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8/13/2019 Module 1 Probability
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Module 1
Probability
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Review of Sets
Definitions Union of Sets (AUB): Elements in either set
Intersection (AB): Elements in both sets
Complement (Ac): Elements not in A
Rules A(BUC) = (AB)U(AC)
AU(BC) = (AUB)(AUC)
n(A) = n(AB) + n(ABc)
n(AUB) = n(A) + n(B)n(AB)
n(AUBUC) = n(A) + n(B) + n(C)n(AB)n(AC)n(BC) + n(ABC)
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Solving Problems
Method
Draw diagram of intersecting sets
Label known quantities Work from the inside out to determine unknowns
Example 1
N = 100, A = 40, B = 30, AB = 20
Find ABc= 4020 = 20
Find (AUB)c= 100(40 + 3020) = 50
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Solving Problems
Example 2 A = 40, B = 30, C = 20
AB = 20, AC = 10, BC = 10
ABC = 5
Calculations ABCc= ABABC = 205 = 15
A
Bc
C = 105 = 5 Ac
B
C = 105 = 5 ABcCc= AABCcABcCABC = 15
AcBCc= 5 AcBcC = 5
AUBUC = (40 + 30 + 20)(20 + 10 + 10) + 5 = 55
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Using Table
Use table if Acis as important as A For instance, not just male, but male and female
Or looking for values such as (AcBc)
Method Draw table with rows representing one variable and
columns representing another.
Systematically calculate desired values using set rules:
n(A) = Nn(Ac)
n(ABc) = n(A)n(AB)
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Example Problem
Males: 500 (Given)
Male & College: 350 (Given) Male & No College
Male & Imm: 100 (Given) 80 (Given) 100 - 80 = 20
M & Non-I: 500 - 100 = 400 350 - 80 = 270 400 - 279 = 130
Imm & College: Not Needed
A survey of 1000 students: 500 students are male 600 plan to attend college
200 are immigrants 350 are males/college
150 are imm/college 100 are immigrant males 80 are male/immigrants/college
How many students are male, non-immigrants, who donot plan to attend college?
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Probability Definitions
Sample Point: Simple outcome of experiment
Rolling a 4 on a die
Sample Space (S): All possible outcomes Set for rolling a die: {1, 2, 3, 4, 5, 6}
Mutually Exclusive: Never at same time (AB)=0
Can not role a 3 and an even number at same time
Exhaustive: Include entire sample space
Even and odd rolls include all possible outcomes
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Probability Definitions (Same as Sets)
Event: Collection of sample points
Event A: Rolling an even number with die: {2, 4, 6}
Union (AUB): Outcomes in events A or B A: {2, 4, 6}, B: {4, 5, 6} AUB = {2, 4, 5, 6}
Intersection (AB): Outcomes in events A and B
A: {2, 4, 6}, B: {4, 5, 6} AB = {4, 6}
Complement (Ac): Outcomes not in A
A: {2, 4, 6} Ac: {1, 3, 5}
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Probability Definition
Probability RulesP(Ac) = 1P(A)
Law of Total Probability
Probability Definition and Rules
Note: Aiare mutually exclusive and exhaustive
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Other useful Rules
DeMorgans Laws (AUB)c= AcBc
(AB)c= AcUBc
Example: Not (even or > 3) = odd and (
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Example
A local survey indicated
80% own auto 60% own house 50% own both
What percentage own an auto or a house,but not both?
What percentage own neither an auto nor a house?
Solution
P(ABc)+P(A
cB) = (.80.50) + (.60.50) = .40
P(AcBc) = P[(AUB)c] = 1(.8 + .6.5) = .1
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Conditional Probability
Definitions
Example
Law of Total Probability
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Sample Problem
A survey of company clients indicated 30% have life insurance 70% medical 20% both
50% of those with only life will renew policy
40% of those with only medical will renew
60% with both will renew
Determine probability that a random policy holder will renew?
Method Draw Venn diagram and calculate regions.
Set up equation for conditional probability.
Use conditional definition to change and probability to conditional.
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Solution
Note: Looking for conditional probability that person will renew(E), given that they hold a policy (AUBUC). A, B, and C aremutually exclusive, so equations simplify.
A = LMc P[A] = P[L]P[LM] = .3 - .2 = .1
B = LcM P[B] = P[M]P[LM] = .7 - .2 = .5
C = LM P[C] = P[LM] = .2
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Bayes Theorem
Useful when swapping conditional probability
Example
Two urns have 5 red and 6 blue balls. A ball fromurn 1 is picked randomly and placed in urn 2.
If a red ball is pulled from urn 2, what is theprobability that a red ball was picked from urn 1?
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Solution
Calculate Probabilities
P[R1] = 5/11 P[B1] = 6/11
P[R2|R1] = 6/12 P[R2|B1] = 5/12
Use Bayes Theorem
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Sample Problem
Given the following data, what is the probability of a personhaving an accident falling into the 2630 bracket?
Age of Driver Probability ofAccident
Portion ofDrivers
Group 1: 18-20 .07 .03Group 2: 21-25 .05 .05
Group 3: 26-30 .03 .10
Group 4: 31-65 .02 .52
Group 5: >65 .05 .30
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Independence
Events are independent if the probability of one isnot affected by the other.
Independence Equations P[AB] = P[A] P[B]
P[A|B] = P[A]
Example
P[A] = P[B] = A&B independent
P[AUB] = P[A] + P[B]P[AB] = P[A]+ P[B]P[A]*P[B]
P[AUB] = + - = 5/8