Introductory Probability
Counting Outcomes Intro
Nicholas Nguyen
[email protected]
Department of Mathematics
UK
January 27, 2017
Agenda
Counting Multistage Outcomes
Announcement: The second homework is due this Saturday.
I
Ice Cream Options
An ice cream shop oers 3 avors (chocolate, strawberry, or
vanilla) and 2 containers (cone or dish). A dessert is a scoop
of one avor served in a container. For example, one option is
Chocolate+Cone.
If you are allowed to choose one avor and one container for
your dessert, how many options do you have?
Table
We can sort the options in a table:
Choco. Straw. Vanilla
Cone C, Cone S, Cone V, Cone
Dish C, Dish S, Dish V, Dish
There are six options. Notice the table has:
I Three columns (1 per avor choice)
I Two rows (1 per container choice)
3×2 = 6
Tree Diagram
C
S
V
Tree Diagram
C
Cone
S
Dish
Cone
V
Dish
Cone
Dish
Tree Diagram Explanation
We can use the tree diagram to count and organize the ways
of making a dessert.
I We have 3 options for the avor.
I Each choice of avor leads to 2 options each for
container.
3×2 = 6
Multistage Experiments and Outcomes
Suppose an experiment consists of a sequence of r stages such
that the number of outcomes at the any stage is not aected
by the outcomes of the other stages. For i = 1, 2, ..., r , the i th
stage has ni (a natural number) outcomes. Then the total
number of outcomes for the entire experiment is
n1 · n2 · ... · nr .
LinkBlue ID's
A LinkBlue ID consists of four letters (A-Z) followed by three
digits (0-9). Picking a random LinkBlue ID involves seven
stages:
I Four stages with 26 outcomes each (letters)
I Three stages with 10 outcomes each (digits)
The total number of possible LinkBlue ID's is
264 · 103
LinkBlue ID's
A LinkBlue ID consists of four letters (A-Z) followed by three
digits (0-9). Picking a random LinkBlue ID involves seven
stages:
I Four stages with 26 outcomes each (letters)
I Three stages with 10 outcomes each (digits)
The total number of possible LinkBlue ID's is
264 · 103
LinkBlue ID's
A LinkBlue ID consists of four letters (A-Z) followed by three
digits (0-9). Now what if the rst digit must be a 2? There
are seven stages:
I Four stages with 26 outcomes each (letters)
I One stage with 1 outcome (rst digit must be a 2)
I Two stages with 10 outcomes each (2nd and 3rd digits)
The total number of possible LinkBlue ID's now is
264 · 1 · 102 = 264 · 102
LinkBlue ID's
A LinkBlue ID consists of four letters (A-Z) followed by three
digits (0-9). Now what if the rst digit must be a 2? There
are seven stages:
I Four stages with 26 outcomes each (letters)
I One stage with 1 outcome (rst digit must be a 2)
I Two stages with 10 outcomes each (2nd and 3rd digits)
The total number of possible LinkBlue ID's now is
264 · 1 · 102 = 264 · 102
Multiple-Choice Options
A multiple-choice quiz has 3 questions. Each question has 4
choices. How many ways can a student answer the quiz
(assuming they do not leave any blank answers)?
This is a task with 3 stages (the questions) with 4 choices
each, so there are
4 × 4 × 4 = 64
ways to answer the quiz.
Multiple-Choice Options
A multiple-choice quiz has 3 questions. Each question has 4
choices. How many ways can a student answer the quiz
(assuming they do not leave any blank answers)?
This is a task with 3 stages (the questions) with 4 choices
each, so there are
4 × 4 × 4 = 64
ways to answer the quiz.
Multiple-Choice Perfection
A multiple-choice quiz has 3 questions. Each question has 4
choices, and only one of those choices is correct. How many
ways can a student get all questions correct?
This is a task with 3 stages (the questions), but this time,
each stage only has 1 choice (the correct one), so there is
1×1×1 = 1
way to get all questions correct.
In general, there is only one way
(in lotteries, exams, etc.).
to get a perfect match
Multiple-Choice Perfection
A multiple-choice quiz has 3 questions. Each question has 4
choices, and only one of those choices is correct. How many
ways can a student get all questions correct?
This is a task with 3 stages (the questions), but this time,
each stage only has 1 choice (the correct one), so there is
1×1×1 = 1
way to get all questions correct.
In general, there is only one way
(in lotteries, exams, etc.).
to get a perfect match
Playing Cards
A playing card can have one of four suits (♥, ♦, ♣, ♠) and
one of thirteen ranks (A, 2-10, J, Q, K), so a deck of cards has
4 · 13 = 52 cards.
Talent Show
There are three competitors (A, B, C) at a talent show. Four
judges (1, 2, 3, 4) review the candidates independently. A
competitor wins if at least three out of four judges choose that
competitor as their favorite. In reality, the judges choose
randomly.
How many possible ways to vote are there? Each of the 4
judges has 3 choices, so there are
3 · 3 · 3 · 3 = 81 total ways to vote.
How many outcomes (ways) are there for a competitor to win
with approval from only 3 judges?
Example outcome:
A picked by judges 1, 2, 3, but judge 4 picked B.
To describe this outcome, we need: the winner, the dissenting
judge, and that judge's choice.
Talent Show
There are three competitors (A, B, C) at a talent show. Four
judges (1, 2, 3, 4) review the candidates independently. A
competitor wins if at least three out of four judges choose that
competitor as their favorite. In reality, the judges choose
randomly.
How many possible ways to vote are there? Each of the 4
judges has 3 choices, so there are
3 · 3 · 3 · 3 = 81 total ways to vote.
