Structure of a model for mineral resource potential mapping

Indian Institute of Technology BombayIndian Institute of Technology Bombay

1

Structure of a model for mineral resource potential mapping

∫Integrating function

• linear or non-linear

• parametersInput predictor maps

• Categoric or numeric

• Binary or multi-class

Output mineral potential map • Grey-scale or

binary


GIS-based mineral resource potential mapping - Modelling approaches

Exploration datasets with homogenous coverage – required for all models

Expert knowledge (a knowledge base) and/or

Model parameters estimated from mineral deposits data(Known deposits required)Brownfields explorationExamples - Weights of evidence, Bayesian classifiers, NN, Logistic Regression

Data-driven Knowledge-driven

Hybrid

Mineral deposit data

Model parameters estimated from both mineral deposits dataand expert knowledge (Known deposits necessary)Semi-brownfields to brownfields explorationExamples – Neuro-fuzzy systems

Training data Expert knowledge

Model parameters estimated from expert knowledge (Known deposits not necessary)Greenfields explorationExamples – Fuzzy systems; Dempster-Shafer belief theory


Probabilistic Model (Weights of Evidence):

- used in the areas where there are already some known deposits

- spatial associations of known deposits/oil well with the geological features are used to determine the probability of occurrence of a mineral deposit (or well) in each unit cell of the study area.

Fuzzy Model:

- used in the areas where there are no known mineral deposits

- each geological feature is assigned a weight based on the expert knowledge, these weights are subsequently combined to determine the probability of occurrence of mineral deposit in each unit cell of the study area.

GIS MODELS FOR MINERAL EXPLORATION

Indian Institute of Technology Bombay

Fuzzy Inference Systems for Mineral Prospectivity Mapping


Artificia

l intellig

ence!


IntroductionFuzzy logic:

• A way to represent imprecision in logic and approximate reasoning• A way to make use of natural language in logic

•Humans say things like :• “If it is cool and dry, I will walk faster on my morning walk”•"If it is overcast and warm and humid, it will rain heavily"

•Linguistic variables:• Speed: {Fast, slow}•Temp: {freezing, cool, warm, hot}•Cloud Cover: {overcast, partly cloudy, sunny}•Humid: {high, average, low}•Rain: {Heavy, moderate, light}


Crisp (Traditional) Variables• Crisp variables represent precise quantities:

Temp = 36 deg CHumidity = 70%

• Some thing like:– If the cloud cover is 90% and temperature is 40

degrees C and humidity is 70%, then the rainfall would be 30 mm.

– If the temperature is 25 deg C, I will walk at 20 km/hr on my morning walk.


Fuzzy Sets• Extension of Classical Sets• Not just a membership value of in the set and out the set, 1 and 0

– but partial membership value, between 1 and 0


Example: Height• Tall people: say taller than or equal to 6 feet

– 6’, 6’1”, 6’3” feet are members of this set– 5’11.9” are not members of this set - Is that

reasonable? (measurement can be inaccurate and/or imprecise )


Example: Weekend days


In fuzzy logic, the truth of any statement becomes a matter of degree.

Q: Is Saturday a weekend day?A: 1 (yes, or true)Q: Is Tuesday a weekend day?A: 0 (no, or false)Q: Is Friday a weekend day?A: 0.8 (for the most part yes, but not completely)Q: Is Sunday a weekend day?A: 0.95 (yes, but not quite as much as Saturday).

BINARY FUZZY


Example: Season

BINARY FUZZY

Summer Rainy Autumn Winter Summer Rainy Autumn Winter


MEMBERSHIP FUNCTIONA membership function (MF) is a curve that defines how each point in the input space is mapped to a membership value (or degree of membership) between 0 and 1.

Example: Membership function of a set of tall people

Crisp set

Fuzzy set:

6x1μ6x0μ

x

x

6 if 156 if 5

5 if 0

xxx

x

x

x

x


Example: Height


Membership Function

A = {x | x > 6}

A = {x, µA

(x) | x ∈ X}

A membership function must vary between 0 and 1.

The function itself can be an arbitrary curve whose shape we can define as a function that suits us from the point of view of simplicity, convenience, speed, and efficiency.

A classical set might be expressed as

A fuzzy set is an extension of a classical set. If X is the universe of discourse and its elements are denoted by x, then a fuzzy set A in X is defined as a set of ordered pairs.

