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Basic notions of probability theory
• Discrete Random Variables
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Contents
o Basic Definitionso Boolean Logico Definitions of probabilityo Probability lawso Random variableso Probability Distributions
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Random variables
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Random variablesExperiment: 𝜀𝜀Sample space: ΩGeneric outcome: 𝜔𝜔
Ω𝜔𝜔
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Random variablesExperiment: 𝜀𝜀Sample space: ΩGeneric outcome: 𝜔𝜔
Ω𝜔𝜔
X(ω) random variable in ℛ
ℛ𝑋𝑋(𝜔𝜔)
Univocal mapping
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Random variable - Example
Ω1 2
X(ω) in ℜ
ℛ
Univocal mapping
34 5 6
1 2 3 4 5 6
Experiment: 𝜀𝜀 = {throwing a die}Sample space: Ω={1,2,3,4,5,6}Generic outcome: 𝜔𝜔
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Random variable - EventExperiment: 𝜀𝜀 = {throwing a die}Sample space: Ω={1,2,3,4,5,6}Event: 𝐸𝐸1= {1,2,3,4}
𝐸𝐸2= ∅
Ω1 2
X(ω)
𝐸𝐸1={X<4.236}𝐸𝐸2={X<0}𝐸𝐸3={X<+∞}
ℛ
Univocal mapping
34 5 6
1 2 3 4 5 6
𝐸𝐸3 = Ω
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Random variables
X(ω) random variable in ℛ
General mathematical models of random behaviours (It is not necessary to speak of the physical process)
They apply to different physical phenomena which behave similarly
Experiment: 𝜀𝜀Sample space: ΩGeneric outcome: 𝜔𝜔
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Probability distributions for reliability, safety and risk
analysis
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Probability functions (I)
o 𝐹𝐹𝑋𝑋 𝑥𝑥 = 𝑃𝑃 𝑋𝑋 ≤ 𝑥𝑥o Properties:
•
•
0)(lim =−∞→
xFXx
1)(lim =+∞→
xFXx
• Cumulative Distribution Function (cdf)
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Probability functions (I)
o 𝐹𝐹𝑋𝑋 𝑥𝑥 = 𝑃𝑃 𝑋𝑋 ≤ 𝑥𝑥o Properties:
•
•
• FX(x) is a non-decreasing function of x• The probability that X takes on a value in the interval [a,b] is:
FX(x)
x
0)(lim =−∞→
xFXx
1)(lim =+∞→
xFXx
{ } )()( aFbFbXaP XX −=≤<
• Cumulative Distribution Function (cdf)
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• discrete probability distributions
Probability distributions for reliability, safety and risk analysis:
• continuous probability distributions
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• 𝑋𝑋 – random variable takes discretevalues 𝑥𝑥𝑖𝑖 , 𝑖𝑖 = 1, … ,𝑛𝑛 Probability mass function:
𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)=P{X= xi}=𝑝𝑝𝑖𝑖
x
0.096
0.384
0 1 2 3
0.008
pX(x) 0.512
• Probability Mass Function (pmf)
Probability functions (II, discrete random variables)
𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)
�𝑖𝑖=1
𝑛𝑛
𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖) = 1
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• 𝑋𝑋 – random variable takes discretevalues 𝑥𝑥𝑖𝑖 , 𝑖𝑖 = 1, … ,𝑛𝑛 Probability mass function:
𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)=P{X= xi}=𝑝𝑝𝑖𝑖
Cumulative distribution function:x
0.096
0.384
0 1 2 3
0.008
pX(x)
0.104
0.616
1.00
0 1 2 3
0.008
FX(x)
0.512
• Probability Mass Function (pmf)
Probability functions (II, discrete random variables)
𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)
𝐹𝐹𝑋𝑋 𝑥𝑥 = 𝑃𝑃 𝑋𝑋 ≤ 𝑥𝑥 =
= �𝑖𝑖: 𝑥𝑥𝑖𝑖≤𝑥𝑥
𝑓𝑓(𝑥𝑥𝑖𝑖)
�𝑖𝑖=1
𝑛𝑛
𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖) = 1
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Summary measures: median, variance, …
• Mean Value (Expected Value):
𝜇𝜇𝑋𝑋 = 𝐸𝐸 𝑥𝑥 = �𝑖𝑖=1
𝑛𝑛
𝑥𝑥𝑖𝑖𝑝𝑝𝑖𝑖
• Variance:
It is a measure of the dispersion of the values around the mean
Where the probability mass is concentrated on average?
