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Introduction To
Probability
What is Probability?
 Probability is the measure of how likely an event is.
 The probability of an event is a number between 0 and 1
that expresses how likely it is to occur.
 “The ratio of the number of favourable cases to the
number of all the cases“ or
P(E) = Number of outcomes favourable to E_____
Number of all possible outcomes of the experiment
Short History of
Probability
French Society in the 1650’s
 Gambling was popular
and fashionable.
 Not restricted by law.
 As the games became
more complicated and
the stakes became larger
there was a need for mathematical methods for
computing chances.
James Bernoulli
Proved that the
frequency method
and classical method
are consistent with
one another in his ars
Conjectandi in 1713.
Abraham De Moivre
Provided many tools
to make the classical
method more useful,
including the
multiplication rule, in
his book the doctrine
of chance in 1718.
Pierre- Simon Laplace
Presented a mathematical
theory of probability with
an emphasis on scientific
applications in his 1812
book theorie analytique
des Probability.
Term used in
Probability
 Event – an event is one occurence of the activity
whose probability is being calculated.
 Outcome - an Outcome is one possible result of an
event.
 Success - an outcome that we want to measure.
 Failure - an outcome that we don’t want to measure.
 Sample Space - The collection of all possible
outcomes is known as Sample Space.
 Equally Likely Outcomes - In this 50-50% chance are
there of an outcome.
Three Types of
Probability
Three Types of Probability
 Theoretical Probability is based on a mathematical
model of probability.
 Empirical Probability ( or experimental probability) is
based on how often an event has occurred in the past.
 Subjective Probability is based on a person’s belief that
an event will occur.
Theoretical Probability
 A Random Experiment is an act whose outcome is
uncertain.
 A sample space S is the set of all possible outcomes
of the random experiment.
 Each outcome is assigned a probability between 0
and 1. the probabilities of all the outcome add up to
1.
Theoretical Probability
 An event is a subset of the sample space (i.e. it is a set
of outcomes).
 The probability of an event is the sum of the events of
the outcomes that comprise the event.
 “Should happen”
Theoretical Probability
 We write P(E) to denote the probability of the event E.
𝑃 𝐸 =
𝑛(𝐸)
𝑛(𝑆)
=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 𝑤ℎ𝑎𝑡 𝑦𝑜𝑢 𝑤𝑎𝑛𝑡
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠
Theoretical Probability
Example:
I have 3 blue candies, 5 red candies and 2
yellow candy. Find the probability of the blue
candies.
Theoretical Probability
Solution:
 P(blue) =
6
10
𝑁𝑜.𝑜𝑓 𝑏𝑙𝑢𝑒
𝑁𝑜.𝑜𝑓 𝑎𝑙𝑙 𝑐𝑎𝑛𝑑𝑖𝑒𝑠 (𝑃𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠)
=
3
5
Empirical Probability
 Empirical ( or experimental ) probability is of an event is an
"estimate" that the event will happen based on how often
the event occurs after collecting data or running an
experiment (in a large number of trials).
 “what did happen?”
Empirical Probability
Empirical Probability =
# 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠
# 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠
Empirical probability only makes sense if the event is
repeatable.
Empirical Probability
Example:
Solution:
E.P. =
4
5
=
1
2
Subjective Probability
 Subjective Probability measures a person’s belief that an
event will occur .
 These probabilities vary from person to person.
 They also change as one learns new information.
Subjective Probability
 No mathematical calculations or proof behind this types
of probability but it could be illustrated the following way:
𝑃 𝑥 = 𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑏𝑒𝑙𝑖𝑒𝑓 𝑡ℎ𝑎𝑡 𝑥 𝑖𝑠 𝑡𝑟𝑢𝑒
Subjective Probability
 Example:
In a car parking area there are totally
32 cars available. In those, 3 are
Innova cars, 8 are Swift cars, 15 are
Accent cars and remaining are Ford
cars. Compute the probability for
choosing i) Swift cars ii) Ford cars?
Subjective Probability
 Solution:
Number of Innova cars = 3
Number of Swift cars = 8
Number of Accent cars = 15
Number of Ford cars = 32 – (3+8+15) = 6
Subjective Probability
Solution:
 Let A be the event of selecting Swift cars.
So, n(A) = 8
= `8/32`
= `1/4`
Let B be the event of selecting Ford cars.