µA(x) is called the membership function (or MF) of x in A. The membership function maps each element of X to a membership value between 0 and 1.

µA(x) is called the membership function (or MF) of x in A. The membership function maps each element of X to

a membership value between 0 and 1.


Membership functions• piece-wise linear functions

• the Gaussian distribution function• the sigmoid curve

• quadratic and cubic polynomial curves


Membership Functions• For each variable value a different membership

function is required• Temp: {Freezing, Cool, Warm, Hot}

50 70 90 1103010

Temp. (F°)

Freezing Cool Warm Hot

0

1

5 20 30 40Temp OC


Fuzzy Operators• How do we use fuzzy membership functions in predicate logic?• Fuzzy logic Connectives:

– Fuzzy Conjunction, – Fuzzy Disjunction,

• Operate on degrees of membership in fuzzy sets


Fuzzy Disjunction (= OR Operator)

• AB max(A, B)• AB = C "Quality C is the disjunction of Quality A and

B"

0

1

0.375

A

0

1

0.75

B

(AB = C) (C = 0.75)


Fuzzy Conjunction (=AND Operator)

• AB min(A, B)• AB = C "Quality C is the conjunction of Quality A and

B"

0

1

0.375

A

0

1

0.75

B

(AB = C) (C = 0.375)


Example: Fuzzy Conjunction

Calculate AB given that A is .4 and B is 20

0

1A

0

1B

.1 .2 .3 .4 .5 .6 .7 .8 .9 1 5 10 15 20 25 30 35 40


Example: Fuzzy Conjunction

Calculate AB given that A is .4 and B is 20

0

1A

0

1B

.1 .2 .3 .4 .5 .6 .7 .8 .9 1 5 10 15 20 25 30 35 40

Determine degrees of membership.


Subjective assignment of fuzzy membership values

• It is also possible to assign the fuzzy membership values subjectively (without using a membership function)

Lithotype

Membership value

Granite 0.2Dolerite 0.7Magnetite quartzite

0.9

Diorite 0.4

Distance to a fault

Membership value

0 – 1 km 0.91 – 2 km 0.72 – 3 km 0.53 – 4 km 0.34 – 5 km 0.1> 5 km 0.001


Fuzzy if-then rulesFuzzy if-then rule statements are used to formulate the conditional statements that comprise fuzzy logic.

A single fuzzy if-then rule assumes the form:

if x is A then y is B

where A and B are linguistic values defined by fuzzy sets on the ranges X and Y, respectively.

The if-part of the rule "x is A" is called the antecedent or premise

The then-part of the rule "y is B" is called the consequent or conclusion.


if x is A then y is B

For example: If FeO is high then gold potential is average

“High” is represented as a number between 0 and 1, and so the antecedent is an interpretation that returns a single number between 0 and 1. “Average” is represented as a fuzzy set, and so the consequent is an assignment that assigns the entire fuzzy set B to the output variable y.

Fuzzy if-then rules


The input to an if-then rule is the current value for the input variable (in this case, FeO) and the output is an entire fuzzy set (in this case, gold potential).

The consequent specifies a fuzzy set be assigned to the output.

The implication function then modifies that fuzzy set to the degree specified by the antecedent. The most common ways to modify the output fuzzy set are truncation using the min function, where the fuzzy set is truncated.

This set has to be defuzzified, assigning one value to the output.

40% 50% 60% 70%0

1

1 tonne 1000 tonnes

0

1

0.3

Fuzzy if-then rules

FeO Gold

Indian Institute of Technology Bombay Fuzzy if then rules can be quite complex:

IF sky is gray AND wind is strong AND humidity is high AND temperature low, THEN rainfall will be heavy.

IF Granite is Proximal AND Fault is Proximal AND FeO is high AND SiO2 is low, THEN Gold potential is high.

Several fuzzy if-then rules are combined to generate a Fuzzy Inference System

Fuzzy if-then rules & inference systems


Fuzzy Inference SystemStep 1: Identify the factors that control a system:For example: Formation of a deposit (or mineral potential of an area) depends on the following factors:1. FeO content of the rocks2. SiO2 content of the rocks3. Closeness (or proximity) to granite4. Closeness (or proximity) to faults

Step 2: Identify the variables for each of the factor:1. FeO content: {high, average, low}2. SiO2content: {high, average, low}3. Proximity to granite :{proximal, intermediate, distal}4. Proximity to faults: {proximal, intermediate, distal}

Step 3: Identify the output variable of the systemMineral potential: {high, average, low}


Fuzzy Inference System Step 4: Decide a fuzzy membership function for each variable

30% 50% 70%0

1 High

30% 50% 70%0

1 Average

FeO

40% 55% 70%0

1 High

40% 55% 70%0

1 Average

SiO2

30% 50% 70%0

1 Low

40% 55% 70%0

1 Low



(Contd.)