𝑉𝑉𝑉𝑉𝑉𝑉 𝑋𝑋 = 𝜎𝜎𝑋𝑋2 = �𝑖𝑖=1
𝑛𝑛
𝑥𝑥𝑖𝑖 − 𝜇𝜇𝑋𝑋 2𝑝𝑝𝑖𝑖
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Exercise. 1 (Probabilty Mass Function)
Ω‘head’ ‘tail’
𝑥𝑥0 1𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)
Experiment: 𝜀𝜀 = {tossing a fair coin}
Questions:1. 1. Draw the probability mass function2. 2. Draw the cumulative distribution
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Ω‘head’ ‘tail’
𝑥𝑥0 1
𝑥𝑥0 1
Probability mass function:𝑓𝑓𝑋𝑋(0) = 𝑃𝑃 𝑋𝑋 = 0 = 0.5𝑓𝑓𝑋𝑋(1) = 𝑃𝑃 𝑋𝑋 = 1 = 0.5
𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)
0.5
1
Experiment: 𝜀𝜀 = {tossing a coin}Exercise. 1 (Probabilty Mass Function)
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Ω‘head’ ‘tail’
𝑥𝑥0 1
𝑥𝑥0 1
𝑥𝑥0 1
Probability mass function:𝑓𝑓𝑋𝑋(0) = 𝑃𝑃 𝑋𝑋 = 0 = 0.5𝑓𝑓𝑋𝑋(1) = 𝑃𝑃 𝑋𝑋 = 1 = 0.5
Cumulative distribution
𝑓𝑓𝑋𝑋(𝑥𝑥𝑖𝑖)
𝐹𝐹𝑋𝑋(𝑥𝑥)
0.5
0.5
1
1
Experiment: 𝜀𝜀 = {tossing a coin}Exercise. 1 (Probabilty Mass Function)
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Exercise 2: Probability Mass Distribution
In the district of the ‘big red lakes’, there are 5 oil wells. Let 𝑋𝑋 indicates the number of operating oil wells. According to some data, the government assumes that the following equation properly describes the number of operating wells.
𝑓𝑓𝑋𝑋(𝑖𝑖) = 𝑃𝑃 𝑋𝑋 = 𝑖𝑖 = 𝐴𝐴 � 𝑖𝑖 with 𝑖𝑖 = 0,1, … , 5
You are required to:a) Compute Ab) What is the probability that only three or less oil wells are operative?c) What is the mean and standard deviation of X?
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Exercise 2: Probability Mass Distribution (Solution)
In the district of the ‘big red lakes’, there are 5 oil wells. Let X indicates the number of operating oil wells. According to some data, the government assumes that the following equation properly describes the number of operating wells.
𝑓𝑓𝑋𝑋(𝑖𝑖) = 𝑃𝑃 𝑋𝑋 = 𝑖𝑖 = 𝐴𝐴 � 𝑖𝑖 with 𝑖𝑖 = 0,1, … , 5
You are required to:a) Compute A
From ∑𝑖𝑖=05 𝑓𝑓𝑋𝑋(𝑖𝑖) = 1 𝐴𝐴 0 + 1 + 2 + 3 + 4 + 5 = 1 𝐴𝐴 = 115
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In the district of the ‘big red lakes’, there are 5 oil wells. Let X indicates the number of operating oil wells. According to some data, the government assumes that the following equation properly describes the number of operating wells:
𝑓𝑓𝑋𝑋(𝑖𝑖) = 𝑃𝑃 𝑋𝑋 = 𝑖𝑖 = 𝐴𝐴 � 𝑖𝑖 with 𝑖𝑖 = 0,1, … , 5
You are required to:a) Compute Ab) What is the probability that only three or less oil wells are operative?
𝑃𝑃 𝑋𝑋 ≤ 3 = 𝑓𝑓𝑋𝑋 0 + 𝑓𝑓𝑋𝑋 1 +𝑓𝑓𝑋𝑋 2 + 𝑓𝑓𝑋𝑋 3 = 𝐴𝐴 0 + 1 + 2 + 3 = 6𝐴𝐴 =6
15
Exercise 2: Probability Mass Distribution (Solution)
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In the district of the ‘big red lakes’, there are 5 oil wells. Let X indicates the number of operating oil wells. According to some data, the government assumes that the following equation properly describes the number of operating wells:
𝑓𝑓𝑋𝑋(𝑖𝑖) = 𝑃𝑃 𝑋𝑋 = 𝑖𝑖 = 𝐴𝐴 � 𝑖𝑖 with 𝑖𝑖 = 0,1, … , 5
You are required to:a) Compute Ab) What is the probability than only three or less oil wells are operative?c) What is the mean and standard deviation of X?
