Probability And Probability Distributions
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This document provides an overview of key concepts in probability and probability distributions. It defines important terms like probability, sample space, events, mutually exclusive events, independent events, and conditional probability. It also covers rules of probability like addition rules, complement rules, and Bayes' theorem. Finally, it introduces discrete and continuous random variables and discusses properties of discrete probability distributions like expected value and standard deviation.
![Discrete Probability Distribution
A list of all possible [ xi , P(xi) ] pairs
xi = Value of Random Variable (Outcome)
P(xi) = Probability Associated with Value
xi’s are mutually exclusive
(no overlap)
xi’s are collectively exhaustive
(nothing left out)
0 £ P(xi) £ 1 for each xi
Σ P(xi) = 1
32](https://image.slidesharecdn.com/probabilityandprobabilitydistributionspptbecdoms-120223033256-phpapp02-2-150320235922-conversion-gate01/85/Probability-And-Probability-Distributions-32-320.jpg)
![Covariance
Covariance between two discrete random
variables:
σxy = Σ [xi – E(x)][yj – E(y)]P(xiyj)
where:
xi = possible values of the x discrete random variable
yj = possible values of the y discrete random variable
P(xi ,yj) = joint probability of the values of xi and yj
occurring
37](https://image.slidesharecdn.com/probabilityandprobabilitydistributionspptbecdoms-120223033256-phpapp02-2-150320235922-conversion-gate01/85/Probability-And-Probability-Distributions-37-320.jpg)
Probability And Probability Distributions
- 1. Probability And Probability Distributions 1 MADE BY:- ROLL NO. SAHIL NAGPAL 232 PRACHI DABAS (LEADER) 492 KARANBIR SINGH 152 GARIMA LONGIANY 1252 PAVAN KUMAR 842 MANOJ KUMAR 902 RAHUL 1462 LALTHANGLEEN SITLHOI 1052 Sri Guru Nanak Dev Khalsa College (University of Delhi)
- 2. Chapter Goals After completing this chapter, you should be able to: Explain three approaches to assessing probabilities Apply common rules of probability Use Bayes’ Theorem for conditional probabilities Distinguish between discrete and continuous probability distributions Compute the expected value and standard deviation for a discrete probability distribution 2
- 3. Important Terms Probability – the chance that an uncertain event will occur (always between 0 and 1) Experiment – a process of obtaining outcomes for uncertain events Elementary Event – the most basic outcome possible from a simple experiment Sample Space – the collection of all possible elementary outcomes 3
- 4. 4 Sample Space The Sample Space is the collection of all possible outcomes e.g. All 6 faces of a die: e.g. All 52 cards of a bridge deck:
- 5. Events Elementary event – An outcome from a sample space with one characteristic Example: A red card from a deck of cards Event – May involve two or more outcomes simultaneously Example: An ace that is also red from a deck of cards 5
- 6. Visualizing Events Contingency Tables Tree Diagrams 6 Red 2 24 26 Black 2 24 26 Total 4 48 52 Ace Not Ace Total Full Deck of 52 Cards Red Card Black Card Not an Ace Ace Ace Not an Ace Sample Space Sample Space 2 24 2 24
- 7. Elementary Events A automobile consultant records fuel type and vehicle type for a sample of vehicles 2 Fuel types: Gasoline, Diesel 3 Vehicle types: Truck, Car, SUV 6 possible elementary events: e1Gasoline, Truck e2 Gasoline, Car e3 Gasoline, SUV e4 Diesel, Truck e5 Diesel, Car e6 Diesel, SUV 7 Gasoline Diesel Car Truck Truck Car SUV SUV e1 e2 e3 e4 e5 e6
- 8. Probability Concepts Mutually Exclusive Events If E1 occurs, then E2 cannot occur E1 and E2 have no common elements 8 Black Cards Red Cards A card cannot be Black and Red at the same time. E1 E2
- 9. Probability Concepts Independent and Dependent Events Independent: Occurrence of one does not influence the probability of occurrence of the other. Dependent: Occurrence of one affects the probability of the other. 9
