can the letters in the word MATH be rearranged to create
four letter arrangements?
Just list them first, to get a feel for how these problems work:
MATH
AMTH
TAMH
HAMT
MAHT
AMHT
TAHM
HATM
MTAH
ATHM
TMAH
HMAT
So, 24 rearrangements.
MTHA
ATMH
TMHA
HMTA
MHAT
AHMT
THAM
HTMA
MHTA
AHTM
THMA
HTAM
Of course we don’t want to list out all the arrangements every time we solve a problem.
By formula: 4 letters of which we want to select 4, where order matters: 4P4 = 4∙3∙2∙1 =24.
Example How many ways can the letters in the word MOM be arranged to create three
letter arrangements?
Using the same logic as the previous example, 3P3 = 3∙2∙1 = 6. Be careful! This is
wrong! List them out again to see where we went wrong:
MOM
OMM
MMO
There are only 3 arrangements!
The problem is that there is a repeated letter, M, that is getting lost. Write them out like
this and you’ll see the problem:
MOm
OMm
mOM
MmO
OmM
mMO
This list is made to keep track of the repeated Ms. It has 6 elements, but 3 of them are
repeats. MOm = mOM, etc. Our permutation formula over-counts the number of
arrangements. To get the correct number, we must divide out all the repeats. In this case,
the correct number is found by (3P3)/2! = 6/2 = 3 arrangements.
In general, to rearrange n letters where one letter is repeated k times use
n
P n n!
= .
k! k !
Example How many ways can all the letters in the word LILLY be arranged?
5! 5  4  3  2  1

 20
3!
3  2 1
5 letters and L is repeated 3 times.
7
Permutations and Word Arrangements (10.2)
Exercises
For each counting problem, first identify the answer as a permutation number (for
example: 15P5), then write the answer as a product (for example: 15 · 14 · 13 · 12 · 11),
and finally find the answer as a plain number (for example: 360,360).
1. A club with 25 members needs to choose four different people to be the president,
vice-president, secretary, and treasurer. How many ways can this be done?
2. How many letter sequences are there consisting of three different capital letters?
(examples of what you’re counting: ALG, QXV; but not KAK)
3. A store has to hire a manager and an assistant manager. Five people are interviewed
for the jobs. How many different ways can the hiring decisions be made?
4. A web site requires a password consisting of 5 capital letters that must be all different
(no repeated letters). How many different passwords are possible?
5. The manager of a softball team needs to put the 9 players on the team into a batting
order. This means: players have to be assigned places from 1st through 9th.
How many different batting orders are possible?
8
Permutations and Word Arrangements (10.2)
6. The letters “MAINE” can be scrambled in many different ways, such as “AMIEN”.
a. How many different ways can all the letters be arranged?
b. How many different arrangements can be made if each new arrangement must
start with “M”?
c. How many different arrangements can be made if the “AI” letters must stay
together and with the A coming before the I?
7. You and 6 friends decide to go to a movie.
a. In how many different arrangements could you and your friends sit in the same
row? Assume there are no empty seats.
b. If you demand to sit along the aisle of the row, how many arrangements are there
now? Assume there are no empty seats.
8. Find the rearrangements of all the letters in each of the following words.
a. IOWA
b. TACOMA
c. SASSY
d. NOON
e. INDIANA
f. MISSISSIPPI
9
Permutations and Word Arrangements (10.2)
9. Challenge: An Unusual Application
A waiter takes dinner orders from patrons in a restaurant. The evening entrée options
are Fish, Chicken, or Beef. At a table of 4, how many different dinner orders could
possibly be given to the waiter? The kitchen does not care who orders what, so an
“order” is just how many of each entrée is needed at a table, i.e. four fish is a possible
order; one fish, one chicken, and two beef is another possible order.
10
Combinations (10.3)
Combinations
A combination problem involves counting the number of ways that some objects can be
selected without choosing an order.
Example A team with 20 players must cut 4 players to get down to the correct number
for a tournament. How many ways can this be done?
Using permutations we could calculate 20 P4  20∙19∙18∙17 = 116280 . This number,
since it is a permutation, would mean it matters what order the players are cut. In this
situation, however, if players A,B,C, and D got cut it shouldn’t matter if they are listed
D,C,A, and B – they’d still be the same four players getting cut. Since there are
4! = 4∙3∙2∙1 = 24 different ways to list any group of four cut players, the
number 20 P4  116280 is 24 times too large. To find the right number you must divide
by 4! = 24:
P4 116280

 4845 different groups of 4 players that could be cut.
4!
24
20
Example There are 15 tracks on a CD. You are asked to choose your 5 favorite tracks,
but you don’t have to specify an order. How many ways can this be done?
Calculate
P5 (15  14  13  12  11)

. The answer is 3,003.
5!
(1  2  3  4  5)
15
Here is a general statement of the Combination counting method:
Definition - Combination: To count the number of ways of selecting r object out of a set
of n objects, without a relevant order, divide n Pr by r!. As a formula:
n Cr 
Pr
r!
n
11
Combinations (10.3)
Exercises
1. From a box of 16 color crayons, a child selects a set of 4 crayons as her favorites.
How many different ways can she make this selection?
2. From a senior class of 450 students, 3 students will be selected to speak at graduation.
How many different ways can these choices be made?
3. A store has to hire two cashiers. Five people are interviewed for the jobs. How many
different ways can the hiring decisions be made?
4. You go to the video store to rent some movies for the weekend. There are 1000
movies available, and you want to select 3 of them. How many ways can you make
your choices?
5. A pizza place has a list of many toppings to put on a pizza. However, they ask that
you choose exactly four of these many toppings per pizza. In their advertizing, they
make a claim that they offer “More than a million different pizza combinations.”
How many toppings must be on their list for this claim to be true? Note: you can’t
have repeated toppings.
12
Combinations (10.3)
Permutation and combination problems together
When you encounter a problem that asks you to count the number of ways that a group
can be selected from a larger group, you need to figure out whether it is a permutation
problem or a counting problem. Here’s how to decide.
 If the problem situation involves the selections being made in a particular order or
for specific positions, it’s a permutation counting problem (nPr).
 If the order in which the selections are made doesn’t matter in the problem situation,
it’s a combination counting problem (nCr).
Think carefully about this distinction as you do the following problems.
6. An ice cream stand carries 5 flavors: vanilla, chocolate, strawberry, orange,
and pineapple. The stand sells two-scoop ice cream cones.
Some examples are shown in the picture. Note that vanilla-chocolate and
chocolate-vanilla are considered to be different cones. In other words, the
order of the scoops matters. Also assume that you cannot have two of the
same flavor on one cone.
v
c
v
c
v
s
a. How many different kinds of two-scoop ice cream cones are there?
b. List all of the possible two-scoop ice cream cones. You can abbreviate just using
the first letter of each flavor.
7. The same ice cream stand also sells two-scoop ice cream dishes.
Some examples are shown in the picture. In a dish, vanilla-chocolate
would be considered the same as chocolate-vanilla, so you shouldn’t
list them both. Also assume that you cannot have two of the same
flavor on one cone.
v
c
v
s
a. How many different kinds of two-scoop ice cream dishes are there?
b. List all of the possible two-scoop ice cream dishes. You can abbreviate just using
the first letter of each flavor.
13
Combinations (10.3)
8. The answer to the ice cream dish Exercise 7a was half of the answer to the ice cream
cone Exercise 6a. Explain why this happened.
9. From our class of 22 students, I need to select a group of 9 students to perform a topsecret mission. How many different ways can this group be selected?
a. Which of these is the correct notation:
22P9
or
22C9
?
b. Now use your calculator to get the answer.
10. From a class of 22 students, I need to select a group of 9 students to play softball.
This means that besides choosing the students, I need to make a batting order (decide
who will bat 1st, 2nd, 3rd, etc.). How many different ways can this be done?
a. Which of these is the correct notation:
22P9
or
22C9
?
b. Now use your calculator to get the answer.
11. A hiker would like to invite 7 friends to go on a trip, but has room for only 4 of them.
In how many different ways can they be chosen?
12. The hiker with 7 friends wants to make a list of her 1st, 2nd, 3rd, and 4th best friends.
How many different possible lists are there?
13. Suppose that your English teacher asks you to choose 2 of these 4 books to read.
Beloved, Hamlet, Joy Luck Club, To Kill a Mockingbird
How many different ways could you choose?
14
Combinations (10.3)
14. At a bank ATM, you must use a four letter pin (a password) to access your account.
The pin numbers can be digits between 0-9. Some examples of valid pin #s are:
{1,2,3,4}, {0015}, {9999}. If you completely forgot the last two digits of your pin
number, how many guesses would it take to try every possibility?
15. In your own words, explain the formula n Cr 
Pr
.
r!
n
16. Equations for n Pr and n Cr .
P4 
10!
6!
P 
20!
5!
C4 
10!
6!4!
a. Verify that
10
b. Verify that
20 5
c. Verify that
10
d. Write formulas for n Pr and n Cr that involve only n, r, and !.
15
Mixed Counting Problems (10.4)
Mixed Counting Problems
For each problem, first express the answer using counting notation or multiplication (such as 5!,
7P3, 12C4, 1048, etc.), then calculate the answer as an ordinary number. Don’t assume that a
problem will require permutation or combination counting. It might, but it might not!
1. Each of 200 students attending a school event is entered in a drawing for some door prizes.
a. If there are 3 identical prizes to be awarded, how many different ways are there to award
the prizes?
b. If there are 3 different prizes to be awarded, how many different ways are there to award
the prizes?
2. a. A railway has 30 stations. On each ticket, the departure station and the destination station
are printed. How many different tickets are possible?
b. If a ticket can be used in either direction between two stations, how many different tickets
are there?
3. Suppose there are 25 students in a class. If each person in class shook hands with every other
person, how many handshakes would occur?
4. A club has 25 members.
a. In how many different ways could the club choose a president and a vice-president?
b. In how many different ways could the club choose two co-presidents?
c. In how many different ways could the club choose a president, vice-president, and
secretary?
d. In how many different ways could the club choose two co-presidents and a secretary?
e. In how many different ways could the club choose a three-member governing committee?
16
Mixed Counting Problems (10.4)
5. There is a series of Massachusetts license plates consisting of 4 digits followed by 2 letters
(example: “2582 AB”). How many different license plates are there in this series?
6. Suppose you are taking a 10-question true-false test. In how many different ways can you
answer the questions?
7. A modern artist produces a series of paintings called “White, Grey, Black.” Here is a picture
of one painting from the series:
Other paintings in the series look the same except that each square may have been given any
of the three colors available. How many different paintings would there be in this series?
8. An automated teller machine (ATM) uses a 4-digit password called a Personal Identification
Number (PIN).
a. How many different PIN’s are there?
b. How many PIN’s are there in which the digits are all different from each other?
c. How many PIN’s are there that contain no zeroes?
d. How many PIN’s are there that contain at least one zero?
e. How many PIN’s are there that contain exactly one zero?
17
Mixed Counting Problems (10.4)
9. a. Show how to calculate the value of the combination number 1000C3 without needing to
know the values of extremely huge factorial numbers such as 1000!
b. Show how to calculate the value of the permutation number 1000P3 without needing to
know the values of extremely huge factorial numbers such as 1000!
10. There are 2 roads from town A to town B, 5 roads from town B to town C, and 4 roads from
town C to town D. A running club is planning a race route that will begin in A, pass through
B and then C, and end at D. How many different routes are possible?
11. a. How many 4-letter “words” can be formed using 4 of the 8 letters in the word
TRIANGLE?
b. How many of these words contain at least 1 vowel? Hint: First determine how many
words contain no vowels.
12. Suppose you are visiting a city, and you need to walk to a theater that is 3 blocks south and
4 blocks east from your current location. The diagram below illustrates one of the many
possible routes you could take.
you
theater
Assuming you want to walk as little as possible, how many different routes could you take?
Try coming up with a counting idea on your own, but if you need a hint, there is one on
the last page of this section.
18
Mixed Counting Problems (10.4)
13. A 6-sided die (plural: dice) is a cube whose faces are labeled with the numbers 1, 2, 3, 4, 5,
and 6. Suppose you roll a red 6-sided die and a blue 6-sided die, and look at the results.
One example of an outcome would be: “red 5, blue 4.” How many different outcomes
are possible?
14. On a 20-question survey, each question requires a response of 1, 2, 3, 4, or 5 (where 1 means
“strongly agree” and 5 means “strongly disagree”). How many different people would have
to take the survey to guarantee that there would be two people with identical answers?
15. Try these without the use of a calculator.
a. Express the permutation number nP n in the simplest possible form.
b. Express the combination number nC n in the simplest possible form.
c. Express the permutation number nP (n–1) in the simplest possible form.
d. Express the combination number nC (n–1) in the simplest possible form.
16. Rewrite the fraction
Pr
in the simplest possible form. Why does the answer make sense?
n Cr
n
19
Mixed Counting Problems (10.4)
17. In a standard deck of 52 playing cards, each card shows a different pairing of a suit
(spades, hearts, diamonds, or clubs) and a value (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, or A).
A “5-card hand” is a set of 5 cards dealt from the deck’s 52 cards. (The order of the 5 cards
does not matter.)
a. Find the total number of different 5-card hands.
b. Find the number of 5-card hands containing the “10 of hearts” card.
c. Find the number of 5-card hands not containing the “10 of hearts” card.
d. Find the number of 5-card hands not containing any of the “10” cards.
e. Find the number of 5-card hands containing all four of the “10” cards.
20
Mixed Counting Problems (10.4)
18. Here are more questions about 5-card hands from a standard deck of 52 cards.
a. Find the number of 5-card hands containing all diamonds.
b. Find the number of 5-card hands containing no diamonds.
c. Find the number of 5-card hands containing exactly one diamond.
d. Find the number of 5-card hands containing exactly three diamonds.
e. Find the number of 5-card hands containing three diamonds and two clubs.
f. Find the number of 5-card hands containing at least one diamond.
Hint for problem 12.
Each route consists of walking along 7 city block segments: 3 going south and 4 going east.
Think about when the choices are made to walk south. For example, the route in the picture
involves walking south as the 2nd, 4th, and 5th segments. This means that choosing a route is
equivalent to picking 3 numbers from {1, 2, 3, 4, 5, 6, 7}; in this case the choices are {2, 4, 5}.
How many different ways can those choices be made?
21
Introduction to Probability (10.5)
Introduction to Probability
Probability is the study of how to use numbers to indicate the likelihood of events. One approach
is to find the experimental probability of an event. To find this we do an experiment a fixed
number of times and record the result of each experiment.
Definition - Experimental Probability Function: Let Pexp(E) be the experimental probability
that event E occurs. Then the Experimental Probability Function is given by:
Pexp ( E ) 
number of experiments with outcome E
total number of experiments performed
Example What is the range of the probability function Pexp(E)?
number with outcome E
0

