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# Probability Concepts Applications

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### Probability Concepts Applications

1. 1. Probability Concepts and Applications
2. 2. Introduction <ul><li>Life is uncertain! </li></ul><ul><li>We must deal with risk ! </li></ul><ul><li>A probability is a numerical statement about the likelihood that an event will occur </li></ul>
3. 3. Basic Statements About Probability <ul><li>The probability, P , of any event or state of nature occurring is greater than or equal to 0 and less than or equal to 1 . </li></ul><ul><li>That is: 0  P(event)  1 </li></ul><ul><li>2. The sum of the simple probabilities for all possible outcomes of an activity must equal 1 . </li></ul>
4. 4. Example <ul><li>Demand for white latex paint at Diversey Paint and Supply has always been 0, 1, 2, 3, or 4 gallons per day. (There are no other possible outcomes; when one outcome occurs, no other can.) Over the past 200 days, the frequencies of demand are represented in the following table: </li></ul>
5. 5. Example - continued <ul><li>Quantity Demanded (Gallons) </li></ul><ul><li>0 </li></ul><ul><li>1 </li></ul><ul><li>2 </li></ul><ul><li>3 </li></ul><ul><li>4 </li></ul><ul><li>Number of Days </li></ul><ul><li>40 </li></ul><ul><li>80 </li></ul><ul><li>50 </li></ul><ul><li>20 </li></ul><ul><li>10 </li></ul><ul><li>Total 200 </li></ul>Frequencies of Demand
6. 6. Example - continued <ul><li>Quant. Freq. </li></ul><ul><li>Demand (days) </li></ul><ul><li>0 40 </li></ul><ul><li>1 80 </li></ul><ul><li>2 50 </li></ul><ul><li>3 20 </li></ul><ul><li>4 10 </li></ul><ul><li>Total days = 200 </li></ul><ul><li>Probability </li></ul><ul><li>(40/200) = 0.20 </li></ul><ul><li>(80/200) = 0.40 </li></ul><ul><li>(50/200) = 0.25 </li></ul><ul><li>(20/200) = 0.10 </li></ul><ul><li>(10/200) = 0.05 </li></ul><ul><li>Total </li></ul><ul><li>Prob = 1.00 </li></ul>Probabilities of Demand
7. 7. Types of Probability <ul><li>Objective probability: </li></ul><ul><li>Determined by experiment or observation: </li></ul><ul><ul><li>Probability of heads on coin flip </li></ul></ul><ul><ul><li>Probably of spades on drawing card from deck </li></ul></ul>occurrences or outcomes of number Total occurs event times of Number ) (  event P
8. 8. Types of Probability <ul><li>Subjective probability: </li></ul><ul><li>Based upon judgement </li></ul><ul><li>Determined by: </li></ul><ul><ul><li>judgement of expert </li></ul></ul><ul><ul><li>opinion polls </li></ul></ul><ul><ul><li>Delphi method </li></ul></ul><ul><ul><li>etc. </li></ul></ul>
9. 9. Mutually Exclusive Events <ul><li>Events are said to be mutually exclusive if only one of the events can occur on any one trial </li></ul>
10. 10. Collectively Exhaustive Events <ul><li>Events are said to be collectively exhaustive if the list of outcomes includes every possible outcome: heads and tails as possible outcomes of coin flip </li></ul>
11. 11. Example <ul><li>Outcome </li></ul><ul><li>of Roll </li></ul><ul><li>1 </li></ul><ul><li>2 </li></ul><ul><li>3 </li></ul><ul><li>4 </li></ul><ul><li>5 </li></ul><ul><li>6 </li></ul><ul><li>Probability </li></ul><ul><li>1/6 </li></ul><ul><li>1/6 </li></ul><ul><li>1/6 </li></ul><ul><li>1/6 </li></ul><ul><li>1/6 </li></ul><ul><li>1/6 </li></ul><ul><li>Total = 1 </li></ul>Rolling a die has six possible outcomes
12. 12. Example <ul><li>Outcome </li></ul><ul><li>of Roll = 5 </li></ul><ul><li>Die 1 Die 2 </li></ul><ul><li>1 4 </li></ul><ul><li>2 3 </li></ul><ul><li>3 2 </li></ul><ul><li>4 1 </li></ul><ul><li>Probability </li></ul><ul><li>1/36 </li></ul><ul><li>1/36 </li></ul><ul><li>1/36 </li></ul><ul><li>1/36 </li></ul>Rolling two dice results in a total of five spots showing. There are a total of 36 possible outcomes.
