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**Introduction to Probability**

Lecture (5) Introduction to Probability Theory and Applications

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**For the “Roll a dice” example: S={1,2,3,4,5,6}.**

Experiments Experiment: a process that generates well-defined outcomes. Experiment outcomes Roll a dice 1,2,3,4,5,6 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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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)).

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Coin Tossing

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**Sometimes the experiment consists of several steps**

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)

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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

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**Rolling two dices (Example)**

event A event Ac Event A – at least one dice shows the number 1 Thus, event Ac – none of the dices shows the number 1

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

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**Rolling two dices (Example)**

event A event B Event A – at least one dice shows the number 1 Event B – the sum of the numbers is at most 4

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**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 both A and B. The intersection is denoted by Sample space S Event A Event B

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**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)

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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

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**Rolling two dices event B event A**

Event A – at least one dice shows the number 1 Event B – the sum of the numbers is bigger than 10

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**Let be the probability of this event.**

Probability is a number expressing the likelihood that a specific event will occur. Let Ei be 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:

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**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

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**Assessing Probability (Cont.)**

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**Long Run Behavior of Coin Tossing**

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**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).

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**Basic Probability Laws**

where S is the set of all the sample points Thus, 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” 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” and and Thus, Thus,

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**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

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**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 Dice #1 Dice #2

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**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

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**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

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**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 event B has occurred? How did you calculate ?

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**Conditional Probability (Cont.)**

Event A Event B Event B with probability Event B has occurred: Event A and B with probability

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**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.

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**The multiplication Rule for Independent Events (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 But the probability of picking red and rolling a 5 is given by:

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**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:

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**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:

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**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.

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**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.

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**Multiplication Rule for Dependent Events (Example) Cont.**

The multiplication rule that include dependent events reads:

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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.

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**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 = = 54.6%

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**Estimating Probability form Cumulative Histogram**

Continuous F(x1) - F(x2) Discrete F(x1) = P(x < x1) Prob that Q < 20,000 = = 54.6% Prob that Q = 20,000, = = 27%

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**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)

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**Bivariate Distributions**

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**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.

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**Bivariate Distributions**

0.42 Example continued p(x,y) X (in) Y (in) 0.21 0.12 0.06 X 0.06 y=0 0.07 0.03 0.02 y=5 0.01 Y y=10 X=0 X=5 X=10

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**Marginal Probabilities**

Example- continued Sum across rows and down columns X Y p(y) p(x) p(0,0) P(Y=5), the marginal probability. p(0,5) p(0,10) The marginal probability P(X=0)

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**Conditional Probability**

X Y p(y) p(x) Example - continued The sum is equal to 1.0

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**Counting Techniques: 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

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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.

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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

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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.

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**Counting Techniques: 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

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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

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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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