 0.
If outcome E never occurs, then Pexp ( E ) 
total number performed total number
If the outcome E occurs every time, then.
number with outcome E total number
Pexp ( E ) 

1
total number performed total number
Therefore the range of Pexp(E) must be 0  Pexp ( E )  1 .
Example Rolling a die
A group of friends each rolled a six-sided die 10 times and kept track of their results. The
zero in Abe’s row means that he did not roll any fours in his 10 rolls.
If the event E is “Betty rolls a three”, then this
event has experimental probability Pexp ( E )  104
since she rolled 4 three in her 10 rolls.
If E  Claire rolls a one, then Pexp ( E )  0 since
Claire did not roll a one.
Abe
Betty
Claire
Diego
Evan
Totals
1
2
1
0
1
2
6
2
3
1
2
1
2
9
3
1
4
0
2
2
9
4
0
2
3
2
1
8
5
1
1
2
1
3
8
6
3
1
3
3
0
10
If E  someone rolls a five, then Pexp ( E )  508 because five came up 8 times in their 50
collective rolls.
22
Introduction to Probability (7.1)
Definition - Theoretical Probability Function: Let Pth(E) be the theoretical probability that
event E occurs. Then the Theoretical Probability Function is given by:
Pth ( E ) 
number of ways to get a specific outcome E
total number of possible outcomes
Example Again consider the die rolling scenario above
If the event E is “Betty rolls a 3”, then the theoretical probability is Pth ( E )  16 because
there is one way for this particular E to happen and six possible outcomes.
If the event E is “Diego rolls an even number” then the theoretical probability is
Pth ( E )  63 because there are three ways to roll an even number and six possible outcomes.
It is usually clear from the context of a problem whether or not it is asking about experimental or
theoretical probability, so the subscripts on Pexp ( E ) and Pth ( E ) will often be omitted.
Definition - Sample Space: The sample space of an experiment is a list of all possible
outcomes.
Example In the experiment “flip a coin”, the sample space consists of 2 equally-likely outcomes:
{Heads, Tails}.
Example In the experiment “flip a coin and roll a die”, the sample space consists of 12 equallylikely outcomes: {1H ,1T , 2H , 2T ,3H ,3T , 4H , 4T ,5H ,5T ,6H ,6T}
Example In the experiment “draw a card out of a standard deck”, the sample space has 52
equally-likely outcomes; one for each card:
2
3
4
5
6
7
8
9
10
J
Q
K
A
2
3
4
5
6
7
8
9
10
J
Q
K
A
2
3
4
5
6
7
8
9
10
J
Q
K
A
2
3
4
5
6
7
8
9
10
J
Q
K
A
23
Introduction to Probability (7.1)
Exercises
1. Flip a penny 10 times
a. Record your individual results: ______ heads in 10 experiments.
b. Record the class results: ______ heads in ______ experiments.
c. Based on this class experiment, what is the experimental probability of flipping a coin
with an outcome of heads? With an outcome of tails? With some other outcome?
d. Your intuition tells you what the probability should be. What would you expect to
happen to Pexp ( H ) if we performed many, many more experiments?
2. Are the outcomes “heads” and “tails” equally likely when flipping coins? What could make
them not be equally likely?
3. Assuming that “heads” and “tails” are equally likely, what is the theoretical probability that
the coin lands on “heads” the first time you flip it?
4. What is the probability that the coin lands on “heads” the second time you flip it? On the
50th time you flip it?
5. Image something bizarre happens: You flip a coin 100 times and every time it lands on
heads! (it could happen, right?). Does this change the probability the coin lands on “heads”
the next time?
24
Introduction to Probability (7.1)
6. What is the sample space of rolling a die? Remember, sample space is just a list.
7. What is Pth(1), i.e. the theoretical probability of rolling a 1?
8. With the same experiment, evaluate these probabilities:
P(2) =
P(even number) =
P(1 or 2) =
P(1 and 2) =
P(7) =
9. Consider the experiment, “A single coin is tossed three consecutive times.”
a. Write all the elements of the sample space for this experiment. Hint: one possible
outcome would be HTT, meaning “heads” followed by “tails” followed by “tails.”
b. What is the probability that all three tosses came up tails?
c. What is the probability that exactly one of the tosses came up tails?
d. What is the probability that at least one of the three tosses came up tails?
25
Union, Intersection and Complement (10.6)
Union, Intersection and Complement
When calculating theoretical probabilities, we are often interested in finding the number of
elements in various subsets of a sample space. The union and intersection of sets are common
ways to create subsets, as is taking the complement of a set.
The intersection (  ) of two sets is the set of all elements in both (i.e. their “overlap”).
P( A  B) is interpreted as the probability of observing an outcome that is part of both event A
and event B.
The union (  ) of two sets is the set of all elements in either one or the other or both.
P( A  B) is interpreted as the probability of observing an outcome that is part of event A or
event B or both.
The complement ( ~ ) of a set is the set of all elements that are not in a given set, but are still
within the sample space of the problem. P( A) is interpreted as the probability of observing an
outcome that is not part of event A.
Example Consider the sets of people defined below:
Let A = {Abe, Bart, Homer, Lisa, Marge}
Let B = {Bart, Lisa, Martin, Milhouse, Nelson,
Ralph}
A  B = {Lisa, Bart}.
A  B = {Abe, Bart, Homer, Lisa, Marge, Martin,
Milhouse, Nelson, Ralph}
A
B
Homer
Milhouse
Lisa
Marge
Nelson
Bart
Abe
Martin
Ralph
~ A = {Martin, Milhouse, Nelson, Ralph}
~ (A  B) = {Abe, Homer, Marge, Martin, Milhouse, Nelson, Ralph} (All people not in both).
Example Consider the experiment “draw a single card out of a standard deck”.
Let A = the card is a heart, B = the card is a face card
The Venn Diagram describes this scenario. Notice how
you can put the number of outcomes in each part of the
diagram instead of listing out all 52 elements of the
sample space in their appropriate places.
A
10
B
3
9
30
22
because there are 10  9  3  22 cards that are part of A or B or both.