So, n(B) = 6
= `6/32`
= `3/16`
GENEREL PROBABILITY
RULES
RULE 1
The probability of an impossible event is 0
while the probability of a certain event is
1. Therefore, the range of all possibilities is
𝟎 ≤ 𝑷 𝑨 ≤ 𝟏.
Example:
Six-Sided Die
It is impossible to roll an eight on a six-
sided die. Thus,
𝑷(𝐫𝐨𝐥𝐥𝐢𝐧𝐠 𝟖) = 𝟎.
It is certain that you will roll a number
between 1 and 6 if you roll a six-sided
die. Thus,
P(any number between 1 and 6) = 1
RULE 2
The sum of all of the
probabilities for possible
events is equal to 1.
Example:
Cards
In a standard 52-card deck there are 26
black cards and 26 red cards. All cards are
either black or red.
P(red)+P(black)=
26
52
+
26
52
= 1
RULE 3
The complement of any outcome is
equal to one minus the outcome. In
other words:
𝑃(𝐴𝑐
) = 1 − 𝑃(𝐴)
It is also true then that
𝑃(𝐴) = 1 − 𝑃(𝐴𝑐
)
Example:
Rain
 According to the weather report, there is a 30%
chance of rain today: 𝑃(𝑅𝑎𝑖𝑛) = .30
Raining and not raining are complements.
Therefore
𝑷(𝑵𝒐𝒕 𝒓𝒂𝒊𝒏) = 𝑷(𝑹𝒂𝒊𝒏𝒄) = 𝟏 − 𝑷(𝑹𝒂𝒊𝒏) = 𝟏−. 𝟑 =. 𝟕𝟎
 Therefore, there is a 70% chance that it will not
rain today.
RULE 4
Addition Rule
 Determining the probability that either one or
both events occur depends on whether the
events are mutually exclusive or not. This is also
known as a union and is represented by ∪.
 𝑷(𝑨 ∪ 𝑩) is read as “probability of A or B.” Note
that this also includes the possibility of both A
and B occurring at the same time.
If two events, A and B, are mutually exclusive, then
the probabilities of the two events can be added.
In other words,
𝑷(𝑨 ∪ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩).
This rule extends beyond two events if all are
mutually exclusive. For example, if A, B, and C are
all mutually exclusive events, then
𝑷(𝑨 ∪ 𝑩 ∪ 𝑪) = 𝑷(𝑨) + 𝑷(𝑩) + 𝑷(𝑪)
Example:
King or Ace
In a standard 52-deck of cards, what is the
probability of randomly selecting a king or ace?
𝑃 𝑘𝑖𝑛𝑔 =
1
13
= .077 𝑃(𝑎𝑐𝑒) =
1
13
= .077
These events are mutually exclusive because a
card cannot be both a king and an ace.
𝑷(𝒌𝒊𝒏𝒈 ∪ 𝒂𝒄𝒆) =
1
13
+
1
13
=. 𝟎𝟕𝟕+. 𝟎𝟕𝟕 =. 𝟏𝟓𝟒
Example:
Political Affiliation
In a given state, 30% are Democrats, 26%
Republicans, and 43% Independent.
If you randomly select a resident from this state,
what is the probability that you have selected an
Independent or Democrat? These events are
mutually exclusive because each resident may
only identify as one category.
𝑃(𝐼𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 ∪ 𝐷𝑒𝑚𝑜𝑐𝑟𝑎𝑡)
= 𝑃(𝐼𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡) + 𝑃(𝐷𝑒𝑚𝑜𝑐𝑟𝑎𝑡) = .43 + .30 = .73
If two events, A and B, are NOT mutually exclusive,
then the probability of A or B is equal to the
probability of A and the probability of B minus the
overlap of the two. In other words,
𝑷(𝑨 ∪ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩) − 𝑷(𝑨 ∩ 𝑩)
The probability of A and B must be subtracted
when there is overlap otherwise this area is
counted twice.
𝑷(𝑨 ∪ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩) − 𝑷(𝑨 ∩ 𝑩)
Example:
Ice Cream
A large-scale survey finds that:
90% of college students enjoy eating chocolate
ice cream,
70% of college students enjoy eating mint
chocolate chip ice cream,
65% of college student enjoy eating both
chocolate and mint chocolate chip ice cream.
What proportion of college students enjoy eating
chocolate or mint chocolate chip ice cream?
These events are NOT mutually exclusive because
it was possible for a participant to enjoy both.
