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Elementary Statistics
Chapter 4:
Probability
4.1 Basic Concepts of
Probability
1
Chapter 4: Probability
4.1 Basic Concepts of Probability
4.2 Addition Rule and Multiplication Rule
4.3 Complements and Conditional Probability, and Bayes’ Theorem
4.4 Counting
4.5 Probabilities Through Simulations (available online)
2
Objectives:
• Determine sample spaces and find the probability of an event, using classical probability or empirical probability.
• Find the probability of compound events, using the addition rules.
• Find the probability of compound events, using the multiplication rules.
• Find the conditional probability of an event.
• Find the total number of outcomes in a sequence of events, using the fundamental counting rule.
• Find the number of ways that r objects can be selected from n objects, using the permutation rule.
• Find the number of ways that r objects can be selected from n objects without regard to order, using the combination rule.
• Find the probability of an event, using the counting rules.
Recall: 3.1 Measures of Center
Measure of Center (Central Tendency)
A measure of center is a value at the center or
middle of a data set.
1. Mean: 𝑥 =
𝑥
𝑛
, 𝜇 =
𝑥
𝑁
, 𝑥 =
𝑓∙𝑥 𝑚
𝑛
2. Median: The middle value of ranked data
3. Mode: The value(s) that occur(s) with the
greatest frequency.
4. Midrange: 𝑀𝑟 =
𝑀𝑖𝑛+𝑀𝑎𝑥
2
5. Weighted Mean: 𝑥 =
𝑤∙𝑥
𝑤
3
4
Recall: 3.2 Measures of Variation
The variance is the average of the squares of the
distance each value is from the mean.
The standard deviation is the square root of the
variance.
The standard deviation is a measure of how spread
out your data are and how much data values deviate
away from the mean.
Empirical Rule (Normal);
Applies to bell shaped distributions
Range Rule of
Thumb for
Standard
Deviation:
𝑠 ≈
𝑅𝑎𝑛𝑔𝑒
4
&
µ ± 2σ
𝑅 = 𝑀𝑎𝑥 − 𝑀𝑖𝑛
𝑠 ≈
𝑅
4
𝐶𝑉 =
𝑠
𝑥
(100%)
𝑥 ± 2𝑠
𝑥 =
𝑥
𝑛
𝜇 =
𝑥
𝑁
𝑠 =
(𝑥 − 𝑥)2
𝑛 − 1
𝜎 =
(𝑥 − 𝜇)2
𝑁
 
2
2
Population Variance:
X
N





 
2
Population Standard Deviation:
X
N





 
 
 
2
2
2
2
Sample Variance:
1
1
X X
X X
s
n
n
n n






 

 
 
 
2
2
2
Sample Standard Deviation
1
1
:
X X
X X
s
n
n
n n






 

TI Calculator:
How to enter data:
1. Stat
2. Edi
3. Highlight & Clear
4. Type in your data in L1, ..
TI Calculator:
Mean, SD, 5-number
summary
1. Stat
2. Calc
3. Select 1 for 1 variable
4. Type: L1 (second 1)
5. Scroll down for 5-
number summary
Key Concept: This section introduces measures of relative standing, which are
numbers showing the location of data values relative to the other values within
the same data set.
Measures of Relative Standing
(Position)
z-score
Percentile: P1, P2, …P99
Quartile: Q1, Q2, and Q3
Deciles separate the data set into 10
equal groups. D1=P10, D4=P40
Outlier: A value < 𝑄1 − 1.5 𝐼𝑄𝑅
A value > 𝑄3+1.5 𝐼𝑄𝑅
Boxplot: 5- Number summary
5
x x
z
s
x
z