P( A  B)  52
40
P(~ B)  52
because there are 10  30  40 cards that are not part of event B
26
Union, Intersection and Complement (10.6)
1. Let A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, B = {1, 3, 5, 7, 9, 11, 13}, and C = {3, 6, 9, 12}.
Evaluate each of the following.
a. A  C =
b. B  C =
c. B  C =
d. C  {0,3,11} =
e. C  {1,2,4} =
2. The ability to taste certain chemicals is sometimes linked to a single gene. Either you can
taste the chemical or you can’t. An experiment is performed to see if a group of people can
taste chemicals X and Y. The results are summarized in the Venn diagram below.
a. How many people were in the group
experiment?
Can taste Y
b. What fraction of people could taste
chemical Y?
3
c. What fraction of people could taste
neither chemical?
Can taste X
4
13
20
d. Assuming that these results were similar to those in the general population. Explain the
meaning and predict the value of P( ~ (X  Y) ).
27
Union, Intersection and Complement (10.6)
Here are a few problems1 where Venn Diagrams may be very helpful.
3. In a class, there are:
 8 students who play football and hockey
 7 students who do not play football or hockey
 13 students who play hockey
 19 students who play football
How many students are there in the class?
4. There are 20 people in a room. Of these, 15 are holding newspapers and 8 are wearing
glasses. Everyone wears glasses or holds a newspaper. How many people are wearing
glasses and holding a newspaper?
5. All the members of a group of 30 teenagers belong to at least one club. There are 3 clubs:
chess, drama, and art.
 6 of the teenagers belong to only the art club
 5 of the teenagers belong to all 3 clubs
 2 of the teenagers belong to the chess and art
clubs but not to the drama club.
 15 of the teenagers belong to the chess club
 2 of the teenagers belong only to the chess club
 3 of the teenagers belong only to the drama
club
a.
P (chess ) 
P (art ) 
P (drama ) 
b.
1
P(chess  (~ art )) 
Source, http://www.cimt.plymouth.ac.uk/projects/mepres/book7/bk7_1.pdf
28
Conditional Probability and Independence (10.7)
Conditional Probability and Independence
When considering the probability of more than one event occurring, it is necessary to
understand the concepts of conditional probability and independence.
Definition: P( B | A) is the conditional probability that event B will happen given that
event A has already happened.
One way to think about conditional probability is since event A already occurred, you are
in a new sample space consisting of only the outcomes that are part of A. In other words:
P( B | A) 
# of outcomes in B that are also in A
total # of outcomes in A
Example Consider the event “Draw a card from a standard deck”.
P(king | face card) is the probability that the card is a king given that it is a face card.
There are 12 face cards, and 4 of them are kings, so P(king | face card)  124 .
P(face card | king) is the probability that the card is a face card given that it is a king.
There are 4 kings, and all of them are face cards, so (as your intuition tells you)
P(face card | king)  44  1.
Example Consider the experiment “Pick a random letter”.
Let A = “the letter is a vowel” and let B = “the letter is before k in the alphabet”. The
Venn diagram describes this scenario.
P( A | B)  103 because there are 10 letters before k
and three of them are vowels.
P( B | A)  53 because there are 5 vowels and three
of them are before k.
A
2
B
3
7
14
P( B | (~ A))  217 because there are 21 letters that are not vowels, and 7 of those 21
come before k.
29
Conditional Probability and Independence (10.7)
Definition: Events A and B are independent if the occurrence of one event does not alter
the probability of the other occurring. Using the notation of this section, we say that the
condition for events A and B to be independent is
P( B | A)  P( B) and/or P( A | B)  P( A)
Example A six-sided die is rolled once. Consider the events described below. Are
events A and B independent? Are events A and C independent?
A = “the number is even” (even)
B = “the number is greater than three” (>3)
C = “the number is less than five” (<5)
First calculate the probabilities of the events occurring on their own:
P( A)  63 because 3 of the 6 numbers are even.
P( B)  63 because 3 of the 6 numbers are greater than three.
P(C )  64 because 4 of the 6 numbers are less than five.
Now calculate the conditional probabilities:
P( A | B)  23 because 2 of the 3 numbers above three are even. We see that if you
know the number is greater than 3, then the probability of it being even changes from
3/6 to 2/3. Thus, events A and B are NOT independent.
P( A | C )  24 because 2 of the 4 numbers below five are even. We see that if you
know the number is less than five, the probability that it is also even does not change.
It is ½ regardless of whether or not you know it is less than five. Thus, events A and
C are independent.
In some (but not all) situations, it is obvious that two events cannot affect one another’s
probabilities and you can just assume from the context of the problem that they are
independent.
Example The events A = “I get heads on my first flip of a penny” and B = “I get heads
on my second flip” can be assumed to be independent since P( B)  12 regardless of what
happens on the first flip.
Example Consider the events A = “I get the ace of spades on my first draw” and B = “I
get the queen of hearts on my second draw”. IF you replace your first draw and reshuffle
the deck, THEN P( B | A)  521 , which is the same as just P( B) . With replacement, the
events are independent. However, if you do not replace your first draw, then
P( B | A)  511 instead of 1/52. Thus, without replacement the events are not independent.
30
Conditional Probability and Independence (10.7)
Exercises
1. In English class each student is randomly assigned a different letter of the alphabet
that they must use to start off every line of a poem. Find the following conditional
probabilities.
 your letter the letters b,d,e,g,t,u, and z 
a. P 

