1

Statistical independence

if E1 and E2 are s.i.

2 1 2( ) ( )P E E P E

1 2 1( ) ( )P E E P Eor

s.i.

2

Example:

E2 = flood in 廣西 on June E3 = flood in 哈爾濱 on June

E1 = flood in 廣東 on June

P(E1) = 0.1; P(E2)=0.1; P(E3) = 0.1

121 3.0| EPEEP E1 and E2 are not s.i.

1.0| 31 EEPE1 and E3 are s.i.

1EP

3

1 2 1 2 2( ) ( ) ( )P E E P E E P E

1( )P Eif E1 and E2 are s.i.

1 2 3 1 2 3( ) ( ) ( ) ( )P E E E P E P E P Eif all are s.i.

4

s.i. and m.e.

if E1 and E2 are m.e.

1 2( ) 0P E E

1 2 1( ) ( )P E E P E

if E1 and E2 are s.i.

5

E2.19

21 EEP

① ②

ring is super ring

steel bar steel bar

P (failure of this bar system) = ?

P(failure)

E1 = bar is weak (under strength)①E2 = bar is weak (under strength)②

2121 EEPEPEP

6

If 5% of bars are weak

1 2 20.05 0.05 ( ) ( )P E E P E s.i.

P(E1)=0.05 0.05

P(failure)

0975.0

05.0105.005.0 if assume perfectly dependent

P(failure)

05.0

P(E1|E2)=1

7

B

C

A

2

1

3

P(E1)=2/5

P(E2)=3/4

P(E3)=2/3

P(E3|E2)=4/5

P(E1|E2E3)=1/2

a) P(go from A to B through C)

32EEP 2 3( ) ( )P E P E

5

3

4

3

5

4

3 2 2( ) ( )P E E P E

E1 : ① is open

8

b)

P(go from A to B)

132 EEEP

132132 EEEPEPEEP

1 2 3 2 3

3 2( | ) ( )

5 5P E E E P E E

7.0 1/2 3/5

9

Theorem of Total Probabilities

P (L = landslides in the next storm) = ?

small rainfall S medium rainfall M heavy rainfall H

Rainfall magnitude (from hydrologist)

P(S) = 2P(M)P(M) = 3P(H)

P(S)+P(M)+P(H) = 1

Example:

H P(L) = 0.9 P(L|H)M P(L) = 0.2 P(L|M)S P(L) = 0.05 P(L|S)

iffrom geotechnical engineer

10

if rainfall magnitude equally likely,P(L) = 1/3(0.9+0.2+0.05)=0.38

P(S) = 0.6 P(M) = 0.3 P(H) = 0.1

P(L) = 0.05×0.6+0.2×0.3+0.9×0.1=0.18

P(L|S) P(S) P(L|M) P(M)P(L|H) P(H)

E1 E2 En

AS

Ei’s are m.e. and c.e.

A = AS

= A(E1E2 … En)

= AE1AE2 … AEn

c.e.

P(A) = P(AE1)+P(AE2) +…+P(AEn)

= P(A|E1)P(E1)+…

rule

m.e.

Example: 4 - way stop intersection

Given information

traffic from E – 60 veh/10min traffic from S - 50 veh/10min traffic from W – 70 veh/10min

traffic from N – 20 veh/10min

P (next vehicle will go east from the intersection) = ?

addition information from similar intersection:

70% of traffic will go straight

20% of traffic will go right turn

10% of traffic will go left turn

E

N

W

S

SPSAPEPEAPAP ||

29.0

what % of traffic will go east after intersection = 29%

next veh. go east

NPNAPWPWAP ||

0

0.7 0.1

0.2200

60

200

50

200

70200

20

14

Bayes Theorem

given the result, ask for the

likelihood of a specific cause.

15

APASP

P(S | A)

AP

SPSAP |

29.0200

502.0 069.0

6.9% of the east bound traffic from the intersection came from the south

what is the probability that it came from the south?

Intersection example

Suppose a vehicle has just gone east

Suppose landslide occurred,

what is the probability that the rain has been just small?

LPLSP

LSP

|

LP

SPSLP |

18.0

6.005.0

18.0

03.0

6

1 167.0

333.018.0

06.0

18.0

3.02.0|

LMP

5.018.0

09.0

18.0

1.09.0|

LHP

Landslide example

T.O.T (Theorem of Total Probabilities)

Bayes theorem

AP

EPEAPAEP jj

j

||

P(A) = P(A|E1)P(E1)+P(A|E2)P(E2)+…+P(A|En)P(En)

Ei’s are m.e. and c.e.