 # of values below
100%
total # of values
X
Percentile  
100 100
k P
L n n   
𝐼𝑄𝑅 = 𝑄3 − 𝑄1
𝑆𝑒𝑚𝑖 𝐼𝑄𝑅 = (𝑄3−𝑄1)/2
𝑀𝑖𝑑 𝑄𝑢𝑎𝑟𝑡𝑖𝑙𝑒 𝑟𝑎𝑛𝑔𝑒: = (𝑄3+𝑄1)/2
10-90 𝑄𝑢𝑎𝑟𝑡𝑖𝑙𝑒 𝑟𝑎𝑛𝑔𝑒: = 𝑃90 − 𝑃10)
Recall: 3.3 Measures of Relative Standing and Boxplots
n: data size
K: percentile d (25th percentile, k = 25.)
L: locator gives the position of a value (For the 12th value in the
sorted list, L = 12.)
Pk kth percentile (P25 is the 25th percentile.)
Notation: P(A) = the probability of event A , 0 ≤ P(A) ≤ 1
The sum of the probabilities of all the outcomes in the sample space is 1: 𝑃𝑖 = 1
Impossible Set: If an event E cannot occur (i.e., the event contains no members in the sample space), its
probability is 0. P(A) = 0
Sure (Certain) Set: If an event E is certain, then the probability of E is 1. P(E) = 1
Complementary Events: 𝐴: consists of all outcomes that are not included in the outcomes of event A.
𝑃 𝐴 = 1 − 𝑃(𝐴)
The actual odds against event A: O 𝐴 =
𝑃 𝐴
𝑃(𝐴)
, (expressed as form of a:b or “a to b”); a and b are integers.
The actual odds in favor of event A: O 𝐴 =
𝑃 𝐴
𝑃 𝐴
(the reciprocal of the actual odds against the even) If the odds
against A are a:b, then the odds in favor of A are b:a.
Payoff odds against event A =
net profit
amount bet
net profit = (Payoff odds against event A)(amount bet)
Value: Give the exact fraction or decimal or round off final decimal results to 3 significant digits.
6
4.1 Basic Concepts of Probability
Basics of Probability
Probability can be defined as the chance of an event occurring. It can
be used to quantify what the “odds” are that a specific event will occur.
A probability experiment is a chance process that leads to well-
defined results called outcomes.
An outcome is the result of a single trial of a probability experiment.
An event consists any collection of results or outcomes of a procedure.
A simple event is an outcome or an event that cannot be further broken
down into simpler components.
A sample space is the set of all possible outcomes or simple events of
a probability experiment.
How to interpret probability values, which are expressed as values between 0
and 1. A small probability, such as 0.001, corresponds to an event that rarely
occurs.
Odds and their relation to probabilities. Odds are commonly used in situations
such as lotteries and gambling.
7
4.1 Basic Concepts of Probability
Possible values
of probabilities
and the more
familiar and
common
expressions of
likelihood
Example 1
8
4.1 Basic Concepts of Probability
Sample Spaces
Experiment Sample Space
Toss a coin Head, Tail
Roll a die 1, 2, 3, 4, 5, 6
Answer a true/false
question
True, False
Toss two coins HH, HT, TH, TT
Example 2 Simple Events and Sample Spaces
Use “b” to denote a baby boy and “g” to denote a baby girl.
9
Procedure /
Experiment
Example of Event
Sample Space: Complete List of
Simple Events
Single birth 1 girl (simple event) {b, g}
3 births 2 boys and 1 girl (bbg, bgb, and gbb are all simple
events resulting in 2 boys and 1 girl)
{bbb, bbg, bgb, bgg, gbb, gbg, ggb,
ggg}
Simple Events:
1 birth: The result of 1 girl is a simple event and so is the result of 1 boy.
3 births: The result of 2 girls followed by a boy (ggb) is a simple event.
3 births: The event of “2 girls and 1 boy” is not a simple event because it can
occur with these different simple events: ggb, gbg, bgg.
3 births: The sample space consists of the eight (23
) different simple events
listed in the above table.
Example 3: Find the sample space
10
a. for rolling one die.
b. for rolling two dice.
n(s) = 6 Outcomes:
1, 2, 3, 4, 5, 6
𝒏 𝒔 = 𝟔 𝟐
= 𝟑𝟔
𝒏 𝒔 = (# 𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔)# 𝒐𝒇 𝒔𝒕𝒂𝒈𝒆𝒔
3 Common Approaches to Finding the Probability of an Event
•Classical probability uses sample spaces to determine the numerical probability that an event will
happen and assumes that all outcomes in the sample space are equally likely to occur. (confirm that the
outcomes are equally likely)
•Empirical probability (Relative Frequency Approximation of Probability) Conduct (or observe) a
procedure and count the number of times that event A occurs. P(A) is then approximated as follows:
•Subjective probability: P(A), the probability of event A, is estimated by using knowledge of the
relevant circumstances. (A subjective probability can be estimated in the absence of historical data.) uses
a probability value based on an educated guess or estimate, employing opinions and inexact information.
Experiments that have neither equally likely outcomes nor the potential of being repeated are assigned by
subjective probability. Examples: weather forecasting, predicting outcomes of sporting events
•Simulations: Sometimes none of the preceding three approaches can be used. A simulation of a
procedure is a process that behaves in the same ways as the procedure itself so that similar results are
produced. Probabilities can sometimes be found by using a simulation. 11
 