 is a vowel have already been assigned 
 your letter is in the your letter comes before 
b. P 

k in the alphabet
 word "probability"

 your letter comes before your letter is in the 
c. P 

k in the alphabet
word "probability" 

2. A computer generates a random number between 1 and 100. Find the following
conditional probabilities.
 # is divisible # is divisible 
a. P 

by 10
by 5


b. P  # is prime # is a positive integer  50  (2 is the first prime number)
 digits sum

# is a
c. P 

to 9
perfect square 

31
Conditional Probability and Independence (10.7)
3. The Venn diagram shows results of a hypothetical survey of Lakeside students.
Event X is “the student has first period free” and Event Y is the student had more than
10 unexcused absences last year”.
a. How many students participated in the
survey?
X
75
Y
25
15
b. P( X )  ______
100
c. P( X | Y )  ______
d. Are these two events independent? Justify your answer.
4. You are assigned a random number between 1 and 12. Decide if each of the
following pairs of events X and Y are independent. You must justify your answer.
a. X = your number is prime, Y = your number is less than or equal to 6.
b. X = your number is odd, Y = your number is a perfect square.
c. X = your number is divisible by 3, Y = your number is even.
32
The Multiplication Principle of Probability (10.8)
The Multiplication Principle of Probability
Finding the probability of event A and event B both occurring, P( A  B) , is an important
and subtle problem. To calculate this type of probability you often must use the
multiplication principle of probability. To derive the multiplication principle of
probability, we must formalize and extend our earlier explanation of conditional
probability:
P( B | A) 
# of outcomes in B that are also in A # of outcomes in A  B

total # of outcomes in A
# of outcomes in A
Now recall from arithmetic that
x

y
P( B | A) 
x
z
y
z
, which leads to
# of outcomes in A B
# total possible outcomes
# of outcomes in A
# total possible outcomes