- 10. Independent vs. Dependent Events 10 Independent Events E1 = heads on one flip of fair coin E2 = heads on second flip of same coin Result of second flip does not depend on the result of the first flip. Dependent Events E1 = rain forecasted on the news E2 = take umbrella to work Probability of the second event is affected by the occurrence of the first event
- 11. Assigning Probability Classical Probability Assessment 11 • Relative Frequency of Occurrence • Subjective Probability Assessment P(Ei) = Number of ways Ei can occur Total number of elementary events Relative Freq. of Ei = Number of times Ei occurs N An opinion or judgment by a decision maker about the likelihood of an event
- 12. Rules of Probability 12 Rules for Possible Values and Sum Individual Values Sum of All Values 0 ≤ P(ei) ≤ 1 For any event ei 1)P(e k 1i i =∑= where: k = Number of elementary events in the sample space ei = ith elementary event
- 13. Addition Rule for Elementary Events The probability of an event Ei is equal to the sum of the probabilities of the elementary events forming Ei. Ei = {e1, e2, e3} then: P(Ei) = P(e1) + P(e2) + P(e3) 13
- 14. Complement Rule The complement of an event E is the collection of all possible elementary events not contained in event E. The complement of event E is represented by E. Complement Rule: 14 P(E)1)EP( −= E E 1)EP(P(E) =+Or,
- 15. Addition Rule for Two Events 15 P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2) E1 E2 P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2) Don’t count common elements twice! ■ Addition Rule: E1 E2+ =
- 16. Addition Rule Example 16 P(Red or Ace) = P(Red) +P(Ace) - P(Red and Ace) = 26/52 + 4/52 - 2/52 = 28/52 Don’t count the two red aces twice! Black Color Type Red Total Ace 2 2 4 Non-Ace 24 24 48 Total 26 26 52
- 17. Addition Rule for Mutually Exclusive Events If E1 and E2 are mutually exclusive, then P(E1 and E2) = 0 So 17 P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2) = P(E1) + P(E2) = 0 E1 E2 if mutually exclusive
- 18. Conditional Probability Conditional probability for any two events E1 , E2: 18 )P(E )EandP(E )E|P(E 2 21 21 = 0)P(Ewhere 2 >
- 19. Conditional Probability Example What is the probability that a car has a CD player, given that it has AC ? i.e., we want to find P(CD | AC) 19 Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both.
- 20. Conditional Probability Example 20 No CDCD Total AC .2 .5 .7 No AC .2 .1 .3 Total .4 .6 1.0 Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both. .2857 .7 .2 P(AC) AC)andP(CD AC)|P(CD === (continued)
- 21. Conditional Probability Example 21 No CDCD Total AC .2 .5 .7 No AC .2 .1 .3 Total .4 .6 1.0 Given AC, we only consider the top row (70% of the cars). Of these, 20% have a CD player. 20% of 70% is about 28.57%. .2857 .7 .2 P(AC) AC)andP(CD AC)|P(CD === (continued)
- 22. For Independent Events: Conditional probability for independent events E1 , E2: 22 )P(E)E|P(E 121 = 0)P(Ewhere 2 > )P(E)E|P(E 212 = 0)P(Ewhere 1 >
- 23. Multiplication Rules Multiplication rule for two events E1 and E2: 23 )E|P(E)P(E)EandP(E 12121 = )P(E)E|P(E 212 =Note: If E1 and E2 are independent, then and the multiplication rule simplifies to )P(E)P(E)EandP(E 2121 =
- 24. Tree Diagram Example 24 Diesel P(E2) = 0.2 Gasoline P(E1) = 0.8 Truck: P(E3 |E1 ) = 0.2 Car: P(E4|E1) = 0.5 SUV: P(E5|E1) = 0.3 P(E1 and E3) = 0.8 x 0.2 = 0.16 P(E1 and E4) = 0.8 x 0.5 = 0.40 P(E1 and E5) = 0.8 x 0.3 = 0.24 P(E2 and E3) = 0.2 x 0.6 = 0.12 P(E2 and E4) = 0.2 x 0.1 = 0.02 P(E3 and E4) = 0.2 x 0.3 = 0.06 Truck: P(E3 |E2 ) = 0.6 Car: P(E4|E2) = 0.1 SUV: P(E5|E2) = 0.3
- 25. Bayes’ Theorem where: Ei = ith event of interest of the k possible events B = new event that might impact P(Ei) Events E1 to Ek are mutually exclusive and collectively exhaustive 25 )E|)P(BP(E)E|)P(BP(E)E|)P(BP(E )E|)P(BP(E B)|P(E kk2211 ii i +++ =
- 26. Bayes' Theorem Example A drilling company has estimated a 40% chance of striking oil for their new well. A detailed test has been scheduled for more information. Historically, 60% of successful wells have had detailed tests, and 20% of unsuccessful wells have had detailed tests. Given that this well has been scheduled for a detailed test, what is the probability that the well will be successful? 26