 
 
# of desired outcomes
, ( )
Total # of possible outcomes
n E s
P E orP A
n S n
  
 
( ) frequency of desired class
( ) ,
(Procedure Repeated) Sum of all frequencies
n A f
P A OrP E
n n
  
Example: P (Even numbers in
roll of a die) = 3/6 = 1/2
Example: P (A randomly selected student
from a Statistics class is a female) =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑒𝑚𝑎𝑙𝑒 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠
𝑇𝑜𝑡𝑎𝑙
12
Example 4
When a single die is rolled, what is
the probability of getting a number
a. less than 7?
b. Larger than 2?
c. Even number?
d. Larger than 7?
6
( 7) 1
6
P number   
The event of getting
a number less than 7
is certain (Sure set).
4 2
( 2)
6 3
P number   
3 1
( )
6 2
P Even  
( 7) 0P number  
Classical probability: Use sample space
 
 
 
( )
n E s
P E orP A
n S n
 
The event of
getting a number
larger than 7 is
impossible.
Example 5: Find the sample space
13
a. for the gender of the children if a family has three children. Use B for boy and
G for girl.
b. Use a tree diagram (a graph of possible outcomes of a procedure) to find
the sample space for the gender of three children in a family.
{BBB BBG
BGB BGG
GBB GBG
GGB GGG}
𝒏 𝒔 = 𝟐 𝟑
= 𝟖
B
G
B
G
B
G
B
G
B
G
B
G
B
G
BBB
BBG
BGB
BGG
GBB
GBG
GGB
GGG
Classical probability:
If a family has three children,
find the probability that two
of the three children are girls.  
 
 
3
8
n E
P E
n S
 
 
 
 
( )
n E s
P E orP A
n S n
 
Example 6
In a recent year, there were about 3,000,000 skydiving jumps and 21 of
them resulted in deaths. Find the probability of dying when making a
skydiving jump.
14
The classical approach cannot be used because the two outcomes (dying,
surviving) are not equally likely.
( ) 21
( )
( ) 3,000,000
n Death
P Death
n Skydiving
 
0.000007
Empirical probability (Relative Frequency):  
f
P E
n

Example 7
Among one high school drivers, it was found that 400 texted while driving
during the previous semester, and 600 did not text while driving during
that same time period .Based on these results, if a high school driver is
randomly selected, find the probability that he or she texted while driving
during the same time period.
15
( ) 400
( )
( ) 1000
n Texters
P Texting
n Drivers
  0.4
Interpretation: There is a 0.4 probability that if a high school driver is randomly
selected, he or she texted while driving during the last semester.
Texted
Did not
text
400 600
Empirical probability (Relative Frequency):  
f
P E
n

Example 8
In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had
type B blood, and 2 had type AB blood. Set up a frequency distribution and
find the following probabilities.
a. A person has type O blood.
b. A person has type A or type B blood.
c. A person has neither type A nor type O blood.
d. A person does not have type AB blood.
16
Empirical probability (Relative Frequency):  
f
P E
n

