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CS6825: Probability An Introduction Definitions An experiment is the process of observing a phenomenon with multiple possible outcomes The sample space of an experiment is all possible outcomes • The sample space may be discrete or continuous An event is a set (collection) of one or more outcomes in the sample space Presenting data Pie and bar charts Body Pixels Background Face Pixels Other Frequency diagram Scatter diagram Taken from “Multidimensional Representation of Concepts as Cognitive Engrams in the Human Brain “ Pie chart A Pie Chart is useful for presenting nominal data. • For each category we calculate the relative frequency of its occurrence. • Then we take a circle and divide (slice) it proportionally to the relative frequency and portions of the circle are allocated for the different groups Body