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CHAPTER 4BASIC PROBABILITY
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Learning Objectives
In this chapter, you learn:
y Basic probability concepts and definitions
y Conditional probability
y Various counting rules
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Important Terms
y Probability the chance that an uncertain eventwill occur (always between 0 and 1)
y Event Each possible outcome of a variabley Sample Space the collection of all possible
events
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Assessing Probability
y There are three approaches to assessing the probabilityof an uncertain event:
1. a prioriclassical probability
2. empirical classical probability
3. subjective probability
an individual judgment or opinion about the probability of occurrence
outcomeselementaryofnumbertotaloccurcaneventthewaysofnumber
TXoccurrenceofyprobabilit !!
observedoutcomesofnumbertotal
observedoutcomesfavorableofnumberoccurrenceofyprobabilit !
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Sample Space
The Sample Space is the collection of all
possible events
e.g. All 6 faces of a die:
e.g. All 52 cards of a bridge deck:
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Events
y Simple event
An outcome from a sample space with one characteristic
e.g., Ared card from a deck of cards
y Complement of an event A (denoted A) All outcomes that are not part of event A
e.g., All cards that are not diamonds
y Joint event
Involves two or more characteristics simultaneously
e.g., An ace that is also red from a deck of cards
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Visualizing Events
y Contingency Tables
y Tree Diagrams
Red 2 24 26
Black 2 24 26
Total 4 48 52
Ace Not Ace Total
Full Deck
of 52 Cards
Sample
Space
Sample
Space2
24
2
24
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Visualizing Events
y Venn Diagrams
Let A= aces
Let B = red cards
A
B
A B = ace and red
A U B = ace or red
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Mutually Exclusive Events
y Mutually exclusive events
Events that cannot occur together
example:
A= queen of diamonds; B = queen of clubs
Events Aand B are mutually exclusive
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Collectively Exhaustive Events
y Collectively exhaustive events One of the events must occur The set of events covers the entire sample space
example:A= aces; B = black cards;C = diamonds; D = hearts
Events A, B, C and D are collectively exhaustive (but notmutually exclusive an ace may also be a heart)
Events B, C and D are collectively exhaustive and alsomutually exclusive
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Probability
y Probability is the numerical measure
of the likelihood that an event will
occur
y The probability of any event must bebetween 0 and 1, inclusively
y The sum of the probabilities of all
mutually exclusive and collectivelyexhaustive events is 1
Certain
Impossible
0.5
1
0
0 P(A) 1 For any event A
1P(C)P( )P(A) !
If A, B, and are mutually exclusive and
collectively exhaustive
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Computing Joint andMarginal Probabilities
y The probability of a joint event, Aand B:
y Computing a marginal (or simple) probability:
Where B1, B2, , Bkare kmutually exclusive and collectivelyexhaustive events
outcomeselementaryofnumbertotal
BandAsatisfyingoutcomesofnumber
)BandA(P!
)BdanP(A)BandP(A)BandP(AP(A) k! .
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Joint Probability Example
P(Red and Ace)
Black
ColorType Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
52
2
cardsofnumbertotal
aceandredarethatcardsofnumber!!
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Marginal Probability Example
P(Ace)
Black
ColorType Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
52
4
52
2
52
2)BlackandAce(P)dReandAce(P !!!
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P(A1 and B2) P(A1)
TotalEvent
Joint Probabilities Using Contingency Table
P(A2 and B1)
P(A1 and B1)
Event
Total 1
Joint Probabilities Marginal (Simple) Probabilities
A1
A2
B1 B2
P(B1) P(B2)
P(A2 and B2) P(A2)
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General Addition Rule
P(A or B) = P(A) + P(B) - P(A and B)
General Addition Rule:
If A and B are mutually exclusive, then
P(A and B) = 0, so the rule can be simplified:
P(A or B) = P(A) + P(B)
For mutually exclusive events A and B
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General Addition Rule Example
P(Red orAce) = P(Red) +P(Ace) - P(Red and Ace)
= 26/52 + 4/52 - 2 /52 = 28/52
Dont countthe two red
aces twice!Black
ColorType Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
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Computing Conditional Probabilities
y Aconditional probabilityis the probability of oneevent, given that another event has occurred:
P( )
)P()|P( !