Type Frequency
A 22
B 5
AB 2
O 21
Total 50
 
21
. O
50
f
a P
n
 
 
22 5 27
. A or B
50 50 50
b P   
 neither A nor O
. 5 2 7
50 50 50
P
c
  
   . not AB 1 ABd P P 
2 48 24
1
50 50 25
   
Complementary Events:
17
   1P A = P AExample 9
Find the complement of each event.
Event Complement of the Event
Rolling a die and getting a 4 Getting a 1, 2, 3, 5, or 6
Selecting a letter of the alphabet
and getting a vowel
Getting a consonant (assume y is a
consonant)
Selecting a month and getting a
month that begins with a J
Getting February, March, April,
May, August, September, October,
November, or December
Selecting a day of the week
and getting a weekday Getting Saturday or Sunday
Complementary Events:
The complement of an eventA, denoted by 𝐴, consists of all outcomes that are not included in the
outcomes of eventA.
18
   1P A = P AExample 10
In a recent year, there were 3,000,000 skydiving jumps and 21 of them resulted
in death. Find the probability of not dying when making a skydiving jump.
Example 11: In reality, more boys are born than girls. In one typical group,
there are 205 newborn babies, 105 of whom are boys. If one baby is randomly
selected from the group, what is the probability that the baby is not a boy?
Re : ( ) 0.000007call P Death 
0.999993( ) 1 ( ) 1 0.000007P Not Dying P Death    
    ( )P G 1P B = P B
100
0.488
205
 
19
4.1 Basic Concepts of Probability, Odds
The actual odds against event A occurring are the ratio O 𝐴 =
𝑃 𝐴
𝑃(𝐴)
, usually
expressed in the form of a:b (or “a to b”), where a and b are integers having no
common factors.
The actual odds in favor event A occurring are the reciprocal of the actual odds
against the event. If the odds against A are a:b, then the odds in favor of A are b:a.
The payoff odds against event A represent the ratio of the net profit (if you win) to the
amount bet.
Payoff odds against event A = (net profit):(amount bet)
Payoff odds against event A =
net profit
amount bet
net profit = (Payoff odds against event A)(amount bet)
Example 12
20
If you bet $5 on the number 13 in roulette, your probability of winning is 1/38, but
your payoff odds are given be casino as 35:1.
a. Find the actual odds against the outcome of 13.
b. How much net profit would you make if you win by betting $5 on 13?
c. If the casino was not operating for profit and the payoff odds were changed to match
the actual odds against 13, how much would you win if the outcome were 13?
O 𝐴 =
𝑃 𝐴
𝑃(𝐴)
O 𝐴 =
𝑃 𝐴
𝑃(𝐴)
=
1 − 1/38
1/38
=
37/38
1/38
=
37
1
b. 35:1 = (net profit):(amount bet)
net profit = Payoff odds against event A amount bet
= 35 $5 = $175
The winner collects: $175 + the original $5 bet = $180, for a net profit of $175.
c. If the casino were not operating for profit, the payoff odds = 37:1 (Net profit of
$37 for each $1 bet) : net profit = 37 $5 = $185
(The casino makes its profit by providing a profit of only $175 instead of the $185
that would be paid with a roulette game that is fair instead of favoring the casino.)
Law of Large Numbers & Identifying Significant Results with Probabilities
The Rare Event Rule for Inferential Statistics
If, under a given assumption, the probability of a particular observed event is very small and the observed event occurs significantly less than or
significantly greater than what we typically expect with that assumption, we conclude that the assumption is probably not correct.
Using Probabilities to Determine When Results Are Significantly High or Significantly Low
Significantly high number of successes: x successes among n trials is a significantly high number of successes if the probability of x or more
successes is unlikely with a probability of 0.05 or less. That is, x is a significantly high number of successes if P(x or more) ≤ 0.05*.
Significantly low number of successes: x successes among n trials is a significantly low number of successes if the probability of x or fewer
successes is unlikely with a probability of 0.05 or less. That is, x is a significantly low number of successes if P(x or fewer) ≤ 0.05*.
*The value 0.05 is not absolutely rigid.
21
Law of Large Numbers:
As a procedure is repeated again and again, the relative frequency probability of an
event tends to approach the actual probability, and it applies to behavior over a large
number of trials, and it does not apply to any one individual outcome.
Example: Observation: A coin is tossed 100 times and we get 98 heads in a row.
Assumption: The coin is fair. (a normal assumption when an ordinary coin is tossed)
If the assumption is true, we have observed a highly unlikely event.
According to The Rare Event Rule , therefore, what we just observed is evidence against the assumption.
On this basis, we conclude the assumption is wrong (the coin is loaded.)

