1. PROBABILITY and IMPORTANCE to RISK MANAGEMENT

2. What is PROBABILITY? It is a quantitative measure of uncertainty.

3. ETYMOLOGY The word PROBABILTY derives from PROBITY

4. HISTORY OF PROBABILITY MIDDLE OF 17 TH CENTURY RICHARD JEFFRY term PROBABLE-means APPROVABLE  1774 PIERRE-SIMON LAPLACE Made the first attempt to deduce a rule for the combinations of observations.

5. INTERPRETATIONS FREQUENTISTS -dealing with experiments that are random and well- defined.  BAYESIANS -individual’s degree of belief

6. APPLICATIONS OF PROBABILITY RISK ASSESSMENT  TRADE ON COMMODITY MARKETS

7. RELATION OF PROBABILITY ON RISK MANAGEMENT

8. Probability of Occurrence 0 - 10%orVery unlikely to occur 11 - 40%orUnlikely to occur 41 - 60%orMay occur about half of the time 61 - 90%orLikely to occur 91 - 100%orVery likely to occur

10. Definition of Terms Is any action or process that generates observations or set of data Random Experiment Is an experiment whose outcome cannot be predicted with certainty,but the experiment is such a nature that the collection of every possible outcomes can be described prior to its performance. Experiment

11. Sample Space Is the set of all possible outcomes of a statistical experiment.It is collections of all conceivable outcome of an experiment. Element Each outcome in sample space also called as sample point.

12. Event Is any collection of outcomes contained in the sample space. It consist exactly one outcome. Simple Event Compound Event It consist more than one outcome.

13. Complement The complement of an event A, A’,with respect to the sample space,S,is the subset of all elements of S that are not in A. The intersection of two events A and B,AΩB,is the event containing all elements that are common to both A and B. Intersection Example: Let S={1,2,3,4,5,6,7,8,9,0} and let R={2,4,6,8,0} R’=(1,3,4,7,9) Example: Let S={1,2,3,4,5,6,7,8,9,0} and let R={2,4,6,8,0} AΩB= (2,4,6,8,0)

14. Mutually Exclusive or Disjoint Two events, A and B are mutually exclusive or disjoint if A ΩB =ø,that is,if A and B have no elements in common. The union of the two events A and B,A U B, is the event containing all the elements that belong to A or B or both. Union Example: Let S={1,3,5,7,9} and let R={2,4,6,8,0} A ΩB =ø Example: Let S={1,2,3,4,5,6,7,8,9,0} and let R={2,4,6,8,0} A U B=( 1,2,3,4,5,6,7,8,9,0)

15. Counting Sample Points Techniques in counting sample points Multiplication ruleGeneralized Multiplication RulePermutationCircular PermutationDistinct PermutationCombination

16. Multiplication Rule Is the fundamental principle of counting Theorem #1 If an operation in n 1 ways,and if for each of these operation can be performed in n 2 ways,then the two operations can be performed in n 1 n 2 ways. Example: How many sample points are in the sample space when a fair coin is tossed twice? Solution: n 1 n 2 =6(6)ways= 36 possible ways

17. Generalized Multiplication Rule Generalized Multiplication Rule It is used when involving operations is more than two. Theorem #2 If an operation in n 1 ways,and if for each of these operation can be performed in n 2 ways,and for each of the first two a third operation can be performed in n 3 ways and so forth then the sequence of k operations can be performed in n 1 n 2 n 3 …….ways. Example: How many 3 digits can be formed from the digits 1, 2, 5, 6, and 9 if each no. can only used once? Solution: n 1 n 2 n 3 =5(4)(3)ways =60 possible ways

18. Permutation Permutation It is the arrangement of all or part of a set of objects in a definite order. Order is important aspect of permutation. Theorem #3 The number of permutation of n distinct objects is n! Example: Four people are to be arrange in a row to have their picture taken. In how many ways can this done? Solution: n= 4 people to be arrange in definite order n!= 4!=4(3)(2)(1)=24 ways

19. Permutation Permutation Theorem #4 The number of permutation of n distinct objects taken r at a time is n P r = n! (n-r)! Example: Five applicants enter an interview room in which there are seven seats.In how many ways may they be seated? Solution: n= 7;r=5 7 P 5 = 7! = 2520 ways (7-5)!