13. 13. Probability : Mutually Exclusive <ul><li>P(event A or event B) = </li></ul><ul><li>P(event A) + P(event B) </li></ul><ul><li>or: </li></ul><ul><li>P(A or B) = P(A) + P(B) </li></ul><ul><li>i.e., </li></ul><ul><li>P(spade or club) = P(spade) + P(club) </li></ul><ul><li> = 13/52 + 13/52 </li></ul><ul><li> = 26/52 = 1/2 = 50% </li></ul>
14. 14. Probability: Not Mutually Exclusive <ul><li>P(event A or event B) = </li></ul><ul><li>P(event A) + P(event B) - </li></ul><ul><li>P(event A and event B both occurring) </li></ul><ul><li>or </li></ul><ul><li>P(A or B) = P(A)+P(B) - P(A and B) </li></ul>
15. 15. P(A and B) (Venn Diagram) P(A) P(B) P(A and B)
16. 16. P(A or B) + - = P(A) P(B) P(A and B) P(A or B)
17. 17. Statistical Dependence <ul><li>Events are either </li></ul><ul><ul><li>statistically independent ( the occurrence of one event has no effect on the probability of occurrence of the other ) or </li></ul></ul><ul><ul><li>statistically dependent ( the occurrence of one event gives information about the occurrence of the other ) </li></ul></ul>
18. 18. Probabilities - Independent Events <ul><li>Marginal probability : the probability of an event occurring: </li></ul><ul><li>[ P(A)] </li></ul><ul><li>Joint probability : the probability of multiple, independent events, occurring at the same time </li></ul><ul><li>P(AB) = P(A)*P(B) </li></ul><ul><li>Conditional probability ( for independent events ) : </li></ul><ul><ul><li>the probability of event B given that event A has occurred P(B|A) = P(B) </li></ul></ul><ul><ul><li>or, the probability of event A given that event B has occurred P(A|B) = P(A) </li></ul></ul>
19. 19. Probability(A|B) Independent Events P(B) P(A) P(A|B) P(B|A)
20. 20. Statistically Independent Events <ul><li>1. P(black ball drawn on first draw) </li></ul><ul><ul><li>P(B) = 0.30 ( marginal probability ) </li></ul></ul><ul><li>2. P(two green balls drawn) </li></ul><ul><ul><li>P(GG) = P(G)*P(G) = 0.70*0.70 = 0.49 ( joint probability for two independent events ) </li></ul></ul>A bucket contains 3 black balls, and 7 green balls. We draw a ball from the bucket, replace it, and draw a second ball
21. 21. Statistically Independent Events - continued <ul><li>1. P(black ball drawn on second draw, first draw was green) </li></ul><ul><ul><li>P(B|G) = P(B) = 0.30 </li></ul></ul><ul><ul><li>( conditional probability ) </li></ul></ul><ul><li>2. P(green ball drawn on second draw, first draw was green) </li></ul><ul><ul><li>P(G|G) = 0.70 </li></ul></ul><ul><ul><li>( conditional probability ) </li></ul></ul>
22. 22. Probabilities - Dependent Events <ul><li>Marginal probability : probability of an event occurring P(A) </li></ul><ul><li>Conditional probability ( for dependent events ) : </li></ul><ul><ul><li>the probability of event B given that event A has occurred P(B|A) = P(AB)/P(A) </li></ul></ul><ul><ul><li>the probability of event A given that event B has occurred P(A|B) = P(AB)/P(B) </li></ul></ul>
23. 23. Probability(A|B) / P(A|B) = P(AB)/P(B) P(AB) P(B) P(A)
24. 24. Probability(B|A) P(B|A) = P(AB)/P(A) / P(AB) P(B) P(A)
25. 25. Statistically Dependent Events <ul><li>Assume that we have an urn containing 10 balls of the following descriptions: </li></ul><ul><ul><li>4 are white (W) and lettered (L) </li></ul></ul><ul><ul><li>2 are white (W) and numbered N </li></ul></ul><ul><ul><li>3 are yellow (Y) and lettered (L) </li></ul></ul><ul><ul><li>1 is yellow (Y) and numbered (N) </li></ul></ul><ul><li>Then : </li></ul><ul><ul><li>P(WL) = 4/10 = 0.40 </li></ul></ul><ul><ul><li>P(WN) = 2/10 = 0.20 </li></ul></ul><ul><ul><li>P(W) = 6/10 = 0.60 </li></ul></ul><ul><ul><li>P(YL) = 3/10 = 0.3 </li></ul></ul><ul><ul><li>P(YN) = 1/10 = 0.1 </li></ul></ul><ul><ul><li>P(Y) = 4/10 = 0.4 </li></ul></ul>
26. 26. Statistically Dependent Events - Continued <ul><li>Then: </li></ul><ul><ul><li>P(L|Y) = P(YL)/P(Y) </li></ul></ul><ul><li>= 0.3/0.4 = 0.75 </li></ul><ul><ul><li>P(Y|L) = P(YL)/P(L) </li></ul></ul><ul><li>= 0.3/0.7 = 0.43 </li></ul><ul><ul><li>P(W|L) = P(WL)/P(L) </li></ul></ul><ul><li>= 0.4/0.7 = 0.57 </li></ul>
27. 27. Joint Probabilities, Dependent Events <ul><li>Your stockbroker informs you that if the stock market reaches the 10,500 point level by January, there is a 70% probability the Tubeless Electronics will go up in value. Your own feeling is that there is only a 40% chance of the market reaching 10,500 by January. </li></ul><ul><li>What is the probability that both the stock market will reach 10,500 points, and the price of Tubeless will go up in value? </li></ul>
28. 28. Joint Probabilities, Dependent Events - continued <ul><li>Then: </li></ul><ul><li>P(MT) =P(T|M)P(M) </li></ul><ul><li> = (0.70)(0.40) </li></ul><ul><li> = 0.28 </li></ul>Let M represent the event of the stock market reaching the 10,500 point level, and T represent the event that Tubeless goes up.
