Download Basic Models of Probability

5/7/2017
Basic Models of Probability
Instructor: Ron S. Kenett
Email: [email protected]
Course Website: www.kpa.co.il/biostat
Course textbook: MODERN INDUSTRIAL STATISTICS,
Kenett and Zacks, Duxbury Press, 1998
(c) 2000, Ron S. Kenett, Ph.D.
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Course Syllabus
•Understanding Variability
•Variability in Several Dimensions
•Basic Models of Probability
•Sampling for Estimation of Population Quantities
•Parametric Statistical Inference
•Computer Intensive Techniques
•Multiple Linear Regression
•Statistical Process Control
•Design of Experiments
(c) 2000, Ron S. Kenett, Ph.D.
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The Paradox of the Chevalier de Mere - 1
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Success = at least one “1”
(c) 2000, Ron S. Kenett, Ph.D.
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The Paradox of the Chevalier de Mere - 2
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Success = at least one “1,1”
(c) 2000, Ron S. Kenett, Ph.D.
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The Paradox of the Chevalier de Mere - 3
P (Success) = P(at least one “1”)
 4 1  2
6
3
P (Success) = P(at least one “1,1”)
 24  1
36
2
3
Experience proved otherwise !
Game A was a better game to play
(c) 2000, Ron S. Kenett, Ph.D.
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The Paradox of the Chevalier de Mere - 4
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The calculations of Pascal and Fermat
P (Failure) = P(no “1”)
 6   .482
 5
4
P (Success) = .518
P (Failure) = P(no “1,1”)

 35

36
24
 .509
P (Success) = .491
What went wrong before?
(c) 2000, Ron S. Kenett, Ph.D.
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P(outcomes add up to “10”) =?
To add or to multiply ?
(c) 2000, Ron S. Kenett, Ph.D.
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1,1
2,1
3,1
4,1
5,1
6,1
1,2
2,2
3,2
4,2
5,2
6,2
1,3
2,3
3,2
4,3
5,3
6,3
1,4
2,4
3,4
4,4
5,4
6,4
1,5
2,5
3,5
4,5
5,5
6,5
1,6
2,6
3,6
4,6
5,6
6,6
P(outcomes add up to “10”) = 3 / 36
(c) 2000, Ron S. Kenett, Ph.D.
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Mutually Exclusive Events
Two events are mutually exclusive (or disjoint) if it is impossible
for them to occur together.
If two events are mutually exclusive, they cannot be independent
and vice versa.
Example:
A subject in a study cannot be both male and female, nor can
they be aged 20 and 30. A subject could however be both male
and 20, both female and 30.
(c) 2000, Ron S. Kenett, Ph.D.
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Independent Events
Two events are independent if the occurrence of one of the
events gives us no information about whether or not the other
event will occur; that is, the events have no influence on each other.
If two events are independent then they cannot be mutually
exclusive (disjoint) and vice versa.
(c) 2000, Ron S. Kenett, Ph.D.
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Example
Suppose that a man and a woman each have a pack of 52 playing
cards. Each draws a card from his/her pack. Find the probability
that they each draw the ace of clubs.
We define the events
A = 'the man draws the ace of clubs'
B = 'the woman draws the ace of clubs'
Clearly events A and B are independent so,
That is, there is a very small chance that the man and the woman
will both draw the ace of clubs.
(c) 2000, Ron S. Kenett, Ph.D.
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Conditional Probability
In many situations, once more information becomes available, we
are able to revise our estimates for the probability of further
outcomes or events happening. For example, suppose you go out
for lunch at the same place and time every Friday and you are
served lunch within 15 minutes with probability 0.9. However,
given that you notice that the restaurant is exceptionally busy, the
probability of being served lunch within 15 minutes may reduce to
0.7. This is the conditional probability of being served lunch
within 15 minutes given that the restaurant is exceptionally busy.
(c) 2000, Ron S. Kenett, Ph.D.
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The usual notation for "event A occurs given that event B has
occurred" is A|B (A given B). The symbol | is a vertical line and
does not imply division. P(A|B) denotes the probability that event
A will occur given that event B has occurred already.
A rule that can be used to determine a conditional probability
from unconditional probabilities is
P(A|B) = P(A and B) / P(B)
where,
P(A|B) = the (conditional) probability that event A will occur given
that event B has occurred already
P(A and B) = the (unconditional) probability that event A and event
B occur
P(B) = the (unconditional) probability that event B occurs
(c) 2000, Ron S. Kenett, Ph.D.
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Binomial Distribution
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X = Number of successes in n trials
n = 6, x = 0
n = 6, x = 1
n = 6, x = 3
P(x) 
(c) 2000, Ron S. Kenett, Ph.D.
0
x
n
n!  p x (1 p)n  x
x!(n  x)!
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Binomial Distribution
p=.25
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p=.75
b(i; 50, p)
p=.5
0.10
0.05
0.00
0
10
20
30
40
50
i
(c) 2000, Ron S. Kenett, Ph.D.
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Binomial Distribution
p=.25
p=.75
b(i; 50, p)
p=.5
0.10
0.05
0.00
0
10
20
30
i
30
20
10
0
7
9
11
13
15
17
19
40
50
13
10
10
11
13
14
18
15
9
13
16
12
12
11
8
11
13
11
12
9
20
20
9
9
4
14
7
8
12
15
11
16
18
11
13
15
11
11
18
12
12
15
12
14
12
12
9
16
15
12
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C1
(c) 2000, Ron S. Kenett, Ph.D.
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Poisson Distribution
X = Number of occurrences of an event
events
x=0
x=2
x=8
0
x
x e –