20. Circular Permutation Circular Permutation Permutations that occur by arranging objects in a circle are called circular permutation. Theorem # 5 The number of permutation of n distinct objects arranged in a circle is (n-1)!. Example: Seven children join hands and form a circle. In how many ways could the circle be formed? Solution: n=7 (n-1)!= (7-1)!= 6! = 720 ways

21. Distinct Permutation Distinct Permutation Theorem # 6 The number of distinct permutation of n things of which n 1 are the same kind, n 2 of a second kind……., n k of a second kind,…..n k of a kth kind is n! n 1 ! n 2 !....... n k Example: Eight books are to be arrange on a shelf. There are 2 math identical books,3 identical English books and 3 identical Physics books. How many distinct arrangement are possible? Solution: n= 8 books n 1 =2 math books n 2 =3 english books n 3 =3 physics books P=8! = 560 arrangements 2! 3! 3!

22. Combination Combination It is arrangement of number of ways of selecting r objects from n without regard to order. Theorem # 7 The number of combination of n distinct objects taken r at a time is n C r= n! r!(n-1)! Example: In how many ways a committee of 3 can be chosen from a group of 8? Solution: n= 8 ; r= 3 8 C 3= 8!= 56 different ways 3!(8-1)!

23. Probability of an event P(A) = The Number Of Ways Event A Can Occur The Total Number Of Possible Outcomes P(A)= n__ S

24. Basic Properties of Probability  Certainty- this property states that the probability that the event is on the sample space is one. P(S)=1  Non-negativity- this property states that every event has a non-negativity probability. The lowest probability of an event is 0 or no chance and the highest probability of an event is 1 or certain. P(E)≤0, for all E c S  Union- the probability of the union of mutually exclusive events is the sum of each event’s probability. P (A U B)= P (A) + P (B)

25. Example 1: A fair dice is tossed once. What is the probability that a. ) 5 will appear b.) an even number will appear? Solution: S={1,2,3,4,5,6}= 6 a.) If A is the event that a 5 will appear A= {5} n(A)= 1 P(A)= n(A) = 1 S 6 b.) If B is the event that an even number will appear B= {2,4,6} n(B)= 3 P(A)= n(A) =3 = 1 S 6 2

26. Example 2: Suppose that a card is drawn at random from an ordinary deck of playing cards,What is the probability of drawing a. ) a heart b.) a red card Solution: S=52 P(heart)= n(heart) = 13 = 1 S 52 4 P(red card)= n(red card) = 26 = 1 S 52 2

27. ADDITIVE RULES

28. It is applicable to union of events THEOREM If A and B are any two events, then P(AυB) = P(A)+P(B)-P(AΩ B) For three events A, B, and C, P(AυBυC) = P(A)+P(B)+P(C)-P(A Ω B)-P(A Ω C)-P(B Ω C)+P(AυBυC) If A and B are complementary events, then P(A)+P(A’)=1

29. EXAMPLES: ADDITIVE RULE 1.The probability that Mel passes Calculus is 2/3 and the probability that she passes statistics is 4/9. If the probability of passing both courses is ¼, what is the probability that Mel will pass at least one of these subjects? 2.If the probabilities are 0.9, 0.15, 0.21, and 0.23 that a person purchasing a new automobile will choose the color green, white, red and blue, what is the probability that a given buyer will purchase a new automobile that comes in one of those colors? (mutually exclusive)

30. Solution #1 Let A be the event that Mel passes Calculus Let B the event that Mel passes Statistics P(A U B)=P(A) + P(B) -P(A Ω B) 2 + 4 -1 3 9 4 =31 36

31. CONDITIONAL PROBABILITY

32. The probability of an event B occurring when it is known that some event A has occurred is called conditional probability and is denoted by P(B/A). The conditional probability of A, given B, denoted by P(A/B), P(A/B) = P(B Ω A) P(B) If P(B)>ø Two events A and B are independent if and only if P(B/A) = P(B) and P(A/B) = P(A). Otherwise, A and B are dependent.

33. EXAMPLES: CONDITIONAL PROBABILITY 1.A fair die is tossed once. Find the probability that a 4 appear, when it is known that a number greater than 2 results in the toss of the die. 2.The probability that a regularly scheduled flight departs on time is P(A)= 0.83, the probability that it arrives on time is P(B) = 0.82; and the probability that it departs and arrive on time is P(A ΩB) = 0.78. Find the probability that a plane a) arrives on time given that it departed on time. b) departed on time given that it has arrived on time.