Granite

0 km 10 km20km

0

1 Proximal

0

1 Intermediate

0 km 10 km20km

0

1 Distal

0 km 10 km20km

Fault

0 km 5 km20km

0

1 Proximal

0

1 Intermediate

0 km 5 km20km

0

1 Distal

0 km 5 km10km



(Contd.)

Mineral potential

0

1 Low

0

1 Average

0

1 High

0t 100t 1000t

0t 100t 1000t

0t 100t 1000t


Fuzzy Inference SystemStep 5: Develop a set of fuzzy if-then rules to explain the behavior of the system

Rule1: IF FeO is high AND SiO2 is low AND Granite is proximal AND Fault is proximal, THEN mineral potential is high.

Rule 2: IF FeO is average AND SiO2 is high AND Granite is intermediate AND Fault is proximal, THEN mineral potential is average.

Rule 3: IF FeO is low AND SiO2 is high AND Granite is distal AND Fault is distal, THEN mineral potential is low.


Input data preparationStep 6: Rasterize the input predictor maps, combine them, and generate feature vectors

Input feature vector[3, 8, 33, 800]

GIS raster layers

MgO%

Rock type

FeO%

Distance to Fault


Step 7: Represent fuzzy if-then rules in terms of membership functions

1: IF FeO is high & SiO2 is low & Granite is prox & Fault is prox, THEN metal is highImplication (Max)

0

1

0

1

=

2: IF FeO is aver & SiO2 is high & Granite is interm & Fault is prox, THEN metal is aver

30% 50% 70%

0

1

40% 55% 70%

0 km 10 km 20km

0 km 5 km 10km

0t 100t 1000t

3: IF FeO is low & SiO2 is high & Granite is dist & Fault is dist, THEN metal is low

FeO = 60% SiO2 = 60% Granite = 5 km Fault = 1 km Metal = ?

0t 100t 1000t

=

=


Step 8: Combine outputs of each rule

Rule 1: Rule 2: Rule 3: Aggregate (Max)

+ + =

Defuzzify (Find centroid)

125 tonnes metal

Formula for centroid

n

ii

n

iii

x

xx

0

0

)(

)(


Weights of Evidence Model for Mineral Prospectivity Mapping


37

Probabilistic model (Weights of Evidence)

• What is needed for the WofE calculations?– A training point layer –

i.e. known mineral deposits;

– One or more predictor maps in raster format.


PROBABILISTIC MODELS (Weights of Evidence or WofE)

Four steps:1. Convert multiclass maps to binary maps2. Calculation of prior probability3. Calculate weights of evidence (conditional probability) for

each predictor map4. Combine weights


• The probability of the occurrence of the targeted mineral deposit type when no other geological information about the area is available or considered.

Study area (S)

Target deposits D

Assuming- 1. Unit cell size = 1 sq km2. Each deposit occupies 1 unit cell

Total study area = Area (S) = 10 km x 10 km = 100 sq km = 100 unit cells

Area where deposits are present = Area (D) = 10 unit cells

Prior Probability of occurrence of deposits = P {D} = Area(D)/Area(S)= 10/100 = 0.1

Prior odds of occurrence of deposits = P{D}/(1-P{D}) = 0.1/0.9 = 0.1110k

10k

1k1k

Calculation of Prior Probability


40

Convert multiclass maps into binary maps

• Define a threshold value, use the threshold for reclassification

Multiclass map

Binary map

Indian Institute of Technology Bombay • How do we define the threshold?

Use the distance at which there is maximum spatial association as the threshold !



• Spatial association – spatial correlation of deposit locations with geological feature.

A

BC

D

A

BC

D

10km

10km

1km

1km

Gold Deposit (D) Study area (S)



A

B C

D

Which polygon has the highest spatial association with D?More importantly, does any polygon has a positive spatial association with D ???