P( A  B)
P( A)
This rearranges to the multiplication principal of probability:
P( A  B)  P( B | A)  P( A) This formula ALWAYS works
A similar development leads to the alternative expression
P( A  B)  P( A | B)  P( B) This formula also ALWAYS works
To calculate the probability of both A and B both occurring, you must know the
probability of one of the events occurring and a conditional probability. If and only if A
and B are independent, you can use a simplified version of this formula:
P( A  B)  P( A)  P( B)
ONLY WORKS IF A AND B ARE INDPENDENT!!!
33
The Multiplication Principle of Probability (10.8)
Example You draw 2 cards form a deck without replacement after the first draw. Let
event A = “the first card is a king” and B = “the second card is a queen”. Find P( A  B) .
These two events are NOT INDEPENDENT, so we must use the general form of the
multiplication principle: P( A  B)  P( B | A)  P( A) . Here P( B | A)  4 / 51 because all
4 queens are still in the remaining stack of 51 cards. P( A)  4 / 52 , so we find that
4
P( A  B)  514  524  663
Example You draw 2 cards form a deck with replacement and shuffling after the first
draw. Let event A = “the first card is a king” and B = “the second card is a queen”. Find
P( A  B ) .
These two events are independent, so you can use the simplified form of the
multiplication principle: P( A  B)  P( B)  P( A) . Here P( B)  P( A)  4 / 52 so we find
1
that P( A  B)  524  524  169
.
Example 20% of all Lakeside students run cross country, 25% of all Lakeside students
specialize in an art discipline, and 30% of cross-country participants specialize in an art
discipline. If you stop a random student on campus, what is the probability that they run
cross country and specialize in an art discipline?
Let CC = “student runs cross country” and SA = “student specializes in art”. From
careful reading of the problem we know P(CC )  .20 , P(SA)  .25 , and
P(SA | CC )  .30 . We are trying to find P(SA  CC ) . Since P(SA | CC )  P(SA) , these
two events are not independent and we must use the general form of the multiplication
principle:
P(SA  CC )  P(SA | CC )  P(CC )  .30 .20   .06
34
The Multiplication Principle of Probability (10.8)
Exercises
1. Suppose that you flip two coins. Find these probabilities.
a. probability that both coins are heads, i.e P(H,H)
b. P(T,T)
c. P(H,T)
d. P(T,H)
e. probability that you get 1 head and 1 tail, coming in either order.
2. A deck of cards has 52 cards, and 13 of them are hearts.
a. You deal one card from the deck. What is the probability that it is a heart?
b. Suppose that the first card dealt was a heart, and that this card is not returned to the
deck. Now you deal a second card from the deck. What is that probability that this
card is a heart?
c. What is the probability of dealing two hearts in a row?
35
The Multiplication Principle of Probability (10.8)
Definition - Mutually Exclusive: Two events are mutually exclusive if when one event
occurs, the other event doesn’t occur.
With a Venn Diagram, mutually exclusive events would be drawn with circles that do not
intersect.
3. Suppose that events X and Y are mutually exclusive. What can you say about the
following probabilities?
a. P(X and Y)
b. P(X or Y)
c. P(X | Y), i.e. the probability of X occurring given Y occurs.
d. P(Y | X), i.e. the probability of Y occurring given X occurs.
e. P(Y | X), i.e. the probability of Y occurring given X occurs.
4. Two cards are dealt from a standard 52-card deck.
a. What is the probability that the first card is a heart and the second card is not a
heart?
b. What is the probability that the first card is not a heart and the second card is a
heart?
c. What is the probability that both cards are not hearts?
5. Suppose that a jar contains 3 red marbles and 4 green marbles. Two marbles are
drawn from the jar (and not returned to the jar).
a. Make a tree diagram. Label every branch with a probability.
b. What is the probability that both marbles are green?
c. What is the probability of drawing a green marble followed by a red marble?
d. What is the probability of drawing a red marble followed by a green marble?
36
The Multiplication Principle of Probability (10.8)
6. Suppose that a jar contains 5 yellow marbles and 5 blue marbles. Two marbles are
drawn from the jar (and not returned to the jar).
a. Make a tree diagram. Label every branch with a probability.
b. What is the probability that both marbles are yellow?
c. What is the probability that both marbles are blue?
d. What is the probability of drawing a blue marble followed by a yellow marble?
e. What is the probability of drawing a yellow marble followed by a blue marble?
The methods you have been using for a sequence of two events can also be applied to three
events. Your tree diagram will have an extra layer, and multiplications will involve three
probabilities.
7. Suppose that a jar contains 5 yellow marbles and 5 blue marbles. Three marbles are
drawn from the jar (and not returned to the jar).
a. Make a tree diagram. Label every branch with a probability.
b. Find the probability of drawing a yellow marble all three times.
c. Find the probability of this sequence of draws: blue-yellow-blue.
d. Find the probability of this sequence of draws: yellow-yellow-blue.
37
The Multiplication Principle of Probability (10.8)
8. A coin is flipped three consecutive times.
a. What is the probability of flipping tails all three times?
b. What is the probability of flipping heads at least one time (Hint: Think
complement.)?
c. What is the probability of this sequence of flips: heads-tails-tails?
9. In the deck of 52 playing cards, 4 cards are aces (A) and the rest are non-aces (N).
Suppose you deal two cards from the deck, and look at whether each card is an ace or not.
a. Make a tree diagram and label it with probabilities.
b. What is the probability of dealing two aces?
c. What is the probability of dealing two non-aces?
d. What is the probability of dealing an ace followed by a non-ace?
e. What is the probability of dealing a non-ace followed by an ace?
f. What is the probability of dealing one ace and one non-ace in either order?
38
The Multiplication Principle of Probability (10.8)
10. Consider the following scenario.
The test for a certain disease is not 100% reliable. If a patient tests positive (P) for a
disease (D), there is a 95% chance they actually have the disease. In 5% of these
positive tests, however, the patient does not have the disease. Similarly, if a patient
tests negative (N) for the disease, the test is 99% accurate. However, they may have
the disease with a 1% likelihood. Several years of tests indicates that around 15% of
all patients who take the test will test positive.
a. Draw a tree diagram for this situation. Hint: the first branching will separate the
outcomes of the test itself: P and N.
b. In medicine two very important results of such tests are called “False Positive” or
“False Negative.” Describe the situation that would result in each of these results.
Which one is more dangerous to the patient? Find the probability of both results.
Exercises 11-14 adapted from Foerster, Algebra and Trigonometry.
11. You and your family drive on a long summer road trip. The probability you will have
a flat tire is 0.1 and the probability of engine trouble is 0.05. What is the probability
you will have:
a. no flat tire?
b. no engine trouble?
c. no flat tire and no engine trouble?
d. both a flat tire and engine trouble?
e. at least one, either a flat tire or engine trouble?
39
The Multiplication Principle of Probability (10.8)
12. Two traffic lights on Broadway operate independently. Your probability of being
stopped at the first one is 0.4 and your probability of being stopped at the second one
is 0.7. What is your probability of being stopped at
a. both lights?
b. neither light?
c. the first but not the second?
d. the second but not the first?
e. exactly one of the lights?
13. Vital systems such as electronic power distribution systems have “back-up”
components in case one component fails. Suppose that two generators each have
98% probability of working. The system will continue to work as long as at least one
of the generators works. What is the probability the system will continue to work?
40
The Multiplication Principle of Probability (10.8)
14. Three basketball teams each play on Friday nights. The probability that they will win
are: varsity, 70%; JV, 60%; freshman, 80%. What is the probability that
a. all three teams win?
b. all three teams lose?
c. at least one team wins?
d. the varsity wins and the other two lose?
41
Additional Exercises
Additional Exercises
10.1 Counting Principles
1. A deli has three kinds of bread, four kinds of meat and 7 kinds of cheese. How many
different sandwiches could I order?
2. A computer password is 5 characters. Each character can be either a digit 0-9 or a capital
letter. It is OK to repeat characters. Example passwords are 12RT5, 22ABC and
AAB66. How many passwords are possible?
3. In a class of 18 students, 12 are right handed, 10 are girls, and 6 are right-handed girls.
How many students are girls or are right handed?
4. You draw a card from a standard deck. How many ways could the card be a face card or
a spade?
10.2 Permutations and Word Arrangements
1. Write the following permutation numbers as a product. The first has been done as an
example.
a.
P  765
b.
7 3
10
c.
P4
P
5 5
d.
P
100 9
2. I have 4 spots for plants in my front yard and there are 6 types of plants that I like. How
many ways could I put the plants in my yard if I want all the plants to be different?
3. How many ways can you deal three cards out of a standard deck if the order matters?
4. How many distinct ways can you rearrange the letters in the following words?
a. STUDY
b. MOON
c. ALGEBRA
d. MISSES
e. BANANA
10.3 Combinations
1. Write the following combination numbers as fraction with products. The first has been
done as an example.
a.
10
C2  10219
b.
7
C3
c.
50
C5
d.
6
C6
2. The local pound has 11 adorable dogs for adoption. How many different ways could you
adopt 3 of these dogs?
3. How many different 3-card hands can be formed from a standard deck of 52?
4. Five people on the 20 player JV soccer team are being pulled up to varsity for the
playoffs. How many ways could this be done?
42
Additional Exercises
10.4 Mixed Counting Problems
1. Each question on a multiple choice test has 4 options: A-D. If there are 10 questions on
the test, how many answer keys are possible?
2. You go to the ice cream store with four friends. There are 7 different flavors of ice
cream. You each get a single cone. How many ways could you 5 get your ice cream?
What if you all decide to get different flavors?
3. You have 3 pairs of shoes. Any of them can be worn with or without socks. You have 5
pairs of socks. How many different ways could you get your feet ready for school?
4. The pound has 4 dogs and 5 cats. You plan to get 2 dogs for yourself, a cat for your
friend Sam and another cat for your friend Larry. How many ways could you distribute
these animals?
10.5 Introduction to Probability
1. A group of friends each rolled a six-sided die 10 times and kept track of their results.
The zero in Abe’s row means that he did not roll any fours in his 10 rolls. The results are
summarized in the table. Find the following probabilities.
 Diego rolls 
a. Pexp 