𝑃 𝐶ℎ𝑜𝑐𝑜𝑙𝑎𝑡𝑒 ∪ 𝑀𝑖𝑛𝑡𝐶𝐶
= 𝑃(𝐶ℎ𝑜𝑐𝑜𝑙𝑎𝑡𝑒) + 𝑃(𝑀𝑖𝑛𝑡𝐶𝐶) − 𝑃(𝐶ℎ𝑜𝑐𝑜𝑙𝑎𝑡𝑒 ∩ 𝑀𝑖𝑛𝑡𝐶𝐶
= .90 + .70 − .65 = .95
The probability that a randomly selected college
student will enjoy chocolate or mint chocolate
chip ice cream is .95
Example:
Ace or Diamond
What is the probability that a randomly selected card from
a standard 52-card deck will be an ace or a diamond? Suit
and face values are NOT mutually exclusive because it is
possible for a card to be an ace and a diamond (i.e., the
ace of diamonds).
𝑃 𝑎𝑐𝑒 = 1/13 = .077 𝑃(𝑑𝑖𝑎𝑚𝑜𝑛𝑑) = 1/4
= .25
𝑃 𝑎𝑐𝑒 ∩ 𝑑𝑖𝑎𝑚𝑜𝑛𝑑 = 1/52 = .019
𝑃(𝑎𝑐𝑒 ∪ 𝑑𝑖𝑎𝑚𝑜𝑛𝑑)
= 𝑃(𝑎𝑐𝑒) + 𝑃(𝑑𝑖𝑎𝑚𝑜𝑛𝑑) − 𝑃(𝑎𝑐𝑒 ∩ 𝑑𝑖𝑎𝑚𝑜𝑛𝑑)
= .077 + .25 − .019 = .308
Rule 5
Multiplication Rule
Determining the probability that both events
occur depends on whether the events are
independent or not. This is known as an
intersection and is represented by ∩.
𝑃(𝐴 ∩ 𝐵) is read as “probability of A and B.”
If two events, A and B, are independent,
then the probability of A and B is equal to
the product of the two. In other words
𝑷 𝑨 ∩ 𝑩 = 𝑷 𝑨 × 𝑷 𝑩 .
This rule extends beyond two events. For
example, if A, B, and C are all independent
of one another, then
𝑷(𝑨 ∩ 𝑩 ∩ 𝑪) = 𝑷(𝑨) × 𝑷(𝑩) × 𝑷(𝑪)
Example:
Ace of Diamonds
What is the probability that you will randomly pull
the ace of diamonds from a standard 52-card
deck? The probability of pulling an ace and the
probability of pulling a diamond are independent
events.
𝑃(𝑎𝑐𝑒) = 4/52 = .077
𝑃(𝑑𝑖𝑎𝑚𝑜𝑛𝑑𝑠) = 13/52 = .25
𝑃(𝑎𝑐𝑒 ∩ 𝑑𝑖𝑎𝑚𝑜𝑛𝑑) = 𝑃(𝑎𝑐𝑒) × 𝑃(𝑑𝑖𝑎𝑚𝑜𝑛𝑑)
= .077 × .25 = .019
Example:
Dog Ownership
In a given town, 52% of the population are men and 48% are women.
We also know that 40% of residents own a dog. If you randomly select
one individual from the population, what is the probability that you will
select a man who owns a dog?
𝑃(𝑚𝑎𝑛 ∩ 𝑑𝑜𝑔𝑜𝑤𝑛𝑒𝑟) = 𝑃(𝑚𝑎𝑛) × 𝑃(𝑑𝑜𝑔𝑜𝑤𝑛𝑒𝑟) = .52 × .40 = .208
If you randomly select one individual from the population, what is the
probability that you will select a woman who does not own a dog?
𝑃(𝑛𝑜𝑑𝑜𝑔) = 𝑃(𝑑𝑜𝑔𝑐) = 1 − 𝑃(𝑑𝑜𝑔) = 1 − .40 = .6
𝑃(𝑤𝑜𝑚𝑎𝑛 ∩ 𝑛𝑜𝑑𝑜𝑔) = 𝑃(𝑤𝑜𝑚𝑎𝑛) × 𝑃(𝑛𝑜𝑑𝑜𝑔) = .48 × .60 = .288
Example:
Flipping a Coin
When flipping a fair coin (i.e., 50% heads and 50% tails)
three times, what is the probability that you will flip heads
all three times?
𝑃(𝐻) = 1/2 = .5
𝑃(𝐻𝐻𝐻) = 𝑃(𝐻) × 𝑃(𝐻) × 𝑃(𝐻) = .5 × .5 × .5 = .125
The probability of flipping heads all three times is .125
What is the probability of flipping heads, tails, tails, in that
order?
