elementary probability theory chapter 5 of the textbook pages 145-164
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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 outcomesExamples 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}
ExampleAssume 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?
SymbolsP() – 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 (Ω))
A, B, … – Events are typically assigned to capital letters
– Complimentary events are the event letter with a bar
Ø – Null (i.e., no solution)
A
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
UNIONA B
INTERSECTIONA B
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
)( Ai iEP
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 ∩ B) = Ø
n
i
EP1
1 1)(
1)(0 AP
** 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
nmAP /)(
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 2*2*2 = 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 (r=2)– AK, AQ, AJ, KQ, KJ, QJ
)!(!
!
rnr
nC n
r
64
24
)!24(!2
!442
C
Permutations RuleUsed 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 (r=2)– AK, AQ, AJ, KQ, KJ, QJ– KA, QA, JA, QK, JK, JQ
)!(
!
rn
nPn
r
122
24
)!24(
!442
P
Hypergeometric RuleThe 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 (r=2) from each set?
– Answer = 36
11
11 * n
rnr CC
Probability Theorems
Addition Theorem
Rule of thumb: Union uses addition
)()()()( BAPBPAPBAP
UNIONA B
Examples
Coin Flip: – P(heads) = 0.5– P(tails) = 0.5 – P(heads ∩ tails) = 0
Cards– P(heart) = 13/52– P(king) = 4/52– P(heart ∩ king) = 1/52
)()()()( BAPBPAPBAP
Probability Theorems
Complementation Theorem
Recall that P(S) = 1
)(1)( APAP
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
)(
)()|(
BP
BAPBAP
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
)()|( APBAP
Probability Theorems
Multiplication Theorem
Rule of thumb: Intersections use multiplication
)(*)|()( BPBAPBAP
INTERSECTIONA B
Statistically Independent Examples
2 Coin Flips – A and B are the probability of getting heads– P(heads) = 1/2– P(heads ∩ heads) = P(A|B) * P(B) = ¼– P(heads | heads on first flip) = P(heads ∩ heads) / P(B) = (¼) / (½) = 1/2
Draw 2 cards– P(heart) = 13/52– P(king) = 4/52– P(heart ∩ king) = 1/52– P(heart | king) = P(heart ∩ king) / P(king) = (1/52) / (4/52) = 13/52 = 1/4– P(king | heart) = P(heart ∩ king) / P(heart) = (1/52) / (13/52) = 4/52 = 1/13
)(
)()|(
BP
BAPBAP
Statistically Dependent ExampleProbability 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(A|B) = P(heart|heart on last draw) = 12/51
– Solution 1: imagine drawing 1 card and then the second: – (A ∩ B) = P(A|B) * 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
78)!213(!2
!13132
C 1326
)!252(!2
!52522
C
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