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Elementary Statistics Chapter 4 Probability Concepts

  • 1. Elementary Statistics Chapter 4: Probability 4.1 Basic Concepts of Probability 1
  • 2. Chapter 4: Probability 4.1 Basic Concepts of Probability 4.2 Addition Rule and Multiplication Rule 4.3 Complements and Conditional Probability, and Bayes’ Theorem 4.4 Counting 4.5 Probabilities Through Simulations (available online) 2 Objectives: • Determine sample spaces and find the probability of an event, using classical probability or empirical probability. • Find the probability of compound events, using the addition rules. • Find the probability of compound events, using the multiplication rules. • Find the conditional probability of an event. • Find the total number of outcomes in a sequence of events, using the fundamental counting rule. • Find the number of ways that r objects can be selected from n objects, using the permutation rule. • Find the number of ways that r objects can be selected from n objects without regard to order, using the combination rule. • Find the probability of an event, using the counting rules.
  • 3. Recall: 3.1 Measures of Center Measure of Center (Central Tendency) A measure of center is a value at the center or middle of a data set. 1. Mean: 𝑥 = 𝑥 𝑛 , 𝜇 = 𝑥 𝑁 , 𝑥 = 𝑓∙𝑥 𝑚 𝑛 2. Median: The middle value of ranked data 3. Mode: The value(s) that occur(s) with the greatest frequency. 4. Midrange: 𝑀𝑟 = 𝑀𝑖𝑛+𝑀𝑎𝑥 2 5. Weighted Mean: 𝑥 = 𝑤∙𝑥 𝑤 3
  • 4. 4 Recall: 3.2 Measures of Variation The variance is the average of the squares of the distance each value is from the mean. The standard deviation is the square root of the variance. The standard deviation is a measure of how spread out your data are and how much data values deviate away from the mean. Empirical Rule (Normal); Applies to bell shaped distributions Range Rule of Thumb for Standard Deviation: 𝑠 ≈ 𝑅𝑎𝑛𝑔𝑒 4 & µ ± 2σ 𝑅 = 𝑀𝑎𝑥 − 𝑀𝑖𝑛 𝑠 ≈ 𝑅 4 𝐶𝑉 = 𝑠 𝑥 (100%) 𝑥 ± 2𝑠 𝑥 = 𝑥 𝑛 𝜇 = 𝑥 𝑁 𝑠 = (𝑥 − 𝑥)2 𝑛 − 1 𝜎 = (𝑥 − 𝜇)2 𝑁   2 2 Population Variance: X N        2 Population Standard Deviation: X N            2 2 2 2 Sample Variance: 1 1 X X X X s n n n n                2 2 2 Sample Standard Deviation 1 1 : X X X X s n n n n          TI Calculator: How to enter data: 1. Stat 2. Edi 3. Highlight & Clear 4. Type in your data in L1, .. TI Calculator: Mean, SD, 5-number summary 1. Stat 2. Calc 3. Select 1 for 1 variable 4. Type: L1 (second 1) 5. Scroll down for 5- number summary
  • 5. Key Concept: This section introduces measures of relative standing, which are numbers showing the location of data values relative to the other values within the same data set. Measures of Relative Standing (Position) z-score Percentile: P1, P2, …P99 Quartile: Q1, Q2, and Q3 Deciles separate the data set into 10 equal groups. D1=P10, D4=P40 Outlier: A value < 𝑄1 − 1.5 𝐼𝑄𝑅 A value > 𝑄3+1.5 𝐼𝑄𝑅 Boxplot: 5- Number summary 5 x x z s x z        # of values below 100% total # of values X Percentile   100 100 k P L n n    𝐼𝑄𝑅 = 𝑄3 − 𝑄1 𝑆𝑒𝑚𝑖 𝐼𝑄𝑅 = (𝑄3−𝑄1)/2 𝑀𝑖𝑑 𝑄𝑢𝑎𝑟𝑡𝑖𝑙𝑒 𝑟𝑎𝑛𝑔𝑒: = (𝑄3+𝑄1)/2 10-90 𝑄𝑢𝑎𝑟𝑡𝑖𝑙𝑒 𝑟𝑎𝑛𝑔𝑒: = 𝑃90 − 𝑃10) Recall: 3.3 Measures of Relative Standing and Boxplots n: data size K: percentile d (25th percentile, k = 25.) L: locator gives the position of a value (For the 12th value in the sorted list, L = 12.) Pk kth percentile (P25 is the 25th percentile.)
  • 6. Notation: P(A) = the probability of event A , 0 ≤ P(A) ≤ 1 The sum of the probabilities of all the outcomes in the sample space is 1: 𝑃𝑖 = 1 Impossible Set: If an event E cannot occur (i.e., the event contains no members in the sample space), its probability is 0. P(A) = 0 Sure (Certain) Set: If an event E is certain, then the probability of E is 1. P(E) = 1 Complementary Events: 𝐴: consists of all outcomes that are not included in the outcomes of event A. 𝑃 𝐴 = 1 − 𝑃(𝐴) The actual odds against event A: O 𝐴 = 𝑃 𝐴 𝑃(𝐴) , (expressed as form of a:b or “a to b”); a and b are integers. The actual odds in favor of event A: O 𝐴 = 𝑃 𝐴 𝑃 𝐴 (the reciprocal of the actual odds against the even) If the odds against A are a:b, then the odds in favor of A are b:a. Payoff odds against event A = net profit amount bet net profit = (Payoff odds against event A)(amount bet) Value: Give the exact fraction or decimal or round off final decimal results to 3 significant digits. 6 4.1 Basic Concepts of Probability