𝑃(𝐻) = 1/2 = .5 𝑎𝑛𝑑 𝑃(𝑇) = 1/2 = .5
𝑃(𝐻𝑇𝑇) = 𝑃(𝐻) × 𝑃(𝑇) × 𝑃(𝑇) = .5 × .5 × .5 = .125
If two events, A and B, are NOT
independent, then the probability of A and
B is equal to the probability of A times the
probability of B given A. In other words
𝑷(𝑨 ∩ 𝑩) = 𝑷(𝑨) × 𝑷(𝑩 ∣ 𝑨).
Similarly, the A and B may be switched for,
𝑷(𝑨 ∩ 𝑩) = 𝑷(𝑩) × 𝑷(𝑨 ∣ 𝑩).
𝑷(𝑩 ∣ 𝑨) is a conditional probability which
we will see in Rule 6, it is read “Probability of
B given A.”
Example:
Gender and Sports Participation
A study by Sabo & Veliz (2008) surveyed 2,132
American children about their participation in
organized and team sports. In the sample, 50.7%
were boys and 49.3% were girls. Of the boys, 73%
currently participate in sports. Of the girls, 63%
currently participate in sports. Gender and sports
participation are NOT independent.
What is the probability that you would
select a girl who currently participates in
sports?
𝑃(𝑔𝑖𝑟𝑙 ∩ 𝑠𝑝𝑜𝑟𝑡𝑠) = 𝑃(𝑔𝑖𝑟𝑙) × 𝑃(𝑠𝑝𝑜𝑟𝑡𝑠 ∣ 𝑔𝑖𝑟𝑙)
= .493 × .63 = .311
What is the probability that a randomly
selected individual from this sample would
be a boy who does not currently
participate in sports?
𝑃(𝑏𝑜𝑦 ∩ 𝑛𝑜𝑠𝑝𝑜𝑟𝑡𝑠) = 𝑃(𝑏𝑜𝑦) × 𝑃(𝑛𝑜𝑠𝑝𝑜𝑟𝑡𝑠
∣ 𝑏𝑜𝑦) = .507 × (1 − .73) = .507 × .27 = .137
Rule 6
Conditional Probability
𝑷 𝑨 𝑩
𝑖𝑠 𝑟𝑒𝑎𝑑 𝑎𝑠 “𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝐴 𝑔𝑖𝑣𝑒𝑛 𝐵. ”
𝑷(𝑨 ∣ 𝑩) =
𝑷(𝑨 ∩ 𝑩)
𝑷(𝑩)
𝑷(𝑩 ∣ 𝑨) =
𝑷(𝑨 ∩ 𝑩)
𝑷(𝑨)
Example:
Gender and Grades
In a large class, the probability of randomly
selecting a woman is .60. The probability of
randomly selecting a student who is a woman and
who earned an A is .20. If you randomly select a
student who is a woman, what is the probability
that she earned an A?
A = earning a grade of A.
B = being a woman.
We want to know the probability of earning an A
given that the student is a female.
In other words, 𝑷(𝑨 ∣ 𝑩)
We are given that
𝑃(𝐴 ∩ 𝐵) = .20 𝑎𝑛𝑑 𝑃(𝐵) = .60
𝑃(𝐴 ∣ 𝐵) =
𝑷(𝑨 ∩ 𝑩)
𝑷(𝑩)
=
. 𝟐𝟎
. 𝟔𝟎
= .333
Given that a randomly selected student is a
woman, there is a 33.3% chance that she earned
an A.
Example:
Clubs
In a standard 52-card deck. What is the probability
that a randomly selected card is a club given that
it is a black card?
𝑃 𝑐𝑙𝑢𝑏 = 1/4 = .25 𝑃(𝑏𝑙𝑎𝑐𝑘) = 2/4 = .50
𝑃(𝑐𝑙𝑢𝑏 ∩ 𝑏𝑙𝑎𝑐𝑘) = 13/52 = .25
𝑃(𝑐𝑙𝑢𝑏 ∣ 𝑏𝑙𝑎𝑐𝑘) =
𝑃(𝑐𝑙𝑢𝑏 ∩ 𝑏𝑙𝑎𝑐𝑘)
𝑃(𝑏𝑙𝑎𝑐𝑘)
=
.25
.50
= .50
Given that a randomly selected card is black,
there is a 50% chance that it's a club.