a1


Abe
Betty
Claire
Diego
Evan
Totals
 Evan rolls an 
b. Pexp 

 even number 
 Evan rolls an 
c. Pth 

 even number 
1
2
1
0
1
2
6
2
3
1
2
1
2
9
3
1
4
0
2
2
9
4
0
2
3
2
1
8
5
1
1
2
1
3
8
6
3
1
3
3
0
10
2. A turn in a hypothetical game involves flipping a coin which comes up Heads or Tails
and spinning a spinner that lands on one of four colors: Blue, Red, Yellow or Purple.
a. What is the sample space?
b. What is P(Red) ?
c. What is P(Tails) ?
3. You draw a card from a standard 52 card deck. Find the following probabilities.
 card is 
a. P 

 an ace 
 card is a 
b. P 

 diamond 
 card is a black 
c. P 

 face card 
43
Additional Exercises
10.6 Union, Intersection and Complement
1. Let A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, B = {1, 3, 5, 7, 9, 11, 13}, and C = {3, 6, 9, 12}.
Evaluate each of the following.
a. A  B = b. B  C = c. A  B = d. C  {3, 6, 9} = e. A  {even integers} =
2. 35 people took a survey about whether or not they liked two different candy bars.
a. Make a Venn Diagram for the following results of this survey:
 15 people like candy bar A
 8 people like both candy bar A and candy bar B
 20 people do not like candy bar A
 35 people took the survey
b. P( B) 
c. P( A  B) 
d. P(~ A) 
10.7 Conditional Probability and Independence
1. Consider the following events for drawing one card from a deck
X = the card is red (hearts or diamonds)
Y = the card is a number (2-10)
Z = the card is a five
a. P( Z )  b. P(Z | X )  c. P(Z | Y ) 
d. Are either Z and X or Z and Y independent? Justify.
10.8 The Multiplication Principle of Probability
1. The numbers 1-10 are in a hat. You pull out two without replacement.
a. What is the probability that the first number is a 1 and the second is a 2?
b. What is the probability the first two numbers are even?
c. What is the probability the first number is even and the second is odd?
d. What is the probability that the first two numbers are both divisible by three?
e. What is the probability that the second number is even given the first number was a 5?
f. What is the probability that the second number is even and the first number is a 5?
44
Additional Exercises
Solutions to Additional Exercises
10.1 Counting Principles
1. 3  4  7  84
3. 12 + 10 – 6 = 16
2. 365 possible passwords
4.16 + 13 – 4 = 25
10.2 Permutations and Word Arrangements
1. a. 7 P3  7  6  5
d.
2.
b.
P  10  9  8  7
c. 5 P5  5  4  3  2 1  5!
10 4
P  100  99  98  97  96  95  94  93  92
100 9
P  65 43
4. a. 5! = 120
b.
4!
2!
P  52  51 50
3.
52 3
 12
c.
6 4
7!
2!
d.
6!
3!
e.
6!
3!2!
10.3 Combinations
1. Write the following combination numbers as fraction with products. The first has been
done as an example.
a.
2.
11
10
C2  10219
b.
 9
C3  1110
321
7
C3  736215
3.
52
c.
50
484746
C5  5049
54321
50
C3  52351
21
d.
6
C6  665544332211  1
181716
4. 20 C5  2019
54321
10.4 Mixed Counting Problems
1. 410
2. 75 with no restrictions or 7 P5  7  6  5  4  3 if all different flavors
3. 3  5  2 options
4. 4 C2  5  4 
43
21
5 4
10.5 Introduction to Probability
 Diego rolls  1
1. a. Pexp 
  10
a1


