Post on 29-Mar-2015
Probability
ProbabilityThe calculated likelihood that a given event will occur
Methods of Determining Probability
Empirical
Experimental observationExample – Process control
TheoreticalUses known elements
Example – Coin toss, die rolling Subjective
AssumptionsExample – I think that . . .
Probability Components
ExperimentAn activity with observable results
Sample SpaceA set of all possible outcomes
EventA subset of a sample space
Outcome / Sample PointThe result of an experiment
ProbabilityWhat is the probability of a tossed coin landing heads up?
Probability Tree
Experiment
Sample Space
Event
Outcome
Probability
A way of communicating the belief that an event will occur.
Expressed as a number between 0 and 1fraction, percent, decimal, odds
Total probability of all possible events totals 1
Relative FrequencyThe number of times an event will occur divided by the number of opportunities
= Relative frequency of outcome x
= Number of events with outcome x
n = Total number of events
xx
nf =
n
Expressed as a number between 0 and 1fraction, percent, decimal, odds
Total frequency of all possible events totals 1
xn
Probability
xx
a
fP =
f
What is the probability of a tossed coin landing heads up?
How many possible outcomes? 2
How many desirable outcomes? 1
1P=
2=.5=50%
Probability Tree
What is the probability of the coin landing tails up?
Probability
xx
a
fP =
f
How many possible outcomes?
How many desirable outcomes? 1
1P=
4
What is the probability of tossing a coin twice and it landing heads up both times?
4
HH
HT
TH
TT
=.25=25%
Probability
xx
a
fP =
f
How many possible outcomes?
How many desirable outcomes? 3
3P=
8
What is the probability of tossing a coin three times and it landing heads up exactly two times?
8
1st
2nd
3rd
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
=.375=37.5%
Binomial Process
Each trial has only two possible outcomesyes-no, on-off, right-wrong
Trial outcomes are independent Tossing a coin does not affect future tosses
Bernoulli Process
P = Probability
x = Number of times for a specific outcome within n trials
n = Number of trials
p = Probability of success on a single trial
q = Probability of failure on a single trial
! = factorial – product of all integers less than or equal
Probability DistributionWhat is the probability of tossing a coin three times and it landing heads up two times?
( )( )( )
x n-x
x
n! p qP =
x! n-x !
Law of Large Numbers
Trial 1: Toss a single coin 5 times H,T,H,H,TP = .600 = 60%
Trial 2: Toss a single coin 500 times
H,H,H,T,T,H,T,T,……TP = .502 = 50.2%
Theoretical Probability = .5 = 50%
The more trials that are conducted, the closer the results become to the theoretical probability
Probability
Independent events occurring simultaneously
Product of individual probabilities
If events A and B are independent, then the probability of A and B occurring is: P(A and B) = PA∙PB
AND (Multiplication)
Probability AND (Multiplication)What is the probability of rolling a 4 on a single die?
How many possible outcomes?
How many desirable outcomes? 16
4
1P =
6
What is the probability of rolling a 1 on a single die?
How many possible outcomes?
How many desirable outcomes? 16 1
1P =
6
What is the probability of rolling a 4 and then a 1 in sequential rolls?
4 1P=(P )(P )1 1
=6 6×
1= =.027 28=
36.78%
Probability
Independent events occurring individually
Sum of individual probabilities
If events A and B are mutually exclusive, then the probability of A or B occurring is:
P(A or B) = PA + PB
OR (Addition)
Probability OR (Addition)What is the probability of rolling a 4 on a single die?
How many possible outcomes?
How many desirable outcomes? 16
4
1P =
6
What is the probability of rolling a 1 on a single die?
How many possible outcomes?
How many desirable outcomes? 16 1
1P =
6
What is the probability of rolling a 4 or a 1 on a single die?
4 1P = P + P1 1
= + 6 6
2= = .333 333 =
6.33%
Probability
Independent event not occurring
1 minus the probability of occurrence
P = 1 - P(A)
NOT
What is the probability of not rolling a 1 on a die?
1P = 1 - P1
= 1 - 6
5= = .833 833 =
6.33%
How many tens are in a deck?
ProbabilityTwo cards are dealt from a shuffled deck. What is the probability that the first card is an ace and the second card is a face card or a ten?
How many cards are in a deck? 52
4
12
4
How many aces are in a deck?
How many face cards are in deck?
Probability
What is the probability that the first card is an ace?
4 1 = = .0769 = 7.69%
52 13
12 4 = = .2353 = 23.53%
51 17
Since the first card was NOT a face, what is the probability that the second card is a face card?
Since the first card was NOT a ten, what is the probability that the second card is a ten?
4 = .0784 = 7.84%
51
ProbabilityTwo cards are dealt from a shuffled deck. What is the probability that the first card is an ace and the second card is a face card or a ten?
A F 10P=P (P +P )
= .0241 = 2.41%
If the first card is an ace, what is the probability that the second card is a face card or a ten? 31.37%
Conditional ProbabilityP(E|A) = Probability of event E, given A
Example: One card is drawn from a shuffled deck. The probability it is a queen is
P(queen) =
However, if I already know it is face card
P(queen | face)=
Conditional ProbabilityProbability of two events A and B both occurring =
P(A and B)
= P(A|B) P(B)
= P(B|A) P(A)
If A and B are independent, then
P(A and B) = P(A) P(B)
Bayes’ TheoremCalculates a conditional probability, based on all the ways the condition might have occurred.
P( A | E ) = probability of A, given we already know the condition E
=
Bayes’ Theorem ExampleLCD screen components for a large cell phone manufacturing company are outsourced to three different vendors. Vendor A, B, and C supply 60%, 30%, and 10% of the required LCD screen components. Quality control experts have determined that .7% of vendor A, 1.4% of vendor B, and 1.9% of vendor C components are defective.
If a cell phone was chosen at random and the LCD screen was determined to be defective, what is the probability that the LCD screen was produced by vendor A?
Bayes’ Theorem Example
Notation Used:
P = Probability
D = Defective
A, B, and C denote vendors
Unknown to be calculated:
P(A|D)= Probability the screen is from A,given that it is defective
?
Bayes’ Theorem Example
P(A)=
P(B)=
P(C)=
Known probabilities:
Probability the screen is from A
Probability the screen is from B
Probability the screen is from C
60%=.60
30%=.30
10%=.10
Bayes’ Theorem Example
P(D|C)=Probability the screen is defective given it is from C
P(D|A)=
P(D|B)=
Probability the screen is defective given it is from A
Probability the screen is defective given it is from B
Known conditional probabilities:
0.7%=.007
1.4%=.014
1.9%=.019
Bayes’ Theorem Example:Defective Part
= P(screen is defective AND from A) P(screen is defective from anywhere)
LCD Screen Example
( )( )( )( ) ( )( ) ( )( )
.60 .007=
.60 .007 + .30 .014 + .10 .019( )P A D
.0042=
.0042+.0042+.0019
.0042=
.0103
= .4078 = 40.78%
LCD Screen Example
If a cell phone was chosen at random and the LCD screen was determined to be defective, what is the probability that the LCD screen was produced by vendor B?
If a cell phone was chosen at random and the LCD screen was determined to be defective, what is the probability that the LCD screen was produced by vendor C?