What is the expected distribution of deposits in each polygon, assuming that they were randomly distributed? What is the observed distribution of deposits in each polygon?

Positive spatial association – more deposits in a polygon than you would expect if the deposits were randomly distributed.

If observed >> expected; positive associationIf observed = expected; no association If observed << expected; negative association



A

B C

D

Area (A) = n(A) = 25; n(D|A) = 2Area (B) = n(A) = 21; n(D|B) = 2Area(C) = n(C) = 7; n(D|C) = 2 Area(D) = n(D) = 47; n(D|D) = 4Area (S) = n(S) = 100; n(D) = 10

OBSERVED DISTRIBUTION



A

B C

DArea (A) = n(A) = 25; n(D|A) = 2.5Area (B) = n(A) = 21; n(D|B) = 2.1Area(C) = n(C) = 7; n(D|C) = 0.7Area(D) = n(D) = 47; n(D|D) = 4.7(Area (S) = n(S) = 100; n(D) = 10)

EXPECTED DISTRIBUTION

Expected number of deposits in A = (Area (A)/Area(S))*Total number of deposits


Indian Institute of Technology BombayA

B C

D

Area (A) = n(A) = 25; n(D|A) = 2.5

Area (B) = n(A) = 21; n(D|B) = 2.1

Area(C) = n(C) = 7; n(D|C) = 0.7

Area(D) = n(D) = 47; n(D|D) = 4.7

(Area (S) = n(S) = 100; n(D) = 10)

EXPECTED DISTRIBUTIONArea (A) = n(A) = 25; n(D|A) = 2

Area (B) = n(A) = 21; n(D|B) = 2

Area(C) = n(C) = 7; n(D|C) = 2

Area(D) = n(D) = 47; n(D|D) = 4

Area (S) = n(S) = 100; n(D) = 10

OBSERVED DISTRIBUTION

Only C has positive association!So, A, B and D are classified as 0; C is classified as 1.

Another way of calculating the spatial association : = Observed proportion of deposits/ Expected proportion of deposits= Proportion of deposits in the polygon/Proportion of the area of the polygon= [n(D|A)/n(D)]/[n(A)/n(S)] • Positive if this ratio >1• Nil if this ratio = 1 • Negative if this ratio is < 1



LA

BC

D

10km

10km

1km

1km

Gold Deposit (D) Study area (S)

Convert multiclass maps into binary maps – Line features


1km

1km

Gold Deposit (D)

10

0

000

0

000

11

112

11

1

1111

111

111

1

11

2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 2 3 4 5 6 7 8

2 3 4 5 6 7 2 3 4 5 6 2 3 4 5 6

2 3 4 5 2 3 4 2 3 4 2 3 4

3 2 3 2 4 3 2 4 3 2

5 4 3 2 5 4 3 2

Distance from

the fault

No. of

pixels

No of deposit

s

Ratio (Observe

d to Expected

)0 9 1 1.11 21 3 1.42 17 0 0.03 16 3 1.94 14 2 1.45 9 0 0.06 6 0 0.07 4 0 0.08 3 1 3.39 1 0 0.0



• Calculate observed vs expected distribution of deposits for cumulative distances

Gold Deposit (D)

10

0

000

0

000

11

112

11

1

1111

111

111

1

11

2 3 4 5 6 7 8 9 2 3 4 5 6 7 8 2 3 4 5 6 7 8

2 3 4 5 6 7 2 3 4 5 6 2 3 4 5 6

2 3 4 5 2 3 4 2 3 4 2 3 4

3 2 3 2 4 3 2 4 3 2

5 4 3 2 5 4 3 2

Distance from the fault

No. of pixels

Cumul No. of pixels

No of deposits

Cumul No. of deposits

Ratio (Observed to Expected)

0 9 9 1 1 1.1

1 21 30 3 4 1.3

2 17 47 0 4 0.9

3 16 63 3 7 1.1

4 14 77 2 9 1.2

5 9 86 0 9 1.0

6 6 92 0 9 1.0

7 4 96 0 9 0.9

8 3 99 1 10 1.0

9 1 100 0 10 1.0

=< 4 – positive association (Reclassified into 1)>4 – negative association (Reclassified into 0)



Weights of evidence ~ quantified spatial associations of deposits with geological features

Study area (S)

10k

Target deposits10k

Unit cell

1k1k

Objective: To estimate the probability of occurrence of D in each unit cell of the study area

Approach: Use BAYES’ THEOREM for updating the prior probability of the occurrence of mineral deposit to posterior probability based on the conditional probabilities (or weights of evidence) of the geological features.