P(x) 
x!
(c) 2000, Ron S. Kenett, Ph.D.
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Poisson Distribution
0.2
p(j; lambda)
lambda=5
lambda=10
lambda=15
0.1
0.0
0
10
20
30
40
50
j
(c) 2000, Ron S. Kenett, Ph.D.
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Negative Binomial
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0.045
0.040
p = 0.2
0.035
0.030
y
0.025
0.020
p= 0.1
0.015
0.010
0.005
0.000
0
50
100
X
(c) 2000, Ron S. Kenett, Ph.D.
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Normal Distribution
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0.4
Sigma = 1
y
0.3
0.2
Sigma = 2
0.1
Sigma = 3
0.0
0
10
20
x
(c) 2000, Ron S. Kenett, Ph.D.
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Normal Distribution
X
-3.0
-2.9
-2.8
-2.7
-2.6
-2.5
-2.4
-2.3
-2.2
-2.1
-2.0
-1.9
-1.8
-1.7
-1.6
-1.5
-1.4
-1.3
-1.2
-1.1
-1.0
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0.0
(c) 2000, Ron S. Kenett, Ph.D.
P(<X)
0.001350
0.001866
0.002555
0.003467
0.004661
0.006210
0.008198
0.010724
0.013903
0.017864
0.022750
0.028717
0.035930
0.044565
0.054799
0.066807
0.080757
0.096800
0.115070
0.135666
0.158655
0.184060
0.211855
0.241964
0.274253
0.308538
0.344578
0.382089
0.420740
0.460172
0.500000
P(Xi< <Xi+1)
0.004432
0.005953
0.007915
0.010421
0.013583
0.017528
0.022395
0.028327
0.035475
0.043984
0.053991
0.065616
0.078950
0.094049
0.110921
0.129518
0.149727
0.171369
0.194186
0.217852
0.241971
0.266085
0.289692
0.312254
0.333225
0.352065
0.368270
0.381388
0.391043
0.396953
0.398942
X
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
P(<X)
0.500000
0.539828
0.579260
0.617911
0.655422
0.691462
0.725747
0.758036
0.788145
0.815940
0.841345
0.864334
0.884930
0.903200
0.919243
0.933193
0.945201
0.955435
0.964070
0.971283
0.977250
0.982136
0.986097
0.989276
0.991802
0.993790
0.995339
0.996533
0.997445
0.998134
0.998650
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N(0,1)
P(Xi< <Xi+1)
0.398942
0.396953
0.391043
0.381388
0.368270
0.352065
0.333225
0.312254
0.289692
0.266085
0.241971
0.217852
0.194186
0.171369
0.149727
0.129518
0.110921
0.094049
0.078950
0.065616
0.053991
0.043984
0.035475
0.028327
0.022395
0.017528
0.013583
0.010421
0.007915
0.005953
0.004432
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Normal Distribution
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0.4
Sigma = 1
y
0.3
0.2
Sigma = 2
0.1
Sigma = 3
0.0
0
10
x
20
10
0
7.9006 11.5151 9.9542 9.4493 8.2387
10.4707 9.4041 9.3517 10.5664 10.9079
10.0077 12.5188 9.6937 10.0757 10.1616
10.2881 9.8560 10.0014 9.8467 11.5006
10.2982 9.6023 9.7238 11.5413 8.4595
9.2372 11.0408 12.8996 9.5590 9.1041
8.9170 9.7734 7.9844 8.3484 11.3703
10.6260 10.0952 11.4019 8.9842 9.3783
9.7574 7.9312 8.1566 9.9305 9.1158
8.6436 10.4689 9.3356 10.8788 7.8790
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7.5 8.0 8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0 12.5
C1
(c) 2000, Ron S. Kenett, Ph.D.
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The t Distribution
0.5
t(x; nu)
0.4
nu=50
0.3
0.2
nu=5
0.1
0.0
-5
0
5
t
(c) 2000, Ron S. Kenett, Ph.D.
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The F Distribution
0.7
0.6
0.5
y
0.4
0.3
0.2
0.1
0.0
0
1
2
3
4
5
6
x
(c) 2000, Ron S. Kenett, Ph.D.
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