34. Solution #1 S=(1,2,3,4,5,6) Let A be the event that 4 appear in a single toss. A=(4) Let B the event that a number is greater that 2 appear. B=(3,4,5,6) A ΩB= (4) 1 P(A/B)= 6 = 1 4 6

35. RANDOM VARIABLES (r.v) is a function whose value is a real number determined by each element in a sample space.

36. DISCRETE SAMPLE SPACE If the sample space contains a finite number of possibilities or a n unending sequence with the elements as there are whole numbers. CONTINUOUS SAMPLE SPACE If a sample space contains an infinite number possibilities equal to the number of points on a line segment.

37. DISCRETE RANDOM VARIABLE If the set of possible outcome is countable and it refers to the count data. CONTINUOUS RANDOM VARIABLE if it takes the values of on a continuous scale and it refers to measured data.

38. Example: 1. A shipment of 8 similar microcomputers to a retail outlet contains 3 that are defective. If a school makes a random purchase of 2 of these computers, find the probability distribution for the number of defectives.

39. Solution: Let X be a random variable whose values are the possible numbers of defective computers purchased by the school. Then X can be any of the numbers 0,1, and 2. f(0)= P(X=0) = 3 C 0 * 5 C 2 = 5 8 C 2 14 f(1)= P(X=1) = 3 C 1 * 5 C 1 = 15 8 C 2 28 f(2)= P(X=2) = 3 C 2 * 5 C 0 = 3 8 C 2 28 Therefore,the probability distribution of X is We can write P(X=x)= f(x) X012 P(X=x)5/1415/283/28

40. Mathematical Expectation Average no. of Heads = Experiment No Heads - 4 One Head- 7 Two Heads- 5 16 0(4)+1(7)+2(5) 16 Average Value is referred to as the Mean or Expected Value or the Mean of the Random variable of x or the Mean of the probability of distribution of x. Mean or Expected Value of X is µ=E(x)= ∑xf(x) if x is discrete µ=E(x)=∫ xf(x)dx if x is continuous x ∞ -∞

41. Mean of the Function on Random Variable Let x =be the random variable Then the variance of x is σ 2 = ∑(x-µ) 2 f(x), if x is discrete σ 2 = ∫ (x-µ) 2 f(x)dx, if x is continuous x ∞ -∞ Variance of the Random Variable Let g(x) =be the function of random variable Then the expected value of x is E (g(x))= ∑ g(x) f(x),if x is discrete E (g(x))= ∫ g(x) f(x) dx, if x is continuous x ∞ -∞

42. Example 1 If the probability mass function of x is given by: f(x)= 1/3,for x=1,2,3 0,elsewhere Find the mean and variance of x. Solution: µ=E(x)= ∑xf(x) = [1(1/3)+2(2/3)+3(1/3) µ = 2 σ 2 = ∑(x-µ) 2 f(x) =(1-2) 2 (1/3)+(2-2) 2 (1/3)+(3-2) 2 (1/3) =1/3+0+1/3 σ 2 =2/3

43. Example 2 Suppose that the probability density function of x is given by: f(x)= 1/2,for -1<x<1 0,elsewhere Find the mean and variance of x. Solution: µ=E(x)= ∫ xf(x)dx = ∫ x(1/2)dx = x 2 /4] µ = 0 σ 2 = ∫ (x-µ) 2 f(x)dx = ∫ (x-0) 2 (0)dx + ∫ (x-0) 2 (1/2)dx + ∫ (x-0) 2 (0)dx = x 3 /6] =1/6-(-1/6) σ 2 =1/3 ∞ -∞ ∞ -∞ 1 1 1 1 ∞1∞1 -∞

44. Example 3 If the probability mass function of x is given by: f(x)= 1/6,for x=1,2,3 0,elsewhere Let g(x)= x 2 Find the mean of g(x). Solution: E (g(x))= ∑ g(x) f(x) = ∑ x 2 (x/6) =[(1) 2 (1/6)+(2) 2 (2/6)+(3) 2 (3/6)] =1/6+8/6+27/6 E (g(x))= 6

45. Example 4 Given the probability density function of x : f(x)= 2(1-X),for 0<x<1 0,elsewhere Let g(x)= x 2 Find the mean of g(x). Solution: E (g(x))= ∫ g(x) f(x) dx E (x 2 ) = ∫ x 2 [2(1-x)]dx = ∫ (2x 2 -2x 3 )dx =2x 3 -2x 4 3 4 = 2/3-1/2 E (x 2 ) =1/6 1010 ∞ -∞ 1010 1010