P( )
)P()|P( !
Where P( ) = joi t prob bility of
P( ) = m rgi l prob bility of
P( ) = m rgi l prob bility of
The co itio l
prob bility of giveth t h s occurre
The co itio l
prob bility of give
th t h s occurre
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Conditional Probability Example
y What is the probability that a car has a CD player,given that it has AC ?
i.e., we want to find P(CD | AC)
Of the cars on a used car lot, 70% have airconditioning (A ) and 40% have a D player( D). 20% of the cars have both.
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Conditional Probability Example
No CDCD TotalAC 0.2 0.5 0.7
No AC 0.2 0.1 0.3
Total 0.4 0.6 1.0
Of the cars on a used car lot, 70% have air conditioning(A ) and 40% have a D player ( D).
20% of the cars have both.
0. 80.
0.
P(A )
A )andP( DA )|P( D !!!
(continued)
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Conditional Probability Example
No CDCD Total
AC 0.2 0.5 0.7
No AC 0.2 0.1 0.3
Total 0.4 0.6 1.0
Given A , we only consider the top row (70% of the cars). Of these,20% have a D player. 20% of 70% is about 2 .57%.
0.2 570.7
0.2
P(A )
A )andP( DA )|P( D !!!
(continued)
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Using Decision Trees
P(A and D) = 0.2
P(A and D) = 0.5
P(A and D) = 0.1
P(A and D) = 0.2
7.
5.
3.
2.
3.
1.
All
Cars
7.
.
Given A or
no A :
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Using Decision Trees
P( D and A ) = 0.2
P( D and A ) = 0.2
P( D and A ) = 0.1
P( D and A ) = 0.5
4.
2.
.
5.
.
1.
All
Cars
4.
.
Given D or
no D:
(continued)
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Statistical Independence
y Two events are independent if and onlyif:
y Events Aand B are independent when the probabilityof one event is not affected by the other event
P(A)B)|P(A !
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Multiplication Rules
y Multiplication rule for two events Aand B:
P(B)B)|P(AB)andP(A !
P(A)B)|P(A !Note: If A and B are independent, then
and the multiplication rule simplifies to
P(B)P(A)B)andP(A !
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Marginal Probability
y Marginal probability for event A:
Where B1, B2, , Bkare kmutually exclusive and collectivelyexhaustive events
)P()|P()P()|P()P()|P(P( ) kk! .
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Counting Rules
y Counting Rule 1:
If there are k1 events on the first trial, k2 events on thesecond trial, and kn events on the n
th trial, the number of
possible outcomes is
Example:
You want to go to a park, eat at a restaurant, and see a movie.There are 3 parks, 4 restaurants, and 6 movie choices. Howmany different possible combinations are there?
Answer: (3)(4)(6) = 72 different possibilities
(k1)(k2)(kn)
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Counting Rules
y Counting Rule 2:
The number of ways that n items can be arranged in order is
Example:
Your restaurant has five menu choices for lunch. Howmany wayscan you order them on your menu?
Answer: 5! = (5)(4)(3)(2)(1) = 120 different possibilities
n! = (n)(n 1)(1)
(continued)
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Counting Rules
y Counting Rule 3:
Permutations: The number of ways of arranging X objectsselected from n objects in order is
Example:
Your restaurant has fivemenu choices, and three are selectedfor daily specials. Howmany different ways can the specialsmenu be ordered?
Answer: different
possibilities
(continued)
X)!(
!Pxn
!
600
3)!(
!
X)!(n
n!nPx !!
!
!
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Counting Rules
y Counting Rule 4: Combinations: The number of ways of selecting X objects from
n objects, irrespective of order, is
Example:
Your restaurant has five menu choices, and three are selected fordaily specials. Howmany different special combinations are there,
ignoring the order in which they are selected?
Answer: different possibilities
(continued)
X)!(nX!n!
xn
!
10(6)(2)
120
3)!(53!
5!
X)!(nX!
n!xn !!
!
!