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group1-151014013653-lva1-app6891.pdf

  • 2. What is Probability?  Probability is the measure of how likely an event is.  The probability of an event is a number between 0 and 1 that expresses how likely it is to occur.  “The ratio of the number of favourable cases to the number of all the cases“ or P(E) = Number of outcomes favourable to E_____ Number of all possible outcomes of the experiment
  • 3.
  • 5. French Society in the 1650’s  Gambling was popular and fashionable.  Not restricted by law.  As the games became more complicated and the stakes became larger there was a need for mathematical methods for computing chances.
  • 6. James Bernoulli Proved that the frequency method and classical method are consistent with one another in his ars Conjectandi in 1713. Abraham De Moivre Provided many tools to make the classical method more useful, including the multiplication rule, in his book the doctrine of chance in 1718. Pierre- Simon Laplace Presented a mathematical theory of probability with an emphasis on scientific applications in his 1812 book theorie analytique des Probability.
  • 8.  Event – an event is one occurence of the activity whose probability is being calculated.  Outcome - an Outcome is one possible result of an event.  Success - an outcome that we want to measure.  Failure - an outcome that we don’t want to measure.  Sample Space - The collection of all possible outcomes is known as Sample Space.  Equally Likely Outcomes - In this 50-50% chance are there of an outcome.
  • 10. Three Types of Probability  Theoretical Probability is based on a mathematical model of probability.  Empirical Probability ( or experimental probability) is based on how often an event has occurred in the past.  Subjective Probability is based on a person’s belief that an event will occur.
  • 11. Theoretical Probability  A Random Experiment is an act whose outcome is uncertain.  A sample space S is the set of all possible outcomes of the random experiment.  Each outcome is assigned a probability between 0 and 1. the probabilities of all the outcome add up to 1.
  • 12. Theoretical Probability  An event is a subset of the sample space (i.e. it is a set of outcomes).  The probability of an event is the sum of the events of the outcomes that comprise the event.  “Should happen”
  • 13. Theoretical Probability  We write P(E) to denote the probability of the event E. 𝑃 𝐸 = 𝑛(𝐸) 𝑛(𝑆) = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑔𝑒𝑡 𝑤ℎ𝑎𝑡 𝑦𝑜𝑢 𝑤𝑎𝑛𝑡 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠
  • 14. Theoretical Probability Example: I have 3 blue candies, 5 red candies and 2 yellow candy. Find the probability of the blue candies.
  • 15. Theoretical Probability Solution:  P(blue) = 6 10 𝑁𝑜.𝑜𝑓 𝑏𝑙𝑢𝑒 𝑁𝑜.𝑜𝑓 𝑎𝑙𝑙 𝑐𝑎𝑛𝑑𝑖𝑒𝑠 (𝑃𝑜𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠) = 3 5
  • 16. Empirical Probability  Empirical ( or experimental ) probability is of an event is an "estimate" that the event will happen based on how often the event occurs after collecting data or running an experiment (in a large number of trials).  “what did happen?”
  • 17. Empirical Probability Empirical Probability = # 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠 # 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠 Empirical probability only makes sense if the event is repeatable.
  • 19. Subjective Probability  Subjective Probability measures a person’s belief that an event will occur .  These probabilities vary from person to person.  They also change as one learns new information.
  • 20. Subjective Probability  No mathematical calculations or proof behind this types of probability but it could be illustrated the following way: 𝑃 𝑥 = 𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑏𝑒𝑙𝑖𝑒𝑓 𝑡ℎ𝑎𝑡 𝑥 𝑖𝑠 𝑡𝑟𝑢𝑒
  • 21. Subjective Probability  Example: In a car parking area there are totally 32 cars available. In those, 3 are Innova cars, 8 are Swift cars, 15 are Accent cars and remaining are Ford cars. Compute the probability for choosing i) Swift cars ii) Ford cars?
  • 22. Subjective Probability  Solution: Number of Innova cars = 3 Number of Swift cars = 8 Number of Accent cars = 15 Number of Ford cars = 32 – (3+8+15) = 6
  • 23. Subjective Probability Solution:  Let A be the event of selecting Swift cars. So, n(A) = 8 = `8/32` = `1/4` Let B be the event of selecting Ford cars. So, n(B) = 6 = `6/32` = `3/16`
  • 25. RULE 1 The probability of an impossible event is 0 while the probability of a certain event is 1. Therefore, the range of all possibilities is 𝟎 ≤ 𝑷 𝑨 ≤ 𝟏.
