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Lecture (5)

Lecture (5). Introduction to Probability Theory and Applications. Experiments. Experiment: a process that generates well-defined outcomes. Sample space: all possible outcomes. Sample point: any particle outcome. For the “Roll a dice” example: S={1,2,3,4,5,6}. In the example {1}.

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Lecture (5)

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  1. Lecture (5) Introduction to Probability Theory and Applications

  2. Experiments Experiment: a process that generates well-defined outcomes. Sample space: all possible outcomes. Sample point: any particle outcome. For the “Roll a dice” example: S={1,2,3,4,5,6}. In the example {1}.

  3. Events An event is a collection of sample points (or, an event is a subset of a sample space) (1) Rolling two dices: (a) the sum of the numbers that come up is odd, (b) the numbers that come up are 3 and 6, etc. (2) Tossing two coins: at least one of them is head (it is a collection of the following sample points (H,T), (T,H) and (H,H)).

  4. Coin Tossing

  5. Sometimes the experiment consists of several steps. In such a case, a tree diagram is handy. Consider the experiment of tossing two coins. S={(H,H), (H,T), (T,H), (T,T)} Sample Point Step 1 First Coin Toss Step 2 Second Coin Toss (H,H) Head Head T Tail (H,T) (T,H) Tail Head Tail (T,T)

  6. Complement of an event: Given an event A, the complement of A is defined to be the event of all sample points that are not in A. The complement of A is denoted by Ac Venn Diagram Sample space S Ac Event A

  7. event Ac event A Rolling two dices(Example) Event A – at least one dice shows the number 1 Thus, event Ac – none of the dices shows the number 1

  8. Union of two events the union of A and B is the event containing all sample points belonging to Aor B or both. The union is denoted by Sample space S Event A Event B

  9. event B event A Rolling two dices (Example) Event A – at least one dice shows the number 1 Event B– the sum of the numbers is at most 4

  10. Intersection of two events Given two events A and B, the intersection of A and B is the event containing the sample points belonging to bothAand B. The intersection is denoted by Sample space S Event A Event B

  11. Rolling two dices (Example) event B event A Event A – at least one dice shows the number 1 Event B– the sum of the numbers is at most 4 The intersection (1,1), (1,2), (1,3), (2,1) and (3,1)

  12. Mutually exclusive events: two events are said to be mutually exclusive if the events has no sample points in common. Sample space S Event A Event B Events Definitions

  13. event B Rolling two dices event A Event A – at least one dice shows the number 1 Event B– the sum of the numbers is bigger than 10

  14. Probability Probability is a number expressing the likelihood that a specific event will occur. Let Eibe a specific event (for example, the sum of the numbers on two dices is 2.) Let be the probability of this event. Axioms of probability:

  15. Assessing Probability The probability of occurrence of a given event can be assessed as the ratio of the number of occurrences to the total number of possible occurrences and non-occurrences of the event. Occurrence of an event is called “ Success” Non-occurrence of the event is called “Failure” The number of successes in N trials = n The relative frequency of successes = n/N

  16. Assessing Probability (Cont.)

  17. Long Run Behavior of Coin Tossing

  18. Probability ( Ex.) Coin: Sample Space {H, T} Do the following experiments with one coin: n=5 n=10 n=20 n=50 n=100 Calculate p(H), p(T). Draw the convergence curve between n and p(H), p(T).

  19. Basic Probability Laws where S is the set of all the sample points Thus, Example (tossing two coins): A is the event “at least one of them is head”, and thus Ac is the event “none of the coins is head” Tossing two coins: A is the event “at least one of them is head”, and thus Ac is the event “none of the coins is head” and and Thus, Thus,

  20. Basic Probability Laws Let’s start with two mutually exclusive events. In this case, and Rolling two dices: Event A – at least one dice shows the number 1, and event B– the sum of the numbers is bigger than 10

  21. Dice #1 Dice #2 Dice #1 Dice #2 Basic Probability Laws (Example Solution) Rolling two dices: Event A – at least one dice shows the number 1, and event B– the sum of the numbers is bigger than 10

  22. Basic Probability Laws Why do we need to subtract when the two events are not mutually exclusive? Because otherwise we would double-count it—both and include it. Rolling two dices: Event A – at least one dice shows the number 1, and event B– the sum of the numbers is at most 4

  23. Basic Probability Laws (Example Solution) Rolling two dices: Event A – at least one dice shows the number 1, and event B– the sum of the numbers is at most 4 Dice #1 Dice #2

  24. How did you calculate ? Conditional Probability One of the most important concepts. It is denoted as which means “the probability of event A given the condition that event B has occurred.” Rolling two dices: Event A – at least one dice shows the number 1, and event B– the sum of the numbers is at most 4. What is the probability of event A? And what is the probability of event A if we know that eventB has occurred?

  25. Event A Event B has occurred: Event A and B with probability Conditional Probability (Cont.) Event B Event B with probability

  26. General Multiplication Law Multiplication Rule for Independent Events Independent events are events in which the occurrence of the events will not affect the probability of the occurrence of any of the other events.

