# Basic notions of probability theory • Discrete Random Variables · 2019. 3. 4. · Probability...

date post

29-Aug-2020Category

## Documents

view

8download

2

Embed Size (px)

### Transcript of Basic notions of probability theory • Discrete Random Variables · 2019. 3. 4. · Probability...

Piero BaraldiPiero Baraldi

Basic notions of probability theory

• Discrete Random Variables

Piero Baraldi

Contents

o Basic Definitionso Boolean Logico Definitions of probabilityo Probability lawso Random variableso Probability Distributions

Piero Baraldi

Random variables

Piero Baraldi

Random variablesExperiment: 𝜀𝜀Sample space: ΩGeneric outcome: 𝜔𝜔

Ω𝜔𝜔

Piero Baraldi

Random variablesExperiment: 𝜀𝜀Sample space: ΩGeneric outcome: 𝜔𝜔

Ω𝜔𝜔

X(ω) random variable in ℛ

ℛ𝑋𝑋(𝜔𝜔)

Univocal mapping

Piero Baraldi

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

Piero Baraldi

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

Piero Baraldi

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

Piero Baraldi

Probability distributions for reliability, safety and risk

analysis

Piero Baraldi

Probability functions (I)

o 𝐹𝐹𝑋𝑋 𝑥𝑥 = 𝑃𝑃 𝑋𝑋 ≤ 𝑥𝑥o Properties:

•

•

0)(lim =−∞→

xFXx

1)(lim =+∞→

xFXx

• Cumulative Distribution Function (cdf)

Piero Baraldi

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)

Piero Baraldi

• discrete probability distributions

Probability distributions for reliability, safety and risk analysis:

• continuous probability distributions

Piero Baraldi

• 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

Piero Baraldi

• 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

Piero Baraldi

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

Piero Baraldi

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

Piero Baraldi

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)

Piero Baraldi

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)

Piero Baraldi

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?

Piero Baraldi

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

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 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)

Piero Baraldi

Univariate discrete probability distributions

Piero Baraldi

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?

Piero Baraldi

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

Piero Baraldi

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

Piero Baraldi

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

Piero Baraldi

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)

Piero Baraldi

Univariate discrete probability distributions:

1) binomial distribution2) geometric distribution3) Poisson distribution

Piero Baraldi

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

Piero Baraldi

Univariate Discrete Distributions: Binomial Distribution (II)

)1(][][

pnpXVarnpXE

−==

x=0,1, 2,…n𝑏𝑏 𝑥𝑥;𝑛𝑛,𝑝𝑝 = 𝑛𝑛𝑥𝑥 𝑝𝑝𝑥𝑥 1 − 𝑝𝑝 𝑛𝑛−𝑥𝑥 with

Piero Baraldi

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?

Piero Baraldi

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

Piero Baraldi

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

Piero Baraldi

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

Piero Baraldi

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

Piero Baraldi

Univariate discrete probability distributions:1) binomial distribution2) geometric distribution3) Poisson distribution

Piero Baraldi

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=

−−=+−+−+=−= −

∞

=∑

Piero Baraldi

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=

−−=+−+−+=−= −

∞

=∑

Piero Baraldi

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

Piero Baraldi

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

Piero Baraldi

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

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

幻灯片编号 1幻灯片编号 2幻灯片编号 3幻灯片编号 4幻灯片编号 5幻灯片编号 6幻灯片编号 7幻灯片编号 8幻灯片编号 9幻灯片编号 10幻灯片编号 11discrete probability distributions�幻灯片编号 13幻灯片编号 14幻灯片编号 15Exercise. 1 (Probabilty Mass Function)Exercise. 1 (Probabilty Mass Function)Exercise. 1 (Probabilty Mass Function)Exercise 2: Probability Mass DistributionExercise 2: Probability Mass Distribution (Solution)Exercise 2: Probability Mass Distribution (Solution)Exercise 2: Probability Mass Distribution (Solution)Univariate discrete probability distributions� 幻灯片编号 24幻灯片编号 25幻灯片编号 26幻灯片编号 27幻灯片编号 28Univariate discrete probability distributions:��1) binomial distribution�2) geometric distribution�3) Poisson distribution� 幻灯片编号 30幻灯片编号 31幻灯片编号 32幻灯片编号 33幻灯片编号 34幻灯片编号 35幻灯片编号 36Univariate discrete probability distributions:�1) binomial distribution�2) geometric distribution�3) Poisson distribution� 幻灯片编号 38幻灯片编号 39幻灯片编号 40幻灯片编号 41幻灯片编号 42幻灯片编号 43Univariate discrete probability distributions:�1) binomial distribution�2) geometric distribution�3) Poisson distribution� 幻灯片编号 45幻灯片编号 46幻灯片编号 47幻灯片编号 48幻灯片编号 49