𝜇𝜇𝑋𝑋 = 𝐸𝐸 𝑥𝑥 = �𝑖𝑖=0
𝑛𝑛
𝑖𝑖 � 𝑝𝑝𝑖𝑖 = �𝑖𝑖=0
𝑛𝑛
𝑖𝑖2 � 𝐴𝐴 = 𝐴𝐴 1 + 4 + 9 + 16 + 25 = 𝐴𝐴 � 55 =5515
= 3,66
𝜎𝜎𝑋𝑋2 = �𝑖𝑖=0
𝑛𝑛
𝑖𝑖 − 𝜇𝜇𝑋𝑋 2𝑝𝑝𝑖𝑖 = �𝑖𝑖=0
𝑛𝑛
𝑖𝑖 − 𝜇𝜇𝑋𝑋 2 � 𝐴𝐴 � 𝑖𝑖 = ⋯ = 1,55 → 𝜎𝜎𝑋𝑋 = 1.25
Exercise 2: Probability Mass Distribution (Solution)
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Univariate discrete probability distributions
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Exercise 3A contractor is planning the purchase of equipment, including bulldozers,
needed for a new project in a remote area. Suppose that from hisprevious experience, he figures there is a 80% chance that eachbulldozer can last at least 6 months without any breakdown.
• If he has purchased 3 bulldozers, what is the probability that there will be only 1 bulldozer left operative in 6 months?
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Exercise 3: Solutiono G = event where a Bulldozer is in good condition after
6 months.o B = event where a Bulldozer is in bad condition after
6 month.
o The possible statuses of the three bulldozers would be: {GGG, GGB, GBG, GBB, BGG, BGB, BBG, BBB}
o The probability of having only 1 bulldozer left operative is:
𝑃𝑃 𝑜𝑜𝑛𝑛𝑜𝑜𝑜𝑜 1 𝑏𝑏𝑏𝑏𝑜𝑜𝑜𝑜𝑏𝑏𝑜𝑜𝑏𝑏𝑏𝑏𝑉𝑉 𝑖𝑖𝑖𝑖 𝑜𝑜𝑝𝑝𝑏𝑏𝑉𝑉𝑉𝑉𝑜𝑜𝑖𝑖𝑜𝑜𝑏𝑏 = 𝑃𝑃 𝐺𝐺𝐺𝐺𝐺𝐺 + 𝑃𝑃 𝐺𝐺𝐺𝐺𝐺𝐺 + 𝑃𝑃 𝐺𝐺𝐺𝐺𝐺𝐺= 3 ∗ 0.8 � 0.2 � 0.2 = 0.096
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Exercise 3 (continue)• Let X be the random variable whose values represent the number of
good bulldozers after 6 months. The probability that a bulldozer will remain operational after 6 months is p = 0.8. Using the above information, you are required to:
1. plot the probability mass function (PMF) as well as the cumulative distribution function (CDF) of X.
2. Compute the following quantities:• Mean of X• Variance of X• Standard deviation of X
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x
0.096
0.384
0 1 2 3
0.008
pX(x)
0.104
0.488
1.00
0 1 2 3
0.008
FX(x)
Exercise 3: Solution
• X = {0, 1, 2, 3}• p=P{no failure in (0,6 months)}=0.8
• PX (0) = (1-p)3=0.008• PX (1) = 3p(1–p)2=0.096• PX (2) = 3p2(1–p)=0.384• PX (3) = p3 = 0.512
0.512
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Exercise 3: Solution
o Mean of X
E(X) = 0(0.008) + 1(0.096) + 2(0.384) + 3(0.512) = 2.40
o Var(X)
Var(X) = 0.008(0 – 2.4)2 + 0.096(1 – 2.4)2 + 0.384(2 – 2.4)2 + 0.512(3 –2.4)2 = 0.48
o Standard deviation
Standard deviation = Var(X)1/2=0.481/2= 0.69
…= 3 � 𝑝𝑝
… = 3 � 𝑝𝑝 �(1- 𝑝𝑝)
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Univariate discrete probability distributions:
1) binomial distribution2) geometric distribution3) Poisson distribution
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Univariate Discrete Distributions: Binomial Distribution (I)
Y = discrete random variable with only two possible outcomes:
• Y=1 (success) with P{Y=1}=p
• Y=0 (failure) with P{Y=0}=1-p
We perform n different trials of the experiment, 𝑌𝑌1, … ,𝑌𝑌𝑛𝑛
X= discrete random variable counting the number of success out of the n trial (independentlyfrom the sequence with which successes appear):
𝑋𝑋 = ∑𝑖𝑖=1𝑛𝑛 𝑌𝑌𝑖𝑖 Ω = 0,1,2, … ,𝑛𝑛
The probability mass function:
𝑛𝑛𝑥𝑥 =binomial coefficient= 𝑛𝑛!