  • 7. Basics of Probability Probability can be defined as the chance of an event occurring. It can be used to quantify what the “odds” are that a specific event will occur. A probability experiment is a chance process that leads to well- defined results called outcomes. An outcome is the result of a single trial of a probability experiment. An event consists any collection of results or outcomes of a procedure. A simple event is an outcome or an event that cannot be further broken down into simpler components. A sample space is the set of all possible outcomes or simple events of a probability experiment. How to interpret probability values, which are expressed as values between 0 and 1. A small probability, such as 0.001, corresponds to an event that rarely occurs. Odds and their relation to probabilities. Odds are commonly used in situations such as lotteries and gambling. 7 4.1 Basic Concepts of Probability Possible values of probabilities and the more familiar and common expressions of likelihood
  • 8. Example 1 8 4.1 Basic Concepts of Probability Sample Spaces Experiment Sample Space Toss a coin Head, Tail Roll a die 1, 2, 3, 4, 5, 6 Answer a true/false question True, False Toss two coins HH, HT, TH, TT
  • 9. Example 2 Simple Events and Sample Spaces Use “b” to denote a baby boy and “g” to denote a baby girl. 9 Procedure / Experiment Example of Event Sample Space: Complete List of Simple Events Single birth 1 girl (simple event) {b, g} 3 births 2 boys and 1 girl (bbg, bgb, and gbb are all simple events resulting in 2 boys and 1 girl) {bbb, bbg, bgb, bgg, gbb, gbg, ggb, ggg} Simple Events: 1 birth: The result of 1 girl is a simple event and so is the result of 1 boy. 3 births: The result of 2 girls followed by a boy (ggb) is a simple event. 3 births: The event of “2 girls and 1 boy” is not a simple event because it can occur with these different simple events: ggb, gbg, bgg. 3 births: The sample space consists of the eight (23 ) different simple events listed in the above table.
  • 10. Example 3: Find the sample space 10 a. for rolling one die. b. for rolling two dice. n(s) = 6 Outcomes: 1, 2, 3, 4, 5, 6 𝒏 𝒔 = 𝟔 𝟐 = 𝟑𝟔 𝒏 𝒔 = (# 𝒐𝒇 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔)# 𝒐𝒇 𝒔𝒕𝒂𝒈𝒆𝒔
  • 11. 3 Common Approaches to Finding the Probability of an Event •Classical probability uses sample spaces to determine the numerical probability that an event will happen and assumes that all outcomes in the sample space are equally likely to occur. (confirm that the outcomes are equally likely) •Empirical probability (Relative Frequency Approximation of Probability) Conduct (or observe) a procedure and count the number of times that event A occurs. P(A) is then approximated as follows: •Subjective probability: P(A), the probability of event A, is estimated by using knowledge of the relevant circumstances. (A subjective probability can be estimated in the absence of historical data.) uses a probability value based on an educated guess or estimate, employing opinions and inexact information. Experiments that have neither equally likely outcomes nor the potential of being repeated are assigned by subjective probability. Examples: weather forecasting, predicting outcomes of sporting events •Simulations: Sometimes none of the preceding three approaches can be used. A simulation of a procedure is a process that behaves in the same ways as the procedure itself so that similar results are produced. Probabilities can sometimes be found by using a simulation. 