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MBA3
Probability distributions
and information
Fred Wenstøp
Discrete probability distributions
• A series of probabilities pi for all possible states of
nature
Spi =1
• A probability distribution can be
– Theoretical, based on simple but fundamental assumptions
• Binomial
• Pascal
• Poisson
– Empirical, based on past experience
– Subjective, based on beliefs
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Fred Wenstøp: MBA3
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The binomial
distribution
n a
na
P ( a )    p (1  p )
a
• How many times will I succeed?
• A binomial process:
– A series of n independent trials
where the outcome each time is
either success or failure and a
constant probability p for success
– The probability of exactly a
successes in a binomial process
– Excel:
• BINOMDIST(a;n;p;0)
T he
=
bino
B IN O MD I
0 . 3
0 . 2 5
0 . 2
0 . 1 5
0 . 1
0 . 0 5
0
01234567891 0
a
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The Pascal
distribution
P ( n )  p (1  p )
• When will it be my turn?
• The probability that it will
take n trials to get the first
success in a binomial process
with probability p.
• Example:
– How many tosses to get on the
board i Ludo?
• p = 1/6
T he
P as
0 . 2 5
0 . 2
0 . 1 5
0 . 1
0 . 0 5
0
0 2 4 6 81 1
01
2
1
4
1
68
1 3 5 7 91 1
1
1
31
51
79
– See graph:
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n 1
n
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The Poisson
distribution
P ( x) 
l
x
e
l
x!
• How often will disasters happen?
– The probability of x occurrences of
an event in a certain period when
the propensity for the event to
occur is constant and equal to l per
period.
• =POISSON(x;l;0)
T he
l ambda
0 . 3
0 . 2 5
0 . 2
• Example
0 . 1 5
– Norway has about two oil spills per
year in coastal waters. The
probability that we will have x
spills in a certain year is
0 . 1
0 . 0 5
0
• =POISSON(x;2;0)
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P ois
01234567891 0
x
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5
=
2
Empirical or subjective distributions
• Based on experience or merely assumed
• Example:
– Future sales of a consumer good
F u t u re
P r oba bi l i ty
0 .2
0 .1
0
N um be r
0
1
2
3
4
5
6
7
8
9
1
1
1
0
1
2
0
0
.1
0
.1
0
.1
0
.1
5
0
.1
5
0
.1
2
0
.0
0
.0
0
.0
8
0
.0
7
0
.0
5
0
.0
3
.0
2
2
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Continuous probability distributions:
Probability densities
– In many situations, any value among an
infinite number can in principle occur
– In practice, the number depends on how
precisely we measure the differences
between them
T he
mean
=
1
0 . 1 2
0 . 1
• Future stock price
• Future employment rates
• Future sales
0 . 0 8
f(x)=density
0 . 0 6
– The probability of a particular value is
therefore zero
– Instead, we use probability densities,
where areas are probabilities
– Example: The normal distribution
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0 . 0 4
0 . 0 2
0
0
5
1 0
1 5
2 0
2 5
x
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Cumulative distributions
• Probability densities can be
represented as cumulative
distributions which make
them easier to handle
=
1
1 . 2
1
0 . 8
0 . 6
• Important parameters
0 . 4
0 . 2
– Median m, F(m) = 0.5
– Fractiles
– The median is the 50% fractile
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mean
F(x)
– The probability of at least x
– y = F(x)
– F(a) = P(x<a)
T he
cu
dis t rib
0
0
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5
1 0
1 5
2 0
2 5
x
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Conditional probability
The value of tests
• A production process produces defect
units with probability 0.1
• If an OK unit is shipped, the reward is
100
• If it is defect, a loss of 160 is incurred
• A unit can be reworked at an expense
of 40 and becomes OK regardless of
previous state
• A test with sensitivity 0.7 and
specificity 0.8 may be performed
before any decision is made
• What should you be willing to pay for
the test?
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Fred Wenstøp: MBA3
Production
P(OK) = 0.9
P(D) = 0.1
Test
P(TD|D) = 0.7
P(TOK|D) = 0.3
P(TOK|OK) = 0.8
P(TD|OK) = 0.2
9
Probability tree to represent
conditional probabilities
0.72
TOK
P(OK) = 0.9
0.8
OK
0.9
0.1
D
TD
0.2
TOK
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P(D) = 0.1
0.18
Test
0.03
P(TD|D) = 0.7
P(TOK|D) = 0.3
0.3
0.7
Production
TD
P(TOK|OK) = 0.8
0.07
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P(TD|OK) = 0.2
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Transforming a probability tree
to a decision oriented tree
0.72
0.72
TOK
OK
0.8
OK
0.9
0.1
D
TD
0.2
0.18
TOK
0.75
D
0.04
0.03
0.03
TOK
0.25 TD
0.3
0.7
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0.96
TD
0.07
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0.18
OK
0.72
0.28
D
0.07
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Decision analysis
Rework
60
60
D: 0.1
Ship
82.2
OK: 0.9
74
-160
100
D: 0.04
OK: 0.96
89.6
100
Ship
Rework
82.2
TOK: 0.75
89.6
Test
27.2
TD: 0.25
60
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60
D: 0.28
Ship
Rework
Fred Wenstøp: MBA3
-160
OK: 0.72
-160
100
60
12