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Elementary Probability Theory

- Chapter 5 of the textbook
- Pages 145-164

Introduction

- Statistical Decision Theory using the

probability of possible outcomes to choose

between several available options - Statistical Inference using samples to infer

the probabilities of the population

Definitions

- Statistical Experiment
- Measuring an elementary outcome that is not known

in advance - Elementary Outcome
- Each possible outcome of a statistical experiment
- If the experiment was to test gender in this

classroom the elementary outcomes would be male

and female - Sample Space
- The set of all possible elementary outcomes

Sample Space Examples

- Definition the set of all possible outcomes of

an experiment. - Examples of sample spaces
- Outcomes of the roll of a die 1, 2, 3, 4, 5, 6
- Outcomes of 2 coin flips HH, HT, TH, TT
- Outcomes of rolling 2 dice

Definitions

- Events
- Subsets of the sample space
- Each event contains 1 or more elementary outcomes
- Event Space
- All the elementary outcomes that constitute an

event - Complimentary Event
- All elementary outcomes not in the event space

Event

- An outcome or a set of outcomes
- Examples of events
- Roll of one die 2
- Roll of one die 2, 5
- Roll of two dice 2 and 4, 4 and 3
- Roll of two dice 1 and 2, 5 and 6
- Flip coin once H
- Flip coin twice HT

Example

- Assume I sampled a people on the bus and asked

their ages and got the following results - 19, 20, 20, 23, 27, 31, 37, 42, 56, 58
- How many elementary outcomes do I have?
- If I break the sample space into events by decade

(e.g., 20s) what are my events? - What is the event space of each event?
- What is the complimentary event of the 50s event?

Symbols

- P() The probability of something (usually an

outcome or an event) - Ei An elementary outcome, note the i which

ranges from 1 to n - (S) The sample space (you may also see (O))
- A, B, Events are typically assigned to

capital letters - Complimentary events are the event letter

with a bar - Ø Null (i.e., no solution)

Relationships Between Events

- Remember each experiment has 1 and only 1

elementary outcome, but an outcome can be in 1 or

more events - Intersection the event space that is shared

(i.e., the outcome is in both (or all) event

spaces) - Example overlapping portion of the Venn Diagram
- Union combination of event spaces, (i.e., the

outcome is in at least 1 event)

Venn Diagrams

Probability Postulates

- P(Ei) probability of an outcome is between 0 and

1 (0 impossible, 1 certain) - P(A) sum of probabilities of all

elementary outcomes in the event space - P(S) 1 certain
- P(Ø) 0 impossible

Rules Derived From Postulates

- The sum of all elementary outcomes is 1 (certain)
- The probability of an event is between 1 and

zero - If A and B are mutually exclusive P(A n B) Ø

Note the book incorrectly uses in this rule

Types of Probability

- Subjective an event probability with

accuracy/validity based on the experience of and

information available to an observer - Objective an event probability determined by the

frequency of elementary outcomes observed during

statistical experimentation

Calculating Probability

- When all outcomes are equally likely, the

probability of an event A - m the number of elementary outcomes in the

event space - n the number of elementary outcomes in the

sample space - In other words.
- P(A) Total number of ways to achieve the event

Total number all possible outcomes

Calculating Probability Example

- Experiment coin toss
- P(heads) 1 / 2 0.5
- The number of elements in the event space (m) 1

(i.e., heads) - The number of elements in the sample space (n)

2 (i.e., heads or tails) - Experiment roll a die.
- P(rolling a 6) 1 / 6 0.166667
- The number of elements in the event space (m) 1

(i.e., a 6) - The number of elements in the sample space (n)

6 (i.e., 1,2,3,4,5,or 6)

Complicating Factors

- What do we do when all outcomes are not equally

likely? - Answer the subset of the sample space that

comprises the event space must be specified the

sum of the elementary outcome probabilities in

the event space will yield the event probability - Conceptually this just means that we back up a

step and calculate the probability of the

outcomes and add them up for each event (think

frequency tables)

A Familiar Example

- Assume I sampled a people on the bus and asked

their ages and got the following results - 19, 20, 20, 23, 27, 31, 37, 42, 56, 58
- What is the probability of getting a result of