  • 26. Example: Six-Sided Die It is impossible to roll an eight on a six- sided die. Thus, 𝑷(𝐫𝐨𝐥𝐥𝐢𝐧𝐠 𝟖) = 𝟎. It is certain that you will roll a number between 1 and 6 if you roll a six-sided die. Thus, P(any number between 1 and 6) = 1
  • 27. RULE 2 The sum of all of the probabilities for possible events is equal to 1.
  • 28. Example: Cards In a standard 52-card deck there are 26 black cards and 26 red cards. All cards are either black or red. P(red)+P(black)= 26 52 + 26 52 = 1
  • 29. RULE 3 The complement of any outcome is equal to one minus the outcome. In other words: 𝑃(𝐴𝑐 ) = 1 − 𝑃(𝐴) It is also true then that 𝑃(𝐴) = 1 − 𝑃(𝐴𝑐 )
  • 30. Example: Rain  According to the weather report, there is a 30% chance of rain today: 𝑃(𝑅𝑎𝑖𝑛) = .30 Raining and not raining are complements. Therefore 𝑷(𝑵𝒐𝒕 𝒓𝒂𝒊𝒏) = 𝑷(𝑹𝒂𝒊𝒏𝒄) = 𝟏 − 𝑷(𝑹𝒂𝒊𝒏) = 𝟏−. 𝟑 =. 𝟕𝟎  Therefore, there is a 70% chance that it will not rain today.
  • 31. RULE 4 Addition Rule  Determining the probability that either one or both events occur depends on whether the events are mutually exclusive or not. This is also known as a union and is represented by ∪.  𝑷(𝑨 ∪ 𝑩) is read as “probability of A or B.” Note that this also includes the possibility of both A and B occurring at the same time.
  • 32. If two events, A and B, are mutually exclusive, then the probabilities of the two events can be added. In other words, 𝑷(𝑨 ∪ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩). This rule extends beyond two events if all are mutually exclusive. For example, if A, B, and C are all mutually exclusive events, then 𝑷(𝑨 ∪ 𝑩 ∪ 𝑪) = 𝑷(𝑨) + 𝑷(𝑩) + 𝑷(𝑪)
  • 33. Example: King or Ace In a standard 52-deck of cards, what is the probability of randomly selecting a king or ace? 𝑃 𝑘𝑖𝑛𝑔 = 1 13 = .077 𝑃(𝑎𝑐𝑒) = 1 13 = .077 These events are mutually exclusive because a card cannot be both a king and an ace. 𝑷(𝒌𝒊𝒏𝒈 ∪ 𝒂𝒄𝒆) = 1 13 + 1 13 =. 𝟎𝟕𝟕+. 𝟎𝟕𝟕 =. 𝟏𝟓𝟒
  • 34. Example: Political Affiliation In a given state, 30% are Democrats, 26% Republicans, and 43% Independent. If you randomly select a resident from this state, what is the probability that you have selected an Independent or Democrat? These events are mutually exclusive because each resident may only identify as one category. 𝑃(𝐼𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 ∪ 𝐷𝑒𝑚𝑜𝑐𝑟𝑎𝑡) = 𝑃(𝐼𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡) + 𝑃(𝐷𝑒𝑚𝑜𝑐𝑟𝑎𝑡) = .43 + .30 = .73
  • 35. If two events, A and B, are NOT mutually exclusive, then the probability of A or B is equal to the probability of A and the probability of B minus the overlap of the two. In other words, 𝑷(𝑨 ∪ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩) − 𝑷(𝑨 ∩ 𝑩) The probability of A and B must be subtracted when there is overlap otherwise this area is counted twice. 𝑷(𝑨 ∪ 𝑩) = 𝑷(𝑨) + 𝑷(𝑩) − 𝑷(𝑨 ∩ 𝑩)
  • 36. Example: Ice Cream A large-scale survey finds that: 90% of college students enjoy eating chocolate ice cream, 70% of college students enjoy eating mint chocolate chip ice cream, 65% of college student enjoy eating both chocolate and mint chocolate chip ice cream.