𝑛𝑛−𝑥𝑥 !𝑥𝑥!
x=0,1, 2,…,n
Bernoulli process
𝑏𝑏 𝑥𝑥;𝑛𝑛,𝑝𝑝 = 𝑛𝑛𝑥𝑥 𝑝𝑝𝑥𝑥 1 − 𝑝𝑝 𝑛𝑛−𝑥𝑥 with
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Univariate Discrete Distributions: Binomial Distribution (II)
)1(][][
pnpXVarnpXE
−==
x=0,1, 2,…n𝑏𝑏 𝑥𝑥;𝑛𝑛,𝑝𝑝 = 𝑛𝑛𝑥𝑥 𝑝𝑝𝑥𝑥 1 − 𝑝𝑝 𝑛𝑛−𝑥𝑥 with
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Exercise 320% of the items produced by a machine are defective. You are required to find:• the probability distribution of the number of healthy items before a
defective item is found. • What is the probability of producing more than 2 consecutive healthy
items?
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Exercise 3 (Solution)20% of items produced by a machine are defective. You are required to find:• the probability distribution of the number of healthy items produced
before a defective item is found.
Probability that an item is defective: 𝑝𝑝 = 0.2 Let T be the number of healthy items produced before a defective item
g 𝑜𝑜 = 0; 𝑝𝑝 = 𝑃𝑃 first item produced is defective = 𝑝𝑝
T1
Defective item
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Exercise 3 (Solution)20% of items produced by a machine are defective. You are required to find:• the probability distribution of the number of healthy items produced
before a defective item is found.
Probability that an item is defective: 𝑝𝑝 = 0.2 Let T be the number of healthy items produced before a defective item
g 𝑜𝑜 = 0; 𝑝𝑝 = 𝑃𝑃 first item produced is defective = 𝑝𝑝
g 𝑜𝑜 = 1; 𝑝𝑝 = 𝑃𝑃 first item produced is healthyand second item produced is defective = (1 − 𝑝𝑝)𝑝𝑝
T1
T1 2
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Exercise 3 (Solution)20% of items produced by a machine are defective. You are required to find:• the probability distribution of the number of healthy items produced
before a defective item is found.
Probability that an item is defective: 𝑝𝑝 = 0.2 Let T be the number of healthy items produced before a defective item
g 𝑜𝑜 = 0; 𝑝𝑝 = 𝑃𝑃 first item produced is 𝑏𝑏𝑏𝑏𝑓𝑓𝑏𝑏𝑑𝑑𝑜𝑜𝑖𝑖𝑜𝑜𝑏𝑏 = 𝑝𝑝
g 𝑜𝑜 = 1; 𝑝𝑝 = 𝑃𝑃 first item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜and second item produced is 𝑏𝑏𝑏𝑏𝑓𝑓𝑏𝑏𝑑𝑑𝑜𝑜𝑖𝑖𝑜𝑜𝑏𝑏 = (1 − 𝑝𝑝)𝑝𝑝
T1
T1 2
T
g 𝑜𝑜;𝑝𝑝 = 𝑃𝑃
first item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜second item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜
…𝑜𝑜 − 1 − th item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜𝑜𝑜 − th item produced is healthy𝑜𝑜 + 1 itemproduced is 𝑏𝑏𝑏𝑏𝑓𝑓𝑏𝑏𝑑𝑑𝑜𝑜𝑖𝑖𝑜𝑜𝑏𝑏
= 1 − 𝑝𝑝 𝑡𝑡 𝑝𝑝
1 2 t t+1
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Exercise 320% of item produced by a machine are defective. You are required to find:• the probability distribution of the number of healthy items before a
defective item is found. • What is the probability of producing more than 2 consecutive healthy items?