11       # of desired outcomes , ( ) Total # of possible outcomes n E s P E orP A n S n      ( ) frequency of desired class ( ) , (Procedure Repeated) Sum of all frequencies n A f P A OrP E n n    Example: P (Even numbers in roll of a die) = 3/6 = 1/2 Example: P (A randomly selected student from a Statistics class is a female) = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑓𝑒𝑚𝑎𝑙𝑒 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 𝑇𝑜𝑡𝑎𝑙
  • 12. 12 Example 4 When a single die is rolled, what is the probability of getting a number a. less than 7? b. Larger than 2? c. Even number? d. Larger than 7? 6 ( 7) 1 6 P number    The event of getting a number less than 7 is certain (Sure set). 4 2 ( 2) 6 3 P number    3 1 ( ) 6 2 P Even   ( 7) 0P number   Classical probability: Use sample space       ( ) n E s P E orP A n S n   The event of getting a number larger than 7 is impossible.
  • 13. Example 5: Find the sample space 13 a. for the gender of the children if a family has three children. Use B for boy and G for girl. b. Use a tree diagram (a graph of possible outcomes of a procedure) to find the sample space for the gender of three children in a family. {BBB BBG BGB BGG GBB GBG GGB GGG} 𝒏 𝒔 = 𝟐 𝟑 = 𝟖 B G B G B G B G B G B G B G BBB BBG BGB BGG GBB GBG GGB GGG Classical probability: If a family has three children, find the probability that two of the three children are girls.       3 8 n E P E n S         ( ) n E s P E orP A n S n  
  • 14. Example 6 In a recent year, there were about 3,000,000 skydiving jumps and 21 of them resulted in deaths. Find the probability of dying when making a skydiving jump. 14 The classical approach cannot be used because the two outcomes (dying, surviving) are not equally likely. ( ) 21 ( ) ( ) 3,000,000 n Death P Death n Skydiving   0.000007 Empirical probability (Relative Frequency):   f P E n 
  • 15. Example 7 Among one high school drivers, it was found that 400 texted while driving during the previous semester, and 600 did not text while driving during that same time period .Based on these results, if a high school driver is randomly selected, find the probability that he or she texted while driving during the same time period. 15 ( ) 400 ( ) ( ) 1000 n Texters P Texting n Drivers   0.4 Interpretation: There is a 0.4 probability that if a high school driver is randomly selected, he or she texted while driving during the last semester. Texted Did not text 400 600 Empirical probability (Relative Frequency):   f P E n 
  • 16. Example 8 In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution and find the following probabilities. a. A person has type O blood. b. A person has type A or type B blood. c. A person has neither type A nor type O blood. d. A person does not have type AB blood. 16 Empirical probability (Relative Frequency):   f P E n  Type Frequency A 22 B 5 AB 2 O 21 Total 50   21 . O 50 f a P n     22 5 27 . A or B 50 50 50 b P     neither A nor O . 5 2 7 50 50 50 P c       . not AB 1 ABd P P  2 48 24 1 50 50 25    
  • 17. Complementary Events: 17    1P A = P AExample 9 Find the complement of each event. Event Complement of the Event Rolling a die and getting a 4 Getting a 1, 2, 3, 5, or 6 Selecting a letter of the alphabet and getting a vowel Getting a consonant (assume y is a consonant) Selecting a month and getting a month that begins with a J Getting February, March, April, May, August, September, October, November, or December Selecting a day of the week and getting a weekday Getting Saturday or Sunday