- 27. Let S = successful well and U = unsuccessful well P(S) = .4 , P(U) = .6 (prior probabilities) Define the detailed test event as D Conditional probabilities: P(D|S) = .6 P(D|U) = .2 Revised probabilities 27 Bayes’ Theorem Example Event Prior Prob. Conditional Prob. Joint Prob. Revised Prob. S (successful) .4 .6 .4*.6 = .24 .24/.36 = .67 U (unsuccessful) .6 .2 .6*.2 = .12 .12/.36 = .33 Sum = .36 (continued)
- 28. Given the detailed test, the revised probability of a successful well has risen to .67 from the original estimate of .4 28 Bayes’ Theorem Example Event Prior Prob. Conditional Prob. Joint Prob. Revised Prob. S (successful) .4 .6 .4*.6 = .24 .24/.36 = .67 U (unsuccessful) .6 .2 .6*.2 = .12 .12/.36 = .33 Sum = .36 (continued)
- 29. Introduction to Probability Distributions Random Variable Represents a possible numerical value from a random event 29 Random Variables Discrete Random Variable Continuous Random Variable
- 30. Discrete Random Variables Can only assume a countable number of values Examples: Roll a die twice Let x be the number of times 4 comes up (then x could be 0, 1, or 2 times) Toss a coin 5 times. Let x be the number of heads (then x = 0, 1, 2, 3, 4, or 5) 30
- 31. Discrete Probability Distribution x Value Probability 0 1/4 = .25 1 2/4 = .50 2 1/4 = .25 31 Experiment: Toss 2 Coins. Let x = # heads. T T 4 possible outcomes T T H H H H Probability Distribution 0 1 2 x .50 .25 Probability
- 32. Discrete Probability Distribution A list of all possible [ xi , P(xi) ] pairs xi = Value of Random Variable (Outcome) P(xi) = Probability Associated with Value xi’s are mutually exclusive (no overlap) xi’s are collectively exhaustive (nothing left out) 0 £ P(xi) £ 1 for each xi Σ P(xi) = 1 32
- 33. Discrete Random Variable Summary Measures Expected Value of a discrete distribution (Weighted Average) E(x) = Σxi P(xi) Example: Toss 2 coins, x = # of heads, compute expected value of x: E(x) = (0 x .25) + (1 x .50) + (2 x .25) = 1.0 33 x P(x) 0 .25 1 .50 2 .25
- 34. Discrete Random Variable Summary Measures Standard Deviation of a discrete distribution where: E(x) = Expected value of the random variable x = Values of the random variable P(x) = Probability of the random variable having the value of x 34 P(x)E(x)}{xσ 2 x −= ∑ (continued)
- 35. Discrete Random Variable Summary Measures Example: Toss 2 coins, x = # heads, compute standard deviation (recall E(x) = 1 35 P(x)E(x)}{xσ 2 x −= ∑ .707.50(.25)1)(2(.50)1)(1(.25)1)(0σ 222 x ==−+−+−= (continued) Possible number of heads = 0, 1, or 2
- 36. Two Discrete Random Variables Expected value of the sum of two discrete random variables: E(x + y) = E(x) + E(y) = Σ x P(x) + Σ y P(y) (The expected value of the sum of two random variables is the sum of the two expected values) 36
- 37. Covariance Covariance between two discrete random variables: σxy = Σ [xi – E(x)][yj – E(y)]P(xiyj) where: xi = possible values of the x discrete random variable yj = possible values of the y discrete random variable P(xi ,yj) = joint probability of the values of xi and yj occurring 37
- 38. Interpreting Covariance Covariance between two discrete random variables: σxy > 0 x and y tend to move in the same direction σxy < 0 x and y tend to move in opposite directions σxy = 0 x and y do not move closely together 38
- 39. Correlation Coefficient The Correlation Coefficient shows the strength of the linear association between two variables where: ρ = correlation coefficient (“rho”) σxy = covariance between x and y σx = standard deviation of variable x σy = standard deviation of variable y 39 yx yx σσ σ ρ =
- 40. Interpreting the Correlation Coefficient The Correlation Coefficient always falls between -1 and +1 ρ = 0 x and y are not linearly related. The farther ρ is from zero, the stronger the linear relationship: ρ = +1 x and y have a perfect positive linear relationship ρ = -1 x and y have a perfect negative linear relationship 40