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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)

Piero Baraldi

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