  • 18. Complementary Events: The complement of an eventA, denoted by 𝐴, consists of all outcomes that are not included in the outcomes of eventA. 18    1P A = P AExample 10 In a recent year, there were 3,000,000 skydiving jumps and 21 of them resulted in death. Find the probability of not dying when making a skydiving jump. Example 11: In reality, more boys are born than girls. In one typical group, there are 205 newborn babies, 105 of whom are boys. If one baby is randomly selected from the group, what is the probability that the baby is not a boy? Re : ( ) 0.000007call P Death  0.999993( ) 1 ( ) 1 0.000007P Not Dying P Death         ( )P G 1P B = P B 100 0.488 205  
  • 19. 19 4.1 Basic Concepts of Probability, Odds The actual odds against event A occurring are the ratio O 𝐴 = 𝑃 𝐴 𝑃(𝐴) , usually expressed in the form of a:b (or “a to b”), where a and b are integers having no common factors. The actual odds in favor event A occurring are the reciprocal of the actual odds against the event. If the odds against A are a:b, then the odds in favor of A are b:a. The payoff odds against event A represent the ratio of the net profit (if you win) to the amount bet. Payoff odds against event A = (net profit):(amount bet) Payoff odds against event A = net profit amount bet net profit = (Payoff odds against event A)(amount bet)
  • 20. Example 12 20 If you bet $5 on the number 13 in roulette, your probability of winning is 1/38, but your payoff odds are given be casino as 35:1. a. Find the actual odds against the outcome of 13. b. How much net profit would you make if you win by betting $5 on 13? c. If the casino was not operating for profit and the payoff odds were changed to match the actual odds against 13, how much would you win if the outcome were 13? O 𝐴 = 𝑃 𝐴 𝑃(𝐴) O 𝐴 = 𝑃 𝐴 𝑃(𝐴) = 1 − 1/38 1/38 = 37/38 1/38 = 37 1 b. 35:1 = (net profit):(amount bet) net profit = Payoff odds against event A amount bet = 35 $5 = $175 The winner collects: $175 + the original $5 bet = $180, for a net profit of $175. c. If the casino were not operating for profit, the payoff odds = 37:1 (Net profit of $37 for each $1 bet) : net profit = 37 $5 = $185 (The casino makes its profit by providing a profit of only $175 instead of the $185 that would be paid with a roulette game that is fair instead of favoring the casino.)
  • 21. Law of Large Numbers & Identifying Significant Results with Probabilities The Rare Event Rule for Inferential Statistics If, under a given assumption, the probability of a particular observed event is very small and the observed event occurs significantly less than or significantly greater than what we typically expect with that assumption, we conclude that the assumption is probably not correct. Using Probabilities to Determine When Results Are Significantly High or Significantly Low Significantly high number of successes: x successes among n trials is a significantly high number of successes if the probability of x or more successes is unlikely with a probability of 0.05 or less. That is, x is a significantly high number of successes if P(x or more) ≤ 0.05*. Significantly low number of successes: x successes among n trials is a significantly low number of successes if the probability of x or fewer successes is unlikely with a probability of 0.05 or less. That is, x is a significantly low number of successes if P(x or fewer) ≤ 0.05*. *The value 0.05 is not absolutely rigid. 21 Law of Large Numbers: As a procedure is repeated again and again, the relative frequency probability of an event tends to approach the actual probability, and it applies to behavior over a large number of trials, and it does not apply to any one individual outcome. Example: Observation: A coin is tossed 100 times and we get 98 heads in a row. Assumption: The coin is fair. (a normal assumption when an ordinary coin is tossed) If the assumption is true, we have observed a highly unlikely event. According to The Rare Event Rule , therefore, what we just observed is evidence against the assumption. On this basis, we conclude the assumption is wrong (the coin is loaded.)