Lecture11
Probability Theory and Statistics
Kasper K. Berthelsen, Dept. For Mathematical [email protected]
Literature:Walpole, Myers, Myers & Ye: Probability and Statistics for Engineers and Scientists, Prentice Hall, 8th ed.
Slides and lecture overview:http://people.math.aau.dk/~kkb/Undervisning/ET610/
Lecture format: 2x45 min lecturing followed by exercises in group rooms
Lecture12
STATISTICSWhat is it good for?
Forecasting:
• Expectations for the future?• How will the stock markets behave??
0
10
20
30
40
50
60
70
80
90
1 2 3 4
A
B
C
Analysis of sales:
• How much do we sell, and when?• Should we change or sales strategy?
Quality control:
• What is my rate of defective products?• How can I best manage my production? • What is the best way to sample?
lecture 13
Probability theorySample space and events
Consider an experiment
Sample space S:
Event A:
Complementary event A´:
A´
VENN DIAGRAM
A
A
Example:A´={2,3,4,5} rolling a dice
Example:S={1,2,…,6} rolling a diceS={plat,krone} flipping a coin
Example:A={1,6} when rolling a dice
S
S
S
lecture 14
Probability theoryEvents
Example:Rolling a dice
4 6 1 3
Union:AB={1,2,3,4,6}
Intersection:AB={2}S={1,2,3,4,5,6}
A={2,4,6}B={1,2,3}
C DDisjoint events: CD = Ø
C={1,3,5} and D={2,4,6} are disjoint
A B
2
5S
S
lecture 15
Probability theoryCounting sample points
Ways of placing your bets: Guess the results of 13 matchesPossible outcomes:
Home winDrawAway win
1 X 2
3 possibilities
3 possibilities
•••
3 possibilities
Answer: 3·3·3· ··· ·3 = 313
The multiplication rule
lecture 16
Probability theoryCounting sample points
Ordering n different objectsNumber of permutations ???
3!= 6
There are• n ways of selecting the first object • n -1 ways of selecting second object
• 1 way of selecting the last object
•••
n · (n -1) · ··· · 1 = n ! ways
The multiplication rule
”n factorial”
lecture 17
Probability theoryCounting sample points
Multiplication rule:If k independent operations can be performed in n1, n2, … , nk ways, respectively, then the k operations can be performed in
n1 · n2 · ··· · nk ways
Tree diagram:
T
TT
T
T
T
TH
H
H
H
H
H
H
23 = 8 possible outcomes
Flipping a coin three times (Head/Tail)
lecture 18
Probability theoryCounting sample points
Number of possible ways of selecting r objects from a set of n destinct elements:
-
Withoutreplacement
Withreplacement
Ordered
Unordered
!
( )!n r
nP
n r
rn
!
!( )!
n n
r r n r
lecture 19
Probability theoryCounting sample points
Example: Ann, Barry, Chris, and Dan should from a committee consisting of two persons, i.e. unordered without replacement.
Number of possible combinations:
Writing it out : AB AC AD BC BD CD
4 4!6
2 2!2!
lecture 110
Probability theoryCounting sample points
Example: Select 2 out of 4 different balls ordered and without replacement
Number of possible combinations: 4 2
4!12
(4 2)!P
Notice: Order matters!
lecture 111
Probability theoryProbability
Let A be an event, then we denote
P(A) the probability for A
It always hold that 0 < P(A) < 1 P(Ø) = 0 P(S) = 1
Consider an experiment which has N equally likely outcomes, and let exactly n of these events correspond to the event A. Then
Example:Rolling a dice
AS
2
1
6
3
(
number) evenP
# successful outcomes # possible outcomes
=N
nAP )(
lecture 112
Probability theoryProbability
Example: Quality controlA batch of 20 units contains 8 defective units.
Select 6 units (unordered and without replacement).
Event A: no defective units in our random sample.
Number of possible samples:
Number of samples without defective units:
20N
6
12n
6
12
6 12!6!14! 77P(A) 0.024
20 6!6!20! 3230
6
(# possible)
(# successful)
lecture 113
Probability theoryProbability
Example: continued
Event B: exactly 2 defective units in our sample
Number of samples with exactly 2 defective units:
12 8n
4 2
12 8
4 2 12!8!6!14!P(B) 0.3576
20 4!8!2!6!20!
6
(# successful)
lecture 114
Probability theoryRules for probabilities
Union:A B
Intersection:A B
P(A B) = P(A) + P(B) - P(A B)
P(B) = P(B A) + P(B A´ )
If A and B are disjoint: P(A B) = P(A) + P(B)
In particular: P( A ) + P(A´ ) = 1
A B
lecture 115
Probability theoryConditional probability
A Bwhere P ( B ) > 0
Conditional probability for A given B:
Bayes’ Rule: P(AB) = P(A|B)P(B) = P(B|A)P(A)
Rewriting Bayes’ rule:
P(B|A)P(A) P(B|A)P(A)P(A|B)
P(B) P(B|A)P(A)+P(B|A´)P(A´)
P(A B)P(A|B)
P(B)
lecture 116
Probability theoryConditional probability
Employed Unemployed Total
Man 460 40 500
Woman 140 260 400
Total 600 300 900
Example page 59:The distribution of employed/unemployed amongst men and women in a small town.
%7.7630
23
600
460
900600
900460
)(
)()|(
/
/
P
PP
employd
employd&manemployedman
%./
/
P
P|P 313
15
2
300
40
900300
90040
)(
)()(
unemployed
unemployed& manunemployedman
lecture 117
Probability theoryBayes’ rule
Example: Lung disease & SmokingAccording to ”The American Lung Association” 7% of the population suffers from a lung disease, and 90% of these are smokers. Amongst people without any lung disease 25.3% are smokers.
Events: Probabilities:A: person has lung disease P(A) = 0.07B: person is a smoker P(B|A) = 0.90
P(B|A´ ) = 0.253
211.093.0253.007.09.0
07.09.0
´)(´)|()()|(
)()|()|(
APABPAPABP
APABPBAP
What is the probability that at smoker suffers from a lung disease?
lecture 118
A1 , … , Ak are a partitioning of S
Probability theoryBayes’ rule – extended version
Bayes’ formel udvidet:
A3
A2
A4 A5
A6
A1
B
S
Law of total probability:
k
iii APABPBP
1
)()|()(
k
iii
rrr
APABP
APABPBAP
1
)()|(
)()|()|(
lecture 119
Probability theoryIndependence
Definition: Two events A and B are said to be independent if and only if
P(B|A) = P(B) or P(A|B) = P(A)
Alternative Definition: Two events A and B are said to be independent if and only if
P(A∩B) = P(A)P(B)
Notice: Disjoint event (mutually exclusive event) are dependent!
lecture 120
Probability theoryConditional probability
Example:
%./
/|P 776
900600
900460)( employedman
%./P 655900500)( man
Employed Unemployed Total
Man 460 40 500
Woman 140 260 400
Total 600 300 900
Conclusion: the two events “man” and “employed” are dependent.
lecture 121
Probability theoryRules for conditional probabilities
)()|()|()( CPCBPCBAPCBAP
)()|()( BPBAPBAP Probability of events A and B happening simultaneously
Probability of events A, B and C happening simultaneously
Proof:)()|()|()()|()( CPCBPCBAPCBPCBAPCBAP
)()|(
)|(
)|()(
1
32
2121
kkk
k
kk
APAAP
AAAP
AAAPAAAP
General rule: