# Probability and Statistics Presented by Carol Dahl

## Presentation on theme: "Probability and Statistics Presented by Carol Dahl"— Presentation transcript:

Probability and Statistics Presented by Carol Dahl
Basic Probability Schaum’s Outlines of Probability and Statistics Chapter 1 Presented by Carol Dahl Examples by Claudianus Adjai

Outline of Topics Topics Covered: Set & Set Operations Probabilities
Powerful tools for analysis under uncertainty Topics Covered: Set & Set Operations Probabilities Counting Rule Conditional Probability Probability of a Sample Permutations & Combinations Independent Events Binomial Bayes’ Theorem

Sample Sets Example: mining company in Chile owns 240 acres of land with copper (Cu) gold (Au) iron (Fe) minerals locations distributed as follows:

Sample Sets Au-1 Cu Fe Au-2 U

Set and Set Operations Universal set U = Total Acreage (rectangle)
U  Au = Land contains Au (subset of U)  = An empty set Fe  Cu = Land contains iron or copper or both Fe  Cu = Land contains both iron and copper Complement of Cu (Cu’): U - Cu = Cu' = Land not contain copper

Probabilities Definition: likelihood that something happens
P(U) = < P(X) < 1 Total of Xi mutually exclusive events for i = 1,2,3,…,n Drill for minerals randomly U = 240 acres Fe = 40 acres Cu = 60 acres Au-1 = 10 acres Au-2 = 10 acres Cu & Fe = 20 acres

Probabilities Au-1 Cu Fe Au-2 U (10 acres) (60 acres) 20 (40 acres)

Probabilities What is the probability of finding: Fe deposits?
Cu deposits? Cu and Fe deposits? Au-1 deposits? Au-2 deposits?

Counting Rule If equally likely outcomes use counting rule:
P(event) = # of items in event  # of total outcomes U = 200 acres => P(U) = (200)/200 = 1 Fe = 40 acres => P(Fe) = (40)/200 = 1/5 Cu = 60 acres => P(Cu) = (60)/200 = 3/10

Counting Rule Au-1 = 10 acres => P(Au-1) = (10)/200 = 1/20
P(Cu only) = (60-20)/200 = 2/10 = 1/5 P(Fe only) = (40-20)/200 = 1/10 Cu & Fe = 20 acres => P(Cu  Fe) = (20)/200 =1/10

Probability of finding nothing: = 1 – 1/10 – 2/10 –1/10 – 1/10 = 5/10 = 1/2 => 50% Probability find copper or iron (addition rule) P(Fe  Cu) = P(Fe) + P(Cu) - P(Fe  Cu) = 2/ /10 – 1/10 = 4/10 = 2/5

Conditional Probabilities
Conditional Probability P(Cu | Fe) = P(Cu  Fe) / P(Fe) = (1/10 ) / (2/10) = 1/2 Probability of a sample Probability of a model given data (A,B), (A,C), (B,C), (B,A), (C,A), (C,B)

Permutations Example: You own three leases (A,B,C) drill two randomly
without replacement how many ways can you choose 2 from 3 (A,B), (A,C), (B,C), (B,A), ( , ), ( , ) (A,B), (A,C), (B,C), (B,A), (C,A), (C,B)

Permutations and Combinations
If order matters choose r from n: Permutations = n!/(n-r)! = 3!/(3-2)! = 3 * 2 * 1/1 = 6 If order doesn't matters choose r from n: Combinations = n!/((n-r)!r!) = 3!/((3-2)!2!) = 3×2/2 = 3

Multiplication Rule and Independence
P(S1 ∩ S2) = P(S1|S2) * P(S1) Independence: P(S1 ∩ S2) = P(S1) * P(S2) Example: Are discovering Fe and Cu independent? P(Fe ∩ Cu) = 1/10 P(Fe) *P(Cu) = (2/10)*(3/10) = 6/10

Implication of independence
P(Cu|Fe) = P(Cu ∩ Fe) / P(Fe) = (P(Fe)P(Cu)) / P(Fe) = P(Cu) marginal probability = conditional

Independent Events Example:
Russian gas company Gazprom exploring 4 gas fields one well per field similar geology – 1/3 chance of success probability you get a success on the first two wells success field independent of success in others P(S1 ∩ S2) = P(S1) *P(S1) = 1/3*1/3 = 1/9

Binomial Notation: Probability of success = p trial = n
without replacement Formula: p(X = x) = n!/((n-x)!x!)(p)n(1-p)(n-x)

Binomial Probability 2 of 4 are successful (S,S,D,D) = 1/3*1/3*2/3*2/3
(S,D,S,D) = 1/3*2/3*2/3*1/3 (D,D,S,S) = 1/3*1/3*2/3*2/3 (S,D,D,S) = 1/3*2/3*2/3*1/3 (D,S,S,D) = 1/3*2/3*1/3*2/3 (D,S,D,S) = 1/3*2/3*1/3*2/3 P(X = 2) = [4!/((4-2)!2!)](1/3)2(2/3)(n-2) = 0.296

Bayes’ Theorem Bayes - Making decisions using new sample information.
Example: Batteries hybrid renewable energy (wind, solar) Three of your plants build the battery E1 E2 E3 Two battery types regular – r heavy duty – h

Bayes’ Theorem Cont. Example: Factory Types Batteries r h
E r 100 h Total 300 E r 150 h Total 200 E r h Total 100 h battery comes back on warrantee Probability battery from plant E2 = P(E2|h)?

Bayes’ Theorem From definition
P(E2|h) = P(h∩E2) => P(h∩E2) = P(E2|h)P(h) P(h) but also P(h∩E2) = P(h|E2)P(E2) Replace in numerator P(E2|h) = P(h|E2 )P(E2 )

Bayes’ Theorem What is P(h) = P(h∩E1) + P(h∩E2 ) + P(h∩E3 ) But
P(h∩E1) = P(h|E1)P(E1) P(h∩E2) = P(h|E2)P(E2) P(h∩E3) = P(h|E3)P(E3)

Bayes’ Theorem Replace in denominator P(E2|h) = P(h|E2)P(E2)
P(h|E1)P(E1)+P(h|E2)P(E2)+P(h|E3)P(E3) = (3/4)(1/3) (1/2)(1/3) + (1/3)(3/4) + (1/6)(1/2) = 1/2

Bayes’ Theorem General formula (h occurs)
P(Ei|h) = P(h|Ei)P(Ei) P(h|E1)P(E1)+P(h|E2)P(E2)+P(h|E3)P(E3) P(Ei|h) = P(h|Ei)P(Ei) i(P(h|Ei)P(Ei)

Bayesian Econometrics
Econometrics = wedding model and data Ei = model, h = data P(model|data) = P(data|model )P(model) P(data) Example: yt =  + ei P(|y) = P(y|)P() P(y)

Prior and Posterior Distributions
P(|y) = P(y|)P() P(y) P(y|) = how well data fits given the model likelihood function pick model to maximize P() = prior beliefs P(y) no model parameters - treat as exogenous P(|y) = posterior  likelihood*prior