  27. The multiplication Rule for Independent Events (Example) Example: Picking a color from a set of crayons, then tossing a die. Separately, each of these events is a simple event and the selection of a color does not affect the tossing of a die. If the set of crayons consists only of red, yellow, and blue, the probability of picking red is . The probability of tossing a die and rolling a 5 is . Butthe probability of picking red and rolling a 5 is given by:

  28. The multiplication Rule for Independent Events (Example) Cont. This can be illustrated using a “tree” diagram. Since there are three choices for the color and six choices for the die, there are eighteen different results. Out of these, only one gives a combination of red and 5. Therefore, the probability of picking a red crayon and rolling a 5 is given by:

  29. The multiplication Rule for Independent Events The general formula states: The multiplication rule for independent events can be stated as:This rule can be extended for more than two independent events:

  30. Multiplication Rule for Dependent Events The occurrence of one event affects the probability of the occurrence of other events. An example of dependent events: Picking a card from a standard deck then picking another card from the remaining cards in the deck. For instance, what is the probability of picking two kings from a standard deck of cards? The probability of the first card being a king is . However, the probability of the second card depends on whether or not the first card was a king.

  31. Multiplication Rule for Dependent Events (Example) Cont. If the first card was a king then the probability of the second card being a king is . If the first card was not a king, the probability of the second card being a king is . Therefore, the selection of the first card affects the probability of the second card.

  32. Multiplication Rule for Dependent Events (Example) Cont. The multiplication rule that include dependent events reads:

  33. Example 2 In a group of 25 people 16 of them are married and 9 are single. What is the probability that if two people are randomly selected from the group, they are both married? If A represents the first person chosen is married and B represents the second person chosen is married then:Here, is now the event of picking another married person from the remaining 15 married persons. The probability for the selection made in B is affected by the selection in A.

  34. Estimating Probability form Histogram 36 27 17.3 1.3 9 1.3 8 Probability that Q is 10,000 to 15, 000 = 17.3% Prob that Q < 20,000 = 1.3 + 17.3 + 36 = 54.6%

  35. Estimating Probability form Cumulative Histogram Continuous F(x1) - F(x2) Discrete F(x1) = P(x < x1) Prob that Q < 20,000 = 1.3 + 17.3 + 36 = 54.6% Prob that Q = 20,000, = 80.6-54.6 = 27%

  36. Bivariate Distributions • The bivariate (or joint) distribution is used when the relationship between two random variables is studied. • The probability that X assumes the value x, and Y assumes the value y is denoted p(x,y) = P(X=x and Y = y)

  37. Bivariate Distributions

  38. Bivariate Distributions • Example • X and Y are two variables. Let X and Y denote the yearly runoff and rainfall (inches) respectively. • The bivariate probability distribution is presented next.

  39. Bivariate Distributions 0.42 p(x,y) Example continued X (in) Y (in) 0 5 10 0 .12 .42 .06 5 .21 .06 .03 10 .07 .02 .01 0.21 0.12 0.06 X 0.06 y=0 0.03 0.07 0.02 y=5 0.01 Y y=10 X=0 X=5 X=10

  40. X Y 0 5 10 p(y) 0 .12 .42 .06 .60 5 .21 .06 .03 .30 10 .07 .02 .01 .10 p(x) .40 .50 .10 1.00 P(Y=5), the marginal probability. Marginal Probabilities • Example- continued • Sum across rows and down columns p(0,0) p(0,5) p(0,10) The marginal probability P(X=0)

  41. X Y 0 5 10 p(y) 0 .12 .42 .06 .60 5 .21 .06 .03 .30 10 .07 .02 .01 .10 p(x) .40 .50 .10 1.00 The sum is equal to 1.0 Conditional Probability Example - continued

  42. Counting Techniques: Permutations Permutations: It is an arbitrary ordering of a number of different objects using all of them. If we want to permute n different objects in r arrangements: The number of permutations of n different objects is the number of different arrangements in which r (r<n) of these objects can be placed, with attention given to the order of the items in each arrangement. If n=r, The number of permutations of n different objects of groups of which n_i are alike, is

  43. Example 1 Example: with objects a, b, and c how many permutations can we do in 2 arrangements?: ab, ac, ba, bc, ca, cb First place: we have 3 choices. Second place we have two choice. Altogether =3!/(3-2)!=6 permutation of three objects in two arrangements.

  44. Example 2 Example: with objects a, b, and c we can produce six permutations: abc, acb, bac, bca, cab, cba First place: we have 3 choices. Second place we two choices. Third place we one choice. Altogether 3*2*1 = 3! =6 permutation of three objects. For n objects we get n*(n-1)*(n-2)………2*1=n! permutations

  45. Example 3 The number of permutations of n objects consisting of groups of which n_i are alike, is The number of permutations of letters in the word “STATISTICS” =50400.

  46. Counting Techniques: Combinations Combinations: The number of combinations of n different objects r at a time (n>r) is the number of different selection that can be made of r objects out of n, without giving attention to the order of arrangement within each selection. Stirling formula

  47. Example 1 Example: with objects a, b, and c we can produce six permutations: abc, acb, bac, bca, cab, cba 3!=3*2*1=6 But three combinations: abc=cba=acb=bca=bac=cab 3!/3!.0!=3*2*1/3*2*1=1

  48. Example 2 Example: Country A: 7 watersheds, and Country B: 4 Watersheds. Five watersheds have to be selected for research 3 from A and 2 from B. How many different ways this choice can be made? Solution: This is a problem of combinations because the order of the watersheds is not important. A: n= 7, r= 3 B: n=4, r=2

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