Calculation of Weights of Evidence

Geological Feature (B1) Geological Feature (B2)


P{D|B} = P{D& B}

P{B}= P{D} P{B|D}

P{B}

P{D|B} = P{D & B}

P{B}= P{D} P{B|D}

P{B}

Posterior probability of D given the presence of B

Posterior probability of D given the absence of B

Bayes’ theorem:D- Deposit

B- Geol. Feature

THE BAYES EQUATION ESTIMATES THE PROBABILTY OF A DEPOSIT GIVEN THE

GEOLOGICAL FEATURE FROM THE PROBABILITY OF THE FEATURE GIVEN THE DEPOSITS

ObservationInference



It has been observed that on an average 100 gold deposits occur in 10,000 sq km area of Archaean geological provinces. In such provinces, 80% of deposits occur in Ultramafic rocks. 9.6% of barren areas also occur in Ultramafic (UM). You are exploring a 1 sq km area of an Archaean geological province with Ultramafic rocks (UM). What is the probability that the area will contain a gold deposit? Assume that a gold deposit occupies 1 sq km area.

EXERCISE

P(D|UM) = P(D) x [P(UM|D) / P(UM)]

P(D) = n(D)/n(S)

P(UM|D) = n(UM & D)/n(D)

P(UM) = n(UM)/n(S)

P(D|UM) =

(100/10000) * [(80/100)/(1030.4/10000)]

= 0.077

n(S) = 10,000

n(D) = 100

n(UM&D) = 80

n(UM) = ?

n(UM) = 80% of 100 + 9.6% (10,000 - 100)

= 80 + 0.096(9900) = 80 + 950.4

= 1030.4


Using odds (P/(1-P)) formulation:

O{D|B} = O{D} P{B|D}

P{B|D}Odds of D given the presence of B

O{D|B} = O{D} P{B|D}

P{B|D}Odds of D given the absence of B

Taking logs on both sides:

Loge (O{D|B}) = Loge(O{D}) + Log of odds of D given the presence of BP{B|D}P{B|D}

loge

Loge (O{D|B}) = Loge(O{D}) + Log of odds of D given the absence of BP{B|D}P{B|D}

loge

+ive weight of evidence (W+)

-ive weight of evidence (W-)


Indian Institute of Technology Bombay Contrast (C) measures the net strength of spatial association between the

geological feature and mineral deposits

Contrast = W+ – W-

+ ive Contrast – net positive spatial association

-ive Contrast – net negative spatial association

zero Contrast – no spatial association

Can be used to test spatial associations

Calculation of contrast


Total number of cells in study area: n(S)Total number of cells occupied by deposits (D): n(D)Total number of cells occupied by the feature (B): n(B)Total number of cells occupied by both feature and deposit: n(B&D)

= n( )/n(D) = n( )/ = n( )/n(D) = n( )/

B & D

B & DP{B|D}P{B|D} B & D

P{B|D}

P{B|D} B & D

n(D)

n(D)

B1

D

B2

B1 D B1 D

B1 D

B1 D

B1

SD

Calculation of Probabilty

P(D) = n(D)/n(S)


Basic quantities for estimating weights of evidenceTotal number of cells in study area: n(S)Total number of cells occupied by deposits (D): n(D)Total number of cells occupied by the feature (B): n(B)Total number of cells occupied by both feature and deposit: n(B&D)

Derivative quantities for estimating weights of evidenceTotal number of cells not occupied by D: n( ) = n(S) – n(D) Total number of cells not occupied by B: n( ) = n(S) – n(B)Total number of cells occupied by B but not D: n( B & D) = n(B) – n( B & D)Total number of cells occupied by D but not B: n(B & D) = n(D) – n(B & D)Total number of cells occupied by neither D but nor B: n( B & D) = n(S) – n(B) – n(D) + n( B & D)

DB

Probabilities are estimated as area (or no. of unit cells) proportions

P{B|D}P{B|D}

loge

W+ = P{B|D}P{B|D}

loge

W- =

= n( )/n(D) = n( )/ = n( )/n(D) = n( )/

B & D


P{B|D}

P{B|D} B & D

n(D)

n(D)