  • 37. What proportion of college students enjoy eating chocolate or mint chocolate chip ice cream? These events are NOT mutually exclusive because it was possible for a participant to enjoy both. 𝑃 𝐶ℎ𝑜𝑐𝑜𝑙𝑎𝑡𝑒 ∪ 𝑀𝑖𝑛𝑡𝐶𝐶 = 𝑃(𝐶ℎ𝑜𝑐𝑜𝑙𝑎𝑡𝑒) + 𝑃(𝑀𝑖𝑛𝑡𝐶𝐶) − 𝑃(𝐶ℎ𝑜𝑐𝑜𝑙𝑎𝑡𝑒 ∩ 𝑀𝑖𝑛𝑡𝐶𝐶 = .90 + .70 − .65 = .95 The probability that a randomly selected college student will enjoy chocolate or mint chocolate chip ice cream is .95
  • 38. Example: Ace or Diamond What is the probability that a randomly selected card from a standard 52-card deck will be an ace or a diamond? Suit and face values are NOT mutually exclusive because it is possible for a card to be an ace and a diamond (i.e., the ace of diamonds). 𝑃 𝑎𝑐𝑒 = 1/13 = .077 𝑃(𝑑𝑖𝑎𝑚𝑜𝑛𝑑) = 1/4 = .25 𝑃 𝑎𝑐𝑒 ∩ 𝑑𝑖𝑎𝑚𝑜𝑛𝑑 = 1/52 = .019 𝑃(𝑎𝑐𝑒 ∪ 𝑑𝑖𝑎𝑚𝑜𝑛𝑑) = 𝑃(𝑎𝑐𝑒) + 𝑃(𝑑𝑖𝑎𝑚𝑜𝑛𝑑) − 𝑃(𝑎𝑐𝑒 ∩ 𝑑𝑖𝑎𝑚𝑜𝑛𝑑) = .077 + .25 − .019 = .308
  • 39. Rule 5 Multiplication Rule Determining the probability that both events occur depends on whether the events are independent or not. This is known as an intersection and is represented by ∩. 𝑃(𝐴 ∩ 𝐵) is read as “probability of A and B.”
  • 40. If two events, A and B, are independent, then the probability of A and B is equal to the product of the two. In other words 𝑷 𝑨 ∩ 𝑩 = 𝑷 𝑨 × 𝑷 𝑩 . This rule extends beyond two events. For example, if A, B, and C are all independent of one another, then 𝑷(𝑨 ∩ 𝑩 ∩ 𝑪) = 𝑷(𝑨) × 𝑷(𝑩) × 𝑷(𝑪)
  • 41. Example: Ace of Diamonds What is the probability that you will randomly pull the ace of diamonds from a standard 52-card deck? The probability of pulling an ace and the probability of pulling a diamond are independent events. 𝑃(𝑎𝑐𝑒) = 4/52 = .077 𝑃(𝑑𝑖𝑎𝑚𝑜𝑛𝑑𝑠) = 13/52 = .25 𝑃(𝑎𝑐𝑒 ∩ 𝑑𝑖𝑎𝑚𝑜𝑛𝑑) = 𝑃(𝑎𝑐𝑒) × 𝑃(𝑑𝑖𝑎𝑚𝑜𝑛𝑑) = .077 × .25 = .019
  • 42. Example: Dog Ownership In a given town, 52% of the population are men and 48% are women. We also know that 40% of residents own a dog. If you randomly select one individual from the population, what is the probability that you will select a man who owns a dog? 𝑃(𝑚𝑎𝑛 ∩ 𝑑𝑜𝑔𝑜𝑤𝑛𝑒𝑟) = 𝑃(𝑚𝑎𝑛) × 𝑃(𝑑𝑜𝑔𝑜𝑤𝑛𝑒𝑟) = .52 × .40 = .208 If you randomly select one individual from the population, what is the probability that you will select a woman who does not own a dog? 𝑃(𝑛𝑜𝑑𝑜𝑔) = 𝑃(𝑑𝑜𝑔𝑐) = 1 − 𝑃(𝑑𝑜𝑔) = 1 − .40 = .6 𝑃(𝑤𝑜𝑚𝑎𝑛 ∩ 𝑛𝑜𝑑𝑜𝑔) = 𝑃(𝑤𝑜𝑚𝑎𝑛) × 𝑃(𝑛𝑜𝑑𝑜𝑔) = .48 × .60 = .288
  • 43. Example: Flipping a Coin When flipping a fair coin (i.e., 50% heads and 50% tails) three times, what is the probability that you will flip heads all three times? 𝑃(𝐻) = 1/2 = .5 𝑃(𝐻𝐻𝐻) = 𝑃(𝐻) × 𝑃(𝐻) × 𝑃(𝐻) = .5 × .5 × .5 = .125 The probability of flipping heads all three times is .125 What is the probability of flipping heads, tails, tails, in that order? 𝑃(𝐻) = 1/2 = .5 𝑎𝑛𝑑 𝑃(𝑇) = 1/2 = .5 𝑃(𝐻𝑇𝑇) = 𝑃(𝐻) × 𝑃(𝑇) × 𝑃(𝑇) = .5 × .5 × .5 = .125
  • 44. If two events, A and B, are NOT independent, then the probability of A and B is equal to the probability of A times the probability of B given A. In other words 𝑷(𝑨 ∩ 𝑩) = 𝑷(𝑨) × 𝑷(𝑩 ∣ 𝑨). Similarly, the A and B may be switched for, 𝑷(𝑨 ∩ 𝑩) = 𝑷(𝑩) × 𝑷(𝑨 ∣ 𝑩). 𝑷(𝑩 ∣ 𝑨) is a conditional probability which we will see in Rule 6, it is read “Probability of B given A.”