g 𝑜𝑜; 𝑝𝑝 = 𝑃𝑃
first item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜second item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜
…𝑜𝑜 − 1 − th item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜𝑜𝑜 − th item produced is ℎ𝑏𝑏𝑉𝑉𝑜𝑜𝑜𝑜ℎ𝑜𝑜
𝑜𝑜 + 1 − th item produced is 𝑏𝑏𝑏𝑏𝑓𝑓𝑏𝑏𝑑𝑑𝑜𝑜𝑖𝑖𝑜𝑜𝑏𝑏
= 1 − 𝑝𝑝 𝑡𝑡 𝑝𝑝
𝑃𝑃 more than 2 consecutive healthy items are produced =
= �𝑡𝑡=3
+∞
𝑔𝑔 𝑜𝑜; 𝑝𝑝 = 1 −�𝑡𝑡=0
2
𝑔𝑔 𝑜𝑜; 𝑝𝑝 = 1 − 𝑝𝑝 + 1 − 𝑝𝑝 𝑝𝑝 + 1 − 𝑝𝑝 2𝑝𝑝 = 0.51
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Univariate discrete probability distributions:1) binomial distribution2) geometric distribution3) Poisson distribution
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Univariate Discrete Distributions, Geometric Distribution
𝑝𝑝 = P{failure}
T= trail of the first experiment whose outcome is “failure”
The probability mass function:
ppptg t 1)1();( −−= t=1, 2,…
Expected value:
pppppppptTE t
t
1)]1(1[
...])1(3)1(21[)1(][ 221
1=
−−=+−+−+=−= −
∞
=∑
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Univariate Discrete Distributions, Geometric Distribution
T= trail of the first failure (or number of trials between two successive occurrences of failure);𝑝𝑝 = P{success}
The probability mass function:
ppptg t 1)1();( −−= t=1, 2,…
Expected value (return period):
pppppppptTE t
t
1)]1(1[
...])1(3)1(21[)1(][ 221
1=
−−=+−+−+=−= −
∞
=∑
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Exercise 4The Christmas tree of New York has 25000 light bulbs. The probability that a single light bulb fails during the first month of service is p=0.004. You are required to find:• the probability that less than 100 light bulbs fail during the first month of
service
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Exercise 4The Christmas tree of the city has 25000 light bulbs. The probability that a single light bulb fails during the first month of service is p=0.004. You are required to find:• the probability that less than 100 light bulbs fail during the first month of
service
x=0,1, 2,…n𝑏𝑏 𝑥𝑥;𝑛𝑛,𝑝𝑝 = 𝑛𝑛𝑥𝑥 𝑝𝑝𝑥𝑥 1 − 𝑝𝑝 𝑛𝑛−𝑥𝑥 with
Binomial distribution:
𝑃𝑃 X<100 = �𝑥𝑥=0
99
𝑏𝑏 𝑥𝑥;𝑛𝑛 = 25000,𝑝𝑝 = 0.004 = �𝑥𝑥=0
99𝑛𝑛𝑥𝑥 𝑝𝑝𝑥𝑥 1 − 𝑝𝑝 𝑛𝑛−𝑥𝑥
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Exercise 4The Christmas tree of the city has 25000 light bulbs. The probability that a single light bulb fails during the first month of service is p=0.004. You are required to find:• the probability that less than 100 light bulbs fail during the first month of
service
x=0,1, 2,…n𝑏𝑏 𝑥𝑥;𝑛𝑛,𝑝𝑝 = 𝑛𝑛𝑥𝑥 𝑝𝑝𝑥𝑥 1 − 𝑝𝑝 𝑛𝑛−𝑥𝑥 with
Binomial distribution:
𝑃𝑃 X<100 = �𝑥𝑥=0
99
𝑏𝑏 𝑥𝑥;𝑛𝑛 = 25000,𝑝𝑝 = 0.004 = �𝑥𝑥=0
99𝑛𝑛𝑥𝑥 𝑝𝑝𝑥𝑥 1 − 𝑝𝑝 𝑛𝑛−𝑥𝑥
Notice that the binomial coefficients 25000x
are too big to be computed (also in Matlab)!
Approximation of the binomial distribution in the case of:
𝑝𝑝 → 0 𝑛𝑛 → ∞ 𝑏𝑏 𝑥𝑥;𝑛𝑛, 𝑝𝑝 =
𝑛𝑛𝑝𝑝𝑥𝑥!