2.
3.
 Evan rolls an  3
b. Pexp 
  10
 even number 
 Evan rolls an 
c. Pth 

 even number 
1
2
a. Sample space = {HB, HR, HY, HP, TB, TR, TY, TP}
b. P(Red)  82  14 ?
c. What is P(Tails)  84  12
 card is  4
a. P 
  52
 an ace 
 card is a  13
b. P 
  52
 diamond 
 card is a black  8
c. P 
  52
 face card 
45
Additional Exercises
10.6 Union, Intersection and Complement
1. a. A  B = {1, 3, 5, 7, 9} b. B  C = {3, 9}
c. A  B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13}
e. A  {even integers} = {2, 4, 6, 8, 10}
d. C  {3, 6, 9} = {3, 6, 9}
2. a.
A
B
7
8
4
16
b. P( B)  12 / 35
c. P( A  B)  19 / 35
d. P(~ A)  20 / 35
10.7 Conditional Probability and Independence
1. a. P(Z )  4 / 52  1/13
b. P(Z | X )  2 / 26  1/13
c. P(Z | Y )  4 / 36  1/ 9
d. Z and X are independent because P(Z | X )  P(Z ) . Z and Y are not independent
because P(Z | Y )  P(Z )
10.8 The Multiplication Principle of Probability
1. a.
1
10
 19
b.
5
10
 94
c.
5
10
 95
d.
3
10
 92
e.
5
9
f.
1
10
 95
46
Chapter 10 Challenge Problems
Challenge Problems
1. A student government class has 6 seniors, 8 juniors, 5 sophomores and 3 freshmen. A 6person dance committee is to be formed. Write an expression that represents the number of
possible committees given the following constraints. The first has been done as an example.
a. There must be at least one senior on the committee.
6  21C5
6 possible seniors
Any 5 of the remaining 21 people can
round out the committee.
b. There must be at least 2 seniors on the committee.
c. There must be at least 2 seniors and 2 juniors on the committee.
d. The committee will have 2 seniors, 2 juniors, 1 sophomore and 1 freshman.
e. The committee will have at least 1 representative from each class.
f. Tom and Todd are both seniors in the class. They don’t get along, so the only restriction
is that they can’t both be on the dance committee together.
47
Chapter 10 Challenge Problems
2. You go to a movie theatre with 5 friends and you all sit next to each other in a single row.
Assuming your best friend is in the group and that everyone sits down randomly, what is the
probability that you will end up sitting next to your best friend?
3. You flip a penny 10 times. What is the probability of getting exactly 5 heads?
4. A bag contains 5 pennies, 3 nickels, 4 dimes and 2 quarters. You pick 4 coins at random
from the bag. What is the probability you will have over $0.50?
48
Chapter 10 Challenge Problems
5. Nine chairs in a row are to be occupied by six students and Professors Alpha, Beta, and
Gamma. These three professors arrive before the six students and decide to choose their
chairs so that each professor will be between two students. In how many ways can
Professors Alpha, Beta, and Gamma choose their chairs?
6. A bag of popping corn contains 2/3 white kernels and 1/3 yellow kernels. Only 1/2 of the
white kernels will pop, whereas 2/3 of the yellow ones will pop. A kernel is selected at
random from the bag, and pops when placed in the popper. What is the probability that the
kernel selected was white?
7. First a is chosen at random from the set {1, 2, 3, … , 99, 100}, and then b is chosen at
random from the same set. Find the probability that the integer 3a + 7b has units digit 8.
49
Chapter 10 Topics Summary
Chapter 10 Topics Summary
Below is a brief summary of the topics covered over the course of this chapter. Note that
all of these topics are covered in greater depth in the individual sections.
10.1 Counting Principles

Multiplication Principle: If event A can occur m ways and event B can occur n
ways, then A and B occur n∙ m ways. In short: “and” = “multiply.”

Addition Principle: If event A can occur n ways and event B can occur m ways,
then A or B occur n + m – (number of ways both A and B occur together) ways.
In short: “or” = “add” but be careful about intersections.
10.2 – 10.3 Permutations and Combinations

Permutations – you are selecting r objects out of a pool of n distinct objects. The
objects are not replaced and the order they are selected is important. If these
conditions hold, the number of ways the objects can be selected is given by
n
Pr  (n)(n  1)(n  2) ... (n  (r  1))
r integers starting at n and
decreasing by one each time

Combinations – you are selecting r objects out of a pool of n distinct objects. The
objects are not replaced and the order they are selected is not important. If these
conditions hold, the number of ways the objects can be selected is given by
n
Cr 
Pr
r!
n
10.4 Mixed Counting Problems

Never assume going in to a problem that the objects you will be counting are
either permutations or combinations. If the objects you are counting do not meet
the conditions for permutations or combinations, you will have to go back to the
general multiplication principle to solve the problem.
10.5 Introduction to Probability
number of experiments with outcome E
total number of experiments performed

Experimental Probability: Pexp ( E ) 

number of ways to get a specific outcome E
total number of possible outcomes
The sample space is the set or list of all possible outcomes in a given scenario.

Theoretical Probability: Pth ( E ) 
50
Chapter 10 Topics Summary
10.6 Union, Intersection and Complement

The intersection (  ) of two sets is the set of all elements in both (i.e. their
“overlap”). P( A  B) is interpreted as the probability of observing an outcome
that is part of both event A and event B.

The union (  ) of two sets is the set of all elements in either one or the other or
both. P( A  B) is interpreted as the probability of observing an outcome that is
part of event A or event B or both.

The complement ( ~ ) of a set is the set of all elements that are not in a given set,
but are still within the sample space of the problem. P( A) is interpreted as the
probability of observing an outcome that is not part of event A.
10.7 Conditional Probability and Independence

P( B | A) is the conditional probability that event B will happen given that event A
has already happened.

Events A and B are independent if the occurrence of one event does not alter the
probability of the other occurring. This idea is specified by the equations
P( B | A)  P( B) and/or P( A | B)  P( A) .
10.8 The Multiplication Principle of Probability

The multiplication principle of probability allows us, in general, to find
P( A  B) ; which is the probability of two events both happening. It is given by
P( A  B)  P( B | A)  P( A) or P( A  B)  P( A | B)  P( B)

If (and ONLY IF) the two events are independent, you can use a simplified
version of the multiplication principle: P( A  B)  P( A)  P( B)
51