18

0.8P G 0.2P G good enough for construction

| 0.9; | 0.1P T G P T G

9.0|;1.0| GTPGTPpositive

E 2.30 aggregate for construction

engineer's judgment based on geology and experience

crude test

reliability (or quality) is as follows:

not a perfect test

19

After 1 successful test, what is P(G)?

TP

GPGTPTGP

||

0.9 0.8

0.9 0.8 0.1 0.2

0.973

( | ) ( )

( | ) ( ) ( | ) ( )

P T G P G

P T G P G P T G P G

20

After another successful independent test, P(G)?

GPGTPGPGTP

GPGTPTGP

||

||

22

22

0.9 0.973

0.9 0.973 0.1 0.027

0.997

21

What if the two tests were performed at the same time?

)()|()()|(

)()|()|(

2121

2121

GPGTTPGPGTTP

GPGTTPTTGP

1 2

1 2 1 2

( | ) ( | ) ( )

( | ) ( | ) ( ) ( | ) ( | ) ( )

P T G P T G P G

P T G P T G P G P T G P T G P G

0.8

0.20.8

0.9 0.9 0.8

0.9 0.9 0.8 0.1 0.1 0.2

0.997

P(G)

UST

HKU

0.8

0.3

After 1 test

0.973

0.77

After 2 tests

0.997

0.965

… 5

1.0000

0.9999

22

R a i n f a l l

Landfill

Clay

Soil stratumGeomembrane

Supplementary Exercise 2-2-6 (on web page)

A landfill containment system

A layer of clay and geomembrane to prevent the contaminants leaking into soil stratum

poorly compacted clay layer holes in geomembrane extremely heavy rainfall

1. during extremely heavy rainfall, and

either the clay was not well compacted or

there were holes in the geomembrane

(Event I).

2. under ordinary rainfalls (i.e. without

extremely heavy rainfall), but when the

clay was not well compacted and the

geomembrane contained holes (Event II).

Leakage will happen:

24

W = clay well compacted; 90%

H = holes in geomembrane; 30%

E =extremely heavy rainfall; 20%

The quality of construction has no effect on the future amount of rainfall.

If the geomembrane contained holes, the probability of a well-compacted clay is reduced to 60%.

E is s.i. of W or H

6.0)|( HWp

25

a) Express Event I and II

II: WHE

I: ( )E H W

26

b) ))(()( WHEPIP

( ) ( )P E P H W

(0.2) ( ) ( ) ( | ) ( )P H P W P W H P H 0.3 0.1 1-0.6 0.3

56.0

)()( WHEPIIP

( ) ( )P E P HW

3.4.8. 096.

27

c1)

W, H m.e.?

( | ) 0.6 0P W H W, H not m.e.

W, H s.i.?

( | ) 0.6P W H ( ) 0.9P W W, H not s.i.

28

I and II m.e.?

Difficult to prove P (I|II) = 0

E

Ē

W H

( )E H WI =

WHEII =

I and II are m.e.

c2)

Method 1.

29

III HWEHWE )(

HWHWEE ))((

Alternatively,

null set

..em

I and II are not c.e.

because they do not together make up the entire sample space

30

d)

)(leakageP

)( IIIP

)()( IIPIP

the probability of leakage

Puzzle: which friend for dinner?

down train

5:00

5:10

5:20

up train

5:08

5:18

5:28

5:00 5:10 5:20

32

Example: damage of bridge

10% of trucks are overloaded

event of overloaded trucks are s.i.

1 2( | ) 30%P D OO

1 2( | ) 5%P D O O

1 2( | ) 5%P D OO

1 2( | ) 0.1%P D O O

at most two trucks

A small old bridge susceptible to damages from heavy trucks

O1 : truck 1 is overloaded

O2 : truck 2 is overloaded

33

a) )(damgeP

)(DP

1 2 1 2 1 2 1 2

1 2 1 2 1 2 1 2

( | ) ( ) ( | ) ( )

( | ) ( ) ( | ) ( )

P D OO P OO P D OO P OO

P D O O P O O P D O O P O O

0.3 (0.1 0.1) 0.05 (0.9 0.1)

0.05 (0.1 0.9) 0.001 (0.9 0.9)

0128.

O1 O2

s

34

b)P(overloaded truck | D)

=1 - P(no overloaded truck | D)

)|(1 21 DOOP

)(

)()|(1

2121

DP

OOPOODP

0128.

9.9.001.1

937.

35

c)alt. I – strengthen bridge

15.3.2

1)|( 21 OODP

P(D) = 0.0064 from 0.0128

alt. II – improve truck inspection

%6%10 overloaded truck from

which is better?

P(D) = 0.0076

and also for others