Where,

B1

D

B2

B2 D B2 D

B2 D

B2 D

B1 D B1 D

B1 D

B1 D

B1

SD

B2

SD



Exercise

B2 D B2 D

B2 D

B2 DB1 D B1 D

B1 D

B1 D

10k

10kB

1

B2

S

B1

SD B

2S

D

Unit cell size = 1 sq km & each deposit occupies 1 unit cell

n(S) = 100n(D) = 10n(B1) = 16n(B2) = 25n(B1 & D) = 4n(B2 & D) = 3

Calculate the weights of evidence (W+ and W-) and Contrast values for B1 and B2

= n( )/n(D) = n( )/ = [n(B) – n( )]/[n(S) –n(D)] = n( )/n(D) = [n(D) – n( )]/n(D) = n( )/ = [n(S) – n(B) – n(D) + n( )]/[n(S) –

n(D)]

B & D


P{B|D}

P{B|D} B & D

n(D)

n(D)

Where,B &

D

B & D

B & D

P{B|D}P{B|D}

loge

W+ = P{B|D}P{B|D}

loge

W- =


Loge (O{D|B}) = Loge(O{D}) + W+B

Loge (O{D|B}) = Loge(O{D}) + W-B

Assuming conditional independence of

the geological features B1 and B2, the

posterior probability of D given B1 and B2

can be estimated using:

Loge (O{D|B1, B2}) = Loge(O{D}) + W+B1 + W+B2

Loge (O{D|B1, B2}) = Loge(O{D}) + W-B1 + W+B2

Loge (O{D|B1, B2}) = Loge(O{D}) + W+B1 + W-B2

Loge (O{D|B1, B2}) =

Loge(O{D}) + W-B1 + W-B2

Probability of D given the presence of B1 and B2

Probability of D given the absence of B1 and presence B2

Probability of D given the presence of B1 and absence B2

Probability of D given the absence of B1 and B2

Loge (O{D|B1, B2, … Bn}) = Loge(O{D}) + ∑W+/-Bii=1

nOr in general, for n geological features,

The sign of W is +ive or -ive, depending on whether the feature is absent or present

The odds are converted back to posterior probability using the relation 0 = P/(1+P)

Combining Weights of Evidence: Posterior Probability

Feature B2 Feature B1

Deposit D



Loge(O{D}) + ∑W+/-Bii=1

n

Calculation of posterior probability (or odds) require:• Calculation of pr prob (or odds) of occurrence of deposits in the study area• Calculation of weights of evidence of all geological features, i.e,

P{B|D}P{B|D}

loge

P{B|D}P{B|D}

loge

W+ =

W- =

&


Indian Institute of Technology Bombay Loge (O{D|B1, B2})

= Loge(O{D}) + W+/-B1 +

W+/-B2 Loge(O{D}) = Loge(0.11) = -

2.2073

Calculate posterior probability given:

1. Presence of B1 and B2;2. Presence of B1 and absence of

B2;3. Absence of B1 and presence of

B2;4. Absence of both B1 and B2

B1

B2

S

Prior Prb = 0.10Prior Odds =

0.11




Loge(O{D}) + W+/-B1 + W+/-B2

Loge (O{D|B1, B2}) = -2.2073 + 1.0988 + 0.2050 = -0.8585

O{D|B1, B2} = Antiloge (-0.8585) = 0.4238 P = O/(1+O) = (0.4238)/(1.4238) = 0.2968

For the areas where both B1 and B2 are present

Loge (O{D|B1, B2}) = -2.2073 + 1.0988 - 0.0763 = -1.1848

O{D|B1, B2} = Antiloge (- 1.1848) = 0.3058 P = O/(1+O) = (0.3058)/(1.3058) = 0.2342

For the areas where B1 is present but B2 is absent

Loge (O{D|B1, B2}) = -2.2073 - 0.3678 + 0.2050 = -2.3701


Loge (O{D|B1, B2}) = -2.2073 - 0.3678 - 0.0763 = -2.6514


For the areas where both B1 and B2 are absent

For the areas where B1 is absent but B2 is present

Loge(O{D}) = Loge(0.11) = -2.2073

Posterior probability0.29680.2342

0.08540.0658

Prospectivity Map

B1

B2

S

Prior Prb = 0.10


Structure of a model for mineral resource potential mapping

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