How many outcomes (ways) are there for a competitor to win
with approval from only 3 judges?
Example outcome:
A picked by judges 1, 2, 3, but judge 4 picked B.
To describe this outcome, we need: the winner, the dissenting
judge, and that judge's choice.
Talent Show
There are three competitors (A, B, C) at a talent show. Four
judges (1, 2, 3, 4) review the candidates independently. A
competitor wins if at least three out of four judges choose that
competitor as their favorite. In reality, the judges choose
randomly.
How many possible ways to vote are there? Each of the 4
judges has 3 choices, so there are
3 · 3 · 3 · 3 = 81 total ways to vote.
How many outcomes (ways) are there for a competitor to win
with approval from only 3 judges?
Example outcome:
A picked by judges 1, 2, 3, but judge 4 picked B.
To describe this outcome, we need: the winner, the dissenting
judge, and that judge's choice.
Talent Show
There are three competitors (A, B, C) at a talent show. Four
judges (1, 2, 3, 4) review the candidates independently. A
competitor wins if at least three out of four judges choose that
competitor as their favorite. In reality, the judges choose
randomly.
How many possible ways to vote are there? Each of the 4
judges has 3 choices, so there are
3 · 3 · 3 · 3 = 81 total ways to vote.
How many outcomes (ways) are there for a competitor to win
with approval from only 3 judges?
Example outcome:
A picked by judges 1, 2, 3, but judge 4 picked B.
To describe this outcome, we need: the winner, the dissenting
judge, and that judge's choice.
There are three competitors (A, B, C) at a talent show. Four
judges (1, 2, 3, 4) review the candidates independently. A
competitor wins if at least three out of four judges choose that
competitor as their favorite. In reality, the judges choose
randomly.
How many possible ways to vote are there? Each of the 4
judges has 3 choices, so there are
3 · 3 · 3 · 3 = 81 total ways to vote.
How many outcomes (ways) are there for a competitor to win
with approval from only 3 judges?
Example outcome:
A picked by judges 1, 2, 3, but judge 4 picked B.
To describe this outcome, we need: the winner, the dissenting
judge, and that judge's choice.
1. There are 3 choices for the winner (A-C).
competitor wins if at least three out of four judges choose that
competitor as their favorite. In reality, the judges choose
randomly.
How many possible ways to vote are there? Each of the 4
judges has 3 choices, so there are
3 · 3 · 3 · 3 = 81 total ways to vote.
How many outcomes (ways) are there for a competitor to win
with approval from only 3 judges?
Example outcome:
A picked by judges 1, 2, 3, but judge 4 picked B.
To describe this outcome, we need: the winner, the dissenting
judge, and that judge's choice.
1. There are 3 choices for the winner (A-C).
2. There are 4 choices for the dissenting judge (1-4).
How many possible ways to vote are there? Each of the 4
judges has 3 choices, so there are
3 · 3 · 3 · 3 = 81 total ways to vote.
How many outcomes (ways) are there for a competitor to win
with approval from only 3 judges?
Example outcome:
A picked by judges 1, 2, 3, but judge 4 picked B.
To describe this outcome, we need: the winner, the dissenting
judge, and that judge's choice.
1. There are 3 choices for the winner (A-C).
2. There are 4 choices for the dissenting judge (1-4).
3. This judge must have voted for one of the other 2
competitors.
How many outcomes (ways) are there for a competitor to win
with approval from only 3 judges?
Example outcome:
A picked by judges 1, 2, 3, but judge 4 picked B.
To describe this outcome, we need: the winner, the dissenting
judge, and that judge's choice.
1. There are 3 choices for the winner (A-C).
2. There are 4 choices for the dissenting judge (1-4).
3. This judge must have voted for one of the other 2
competitors.
Thus, there are
3 · 4 · 2 = 24
ways for a competitor to win with approval from only 3 judges.
Example outcome:
A picked by judges 1, 2, 3, but judge 4 picked B.
To describe this outcome, we need: the winner, the dissenting
judge, and that judge's choice.
1. There are 3 choices for the winner (A-C).
2. There are 4 choices for the dissenting judge (1-4).
3. This judge must have voted for one of the other 2
competitors.
Thus, there are
3 · 4 · 2 = 24
ways for a competitor to win with approval from only 3 judges.
What is the probability of this happening? Assuming each
outcome is equally likely, the probability is
24/81=8/27.
Talent Show
Here are the 24 ways for a competitor to win with 3 out of 4
votes, sorted by winner (A-C), dissenting judge (4-1), and
then by that judge's choice. Each outcome is written as a list
of 4 competitors representing the choice of the 4 judges:
1. A wins:
AAAB, AAAC , AABA, AACA,
ABAA, ACAA, BAAA, CAAA.
2. B wins:
BBBA, BBBC , BBAB, BBCB,
BABB, BCBB, ABBB, CBBB.
3. C wins:
CCCA, CCCB, CCAC , CCBC ,
CACC , CBCC , ACCC , BCCC .
@Home: Reading
Note: page numbers refer to printed version. Add 8 to get
page numbers in a PDF reader.
I Take a look at the tree diagrams on pages 76 and 77,
which illustrate the counting technique introduced at the
beginning of this lecture.
I Please read the Birthday Problem starting on page 77 for
an example of using the counting technique and an
example of using permutations (we will look at these
later) to nd the number of ways for r people to have no
birthdays in common, and the probability of that
happening.
Homework
I
I
I
Please read Sections 3.1-3.2 (you can skip the historical
remarks).
We will study permutations.
The second homework is due this Saturday.