  • 45. Example: Gender and Sports Participation A study by Sabo & Veliz (2008) surveyed 2,132 American children about their participation in organized and team sports. In the sample, 50.7% were boys and 49.3% were girls. Of the boys, 73% currently participate in sports. Of the girls, 63% currently participate in sports. Gender and sports participation are NOT independent.
  • 46. What is the probability that you would select a girl who currently participates in sports? 𝑃(𝑔𝑖𝑟𝑙 ∩ 𝑠𝑝𝑜𝑟𝑡𝑠) = 𝑃(𝑔𝑖𝑟𝑙) × 𝑃(𝑠𝑝𝑜𝑟𝑡𝑠 ∣ 𝑔𝑖𝑟𝑙) = .493 × .63 = .311 What is the probability that a randomly selected individual from this sample would be a boy who does not currently participate in sports? 𝑃(𝑏𝑜𝑦 ∩ 𝑛𝑜𝑠𝑝𝑜𝑟𝑡𝑠) = 𝑃(𝑏𝑜𝑦) × 𝑃(𝑛𝑜𝑠𝑝𝑜𝑟𝑡𝑠 ∣ 𝑏𝑜𝑦) = .507 × (1 − .73) = .507 × .27 = .137
  • 47. Rule 6 Conditional Probability 𝑷 𝑨 𝑩 𝑖𝑠 𝑟𝑒𝑎𝑑 𝑎𝑠 “𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝐴 𝑔𝑖𝑣𝑒𝑛 𝐵. ” 𝑷(𝑨 ∣ 𝑩) = 𝑷(𝑨 ∩ 𝑩) 𝑷(𝑩) 𝑷(𝑩 ∣ 𝑨) = 𝑷(𝑨 ∩ 𝑩) 𝑷(𝑨)
  • 48. Example: Gender and Grades In a large class, the probability of randomly selecting a woman is .60. The probability of randomly selecting a student who is a woman and who earned an A is .20. If you randomly select a student who is a woman, what is the probability that she earned an A? A = earning a grade of A. B = being a woman.
  • 49. We want to know the probability of earning an A given that the student is a female. In other words, 𝑷(𝑨 ∣ 𝑩) We are given that 𝑃(𝐴 ∩ 𝐵) = .20 𝑎𝑛𝑑 𝑃(𝐵) = .60 𝑃(𝐴 ∣ 𝐵) = 𝑷(𝑨 ∩ 𝑩) 𝑷(𝑩) = . 𝟐𝟎 . 𝟔𝟎 = .333 Given that a randomly selected student is a woman, there is a 33.3% chance that she earned an A.
  • 50. Example: Clubs In a standard 52-card deck. What is the probability that a randomly selected card is a club given that it is a black card? 𝑃 𝑐𝑙𝑢𝑏 = 1/4 = .25 𝑃(𝑏𝑙𝑎𝑐𝑘) = 2/4 = .50 𝑃(𝑐𝑙𝑢𝑏 ∩ 𝑏𝑙𝑎𝑐𝑘) = 13/52 = .25 𝑃(𝑐𝑙𝑢𝑏 ∣ 𝑏𝑙𝑎𝑐𝑘) = 𝑃(𝑐𝑙𝑢𝑏 ∩ 𝑏𝑙𝑎𝑐𝑘) 𝑃(𝑏𝑙𝑎𝑐𝑘) = .25 .50 = .50 Given that a randomly selected card is black, there is a 50% chance that it's a club.