29. 29. Revising Probabilities: Bayes’ Theorem <ul><li>Bayes’ theorem can be used to calculate revised or posterior probabilities </li></ul>Prior Probabilities Bayes’ Process Posterior Probabilities New Information
30. 30. General Form of Bayes’ Theorem
31. 31. Posterior Probabilities <ul><li>A cup contains two dice identical in appearance. One, however, is fair (unbiased), the other is loaded (biased). The probability of rolling a 3 on the fair die is 1/6 or 0.166. The probability of tossing the same number on the loaded die is 0.60. </li></ul><ul><li>We have no idea which die is which, but we select one by chance, and toss it. The result is a 3. </li></ul><ul><li>What is the probability that the die rolled was fair? </li></ul>
32. 32. Posterior Probabilities Continued <ul><li>We know that: </li></ul><ul><ul><li>P(fair) = 0.50 P(loaded) = 0.50 </li></ul></ul><ul><li>And: </li></ul><ul><li>P(3|fair) = 0.166 P(3|loaded) = 0.60 </li></ul><ul><li>Then: </li></ul><ul><ul><li>P(3 and fair) = P(3|fair)P(fair) = (0.166)(0.50) </li></ul></ul><ul><ul><li> = 0.083 </li></ul></ul><ul><ul><li>P(3 and loaded) = P(3|loaded)P(loaded) </li></ul></ul><ul><ul><li>= (0.60)(0.50) </li></ul></ul><ul><ul><li>= 0.300 </li></ul></ul>
33. 33. Posterior Probabilities Continued <ul><li>A 3 can occur in combination with the state “fair die” or in combination with the state ”loaded die.” The sum of their probabilities gives the unconditional or marginal probability of a 3 on a toss: </li></ul><ul><li>P(3) = 0.083 + 0.0300 = 0.383 . </li></ul><ul><li>Then, the probability that the die rolled was the fair one is given by: </li></ul>
34. 34. Further Probability Revisions <ul><li>To obtain further information as to whether the die just rolled is fair or loaded, let’s roll it again. </li></ul><ul><li>Again we get a 3. </li></ul><ul><li>Given that we have now rolled two 3s, what is the probability that the die rolled is fair? </li></ul>
35. 35. Further Probability Revisions - continued <ul><li>P(fair) = 0.50, P(loaded) = 0.50 as before </li></ul><ul><li>P(3,3|fair) = (0.166)(0.166) = 0.027 </li></ul><ul><li>P(3,3|loaded) = (0.60)(0.60) = 0.36 </li></ul><ul><li>P(3,3 and fair) = P(3,3|fair)P(fair) </li></ul><ul><li> = (0.027)(0.05) </li></ul><ul><li>= 0.013 </li></ul><ul><li>P(3,3 and loaded) = P(3,3|loaded)P(loaded) </li></ul><ul><li> = (0.36)(0.5) </li></ul><ul><li>= 0.18 </li></ul><ul><li>P(3,3) = 0.013 + 0.18 = 0.193 </li></ul>
36. 36. Further Probability Revisions - continued
37. 37. <ul><li>To give the final comparison: </li></ul><ul><li>P(fair|3) = 0.22 </li></ul><ul><li>P(loaded|3) = 0.78 </li></ul><ul><li>P(fair|3,3) = 0.067 </li></ul><ul><li>P(loaded|3,3) = 0.933 </li></ul>Further Probability Revisions - continued