𝑥𝑥
𝑏𝑏−𝑛𝑛𝑛𝑛 P(X<100)= 0.53
Piero Baraldi
Approximation of the binomial distribution in the case of: 𝑝𝑝 → 0 𝑛𝑛 → ∞
It depends from only one parameter: 𝜇𝜇 = 𝑛𝑛𝑝𝑝=100 = E[X]
which can be interpreted as the average number of faulty light bulbs in 1 month.
𝑏𝑏 𝑥𝑥;𝑛𝑛, 𝑝𝑝 =𝑛𝑛𝑝𝑝𝑥𝑥!
𝑥𝑥
𝑏𝑏−𝑛𝑛𝑛𝑛
𝑏𝑏 𝑥𝑥;𝑛𝑛, 𝑝𝑝 =𝑛𝑛𝑝𝑝𝑥𝑥!
𝑥𝑥
𝑏𝑏−𝑛𝑛𝑛𝑛 → 𝜋𝜋 𝑥𝑥; 𝜇𝜇 =𝜇𝜇𝑥𝑥!
𝑥𝑥
𝑏𝑏−𝜇𝜇 𝑝𝑝 → 0 𝑛𝑛 → ∞
From the binomial to the poisson distribution
Piero Baraldi
Univariate discrete probability distributions:1) binomial distribution2) geometric distribution3) Poisson distribution
Piero Baraldi
k=0,1, 2,…
Univariate Discrete Distributions, Poisson Distribution
Stochastic events that occur in a (continuous) period of time (e.g. failures, earthquakes,…):
• Rate of occurrence, λ, is constant
• Discrete Random Variable:
𝐾𝐾 = number of events in the period of observation (0, 𝑜𝑜)
• Probability mass function: tk
ekttkp λλλ −=!)()),,0(;(
tKVartKEλ
λ=
=][
][
Piero Baraldi
Exercise 5
The occurrences of flood may be modelled by a Poisson process with constant rate of occurrence υ. Let p(k; t, υ) denote the probability of kflood occurrences in t years.
If the mean occurrence rate of floods for a certain region A is once every 8 years, you are required to find:• the probability of no floods in a 10-year period• of 1 flood in a 10-year period• of more than 3 floods in a 10-year period.
Piero Baraldi
Exercise 5: Solution
P{k = 0 in 10 years}= e-1.25 = 0.286P{k = 1 in 10 years}= 1.25(e-1.25) = 0.3525P{k > 3 in 10 years}= 1–P{k ≤ 3 in 10 years}=
= 0.0394
2 331.25 1.25
0
1.25 1.251 ( ; 10, 0.125) 1 0.286 0.35752! 3!k
p k t e eυ − −
=
= − = = = − − − −∑
Mean occurrence rate of floods: once every 8 yearsBeing 𝐸𝐸 𝑘𝑘 = 𝜈𝜈 � 𝑜𝑜 → 1 = 𝜈𝜈 � 8 𝑜𝑜 → 𝜈𝜈 = 1
8𝑜𝑜−1
0.3525
Piero Baraldi
Exercise 5 (continue)
The occurrences of flood may be modelled by a Poisson process with rate υ. Let p(k; t, υ) denote the probability of k flood occurrences in t years.
If the mean occurrence rate of floods for a certain region A is once every 8 years, you are required to find• the probability of no floods in a 10-year period• of 1 flood in a 10-year period• of more than 3 floods in a 10-year period.
Consider a condenser of an electricity production plant in region A. If a flood occurs, the probability that it fails is 0.05. Compute:• the probability that the condenser will not fail over the 10-year
period.
Piero Baraldi
Exercise 5 (continue): SolutionP{condenser fails | flood} = 0.05 P{condenser survives | flood} = 0.95
condenser survives over 10 𝑜𝑜 =
= �𝑘𝑘=0
+∞
(condenser survives over 10 y) AND (𝑘𝑘 floods over 10 years)
P{condenser survives AND 0 flood}= P{0 flood} P{condenser survives | 0 flood}= 0.286 (1) = 0.286
P{condenser survives AND 1 flood}= P{1 flood} P{condenser survives | 1 flood}= (0.3525)(0.95)= 0.3396
P{condenser survives AND k independent floods}=P{k floods}P{condenser survives|k independent floods}=
P{condenser survives over 10 years} =
kk
ek
)95,0(!
25.1 25.1−
9394.0!
1875.1!
1875.1)95,0(!
25.1
1875.1
0
25.125.1
0
25.1
0
25.1
===
==
∑
∑∑∞+
=
−−
+∞
=
−+∞
=
−
eek
e
ek
ek
k
kk
k
k
kk
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