31? - P(answer of 31) 1 / 10 0.1
- What is the probability of getting a result in

the 30s? - P(30s) P(answer of 31) P(answer of 37) 0.1

0.1 0.2 - What is the probability of getting a result in

the 20s? - P(20s) P(answer of 20) P(answer of 23)

P(answer of 27) 0.2 0.1 0.1 0.4

Counting Rules

- These are some useful rules for determining the

elementary outcome counts (which are used to

determine probabilities) - These are useful for many applications beyond

just calculating probability - Symbol Confusion
- The book uses r in two different ways
- For the product rule each r is a group and each

group has n elements (i.e., r1 has n1 elements,

r2 has n2 elements) - For the combinations and permutations rules each

r is a subset of a larger group and r

indicates the size (i.e. the number of elements)

in the group being formed

Product Rule

- Used to calculate all possible combinations

available when selecting one member from each

available group - Number of possible combinations n1 n2 .
- Example Flipping a coin 3 times
- Each flip is a group and each flip has 2 possible

outcomes (n1 n2 n3 2) - The number of possible outcomes is 222 8
- Outcomes HHH, HHT, HTH, HTT, THH,

THT, TTH, TTT

Combinations Rule

- Used to select all the possible groups of size r

from the sample space - Since the sample space has n outcomes, r n
- Example
- 4 cards - A, K, Q, J
- How many combinations can you have if you pick 2

cards (r2) - AK, AQ, AJ, KQ, KJ, QJ

Permutations Rule

- Used to select all the possible groups of size r

from the sample space including the order of the

elements - Since the sample space has n outcomes, r n
- Example
- 4 cards - A, K, Q, J
- How many combinations can you have if you pick 2

cards (r2) - AK, AQ, AJ, KQ, KJ, QJ
- KA, QA, JA, QK, JK, JQ

Hypergeometric Rule

- The combination of the product rule and the

combination rule - Since the sample space has n outcomes, r n
- Example
- 2 sets of 4 cards (A, K, Q, J) and (1, 2, 3, 4)
- How many combinations can you have if you pick 2

cards (r2) from each set? - Answer 36

Probability Theorems

- Addition Theorem
- Rule of thumb Union uses addition

Examples

- Coin Flip
- P(heads) 0.5
- P(tails) 0.5
- P(heads n tails) 0
- Cards
- P(heart) 13/52
- P(king) 4/52
- P(heart n king) 1/52

Probability Theorems

- Complementation Theorem
- Recall that P(S) 1

Probability Theorems

- Conditional Probability
- Think of this as the probability of X given Y

where both X and Y have their own probability - Intuition should tell you that this will hinge on

the intersection of X and Y

Probability Theorems

- Statistically Independent Events the

probability of an event remains the same despite

the occurrence of another event - Example The probability of a coin flip being

heads is ½ regardless of what the last coin flip

was - Based on conditional probability

Probability Theorems

- Multiplication Theorem
- Rule of thumb Intersections use multiplication

Statistically Independent Examples

- 2 Coin Flips
- A and B are the probability of getting heads
- P(heads) 1/2
- P(heads n heads) P(AB) P(B) ¼
- P(heads heads on first flip) P(heads n heads)

/ P(B) (¼) / (½) 1/2 - Draw 2 cards
- P(heart) 13/52
- P(king) 4/52
- P(heart n king) 1/52
- P(heart king) P(heart n king) / P(king)

(1/52) / (4/52) 13/52 1/4 - P(king heart) P(heart n king) / P(heart)

(1/52) / (13/52) 4/52 1/13

Statistically Dependent Example

- Probability of drawing 2 hearts
- Drawing single cards from a complete deck would

equate to - P(A) P(heart) 13/52
- P(B) P(heart) 13/52
- P(AB) P(heartheart on last draw) 12/51
- Solution 1 imagine drawing 1 card and then the

second - (A n B) P(AB) P(B) 0.589
- Solution 2 imagine drawing both cards at once
- Remember P(event) m/n
- n number of all combinations (full sample

space) - m number of possible combinations of 2 hearts

(event space) - Both m and n are calculated using the

combinations rule - m n
- P(drawing 2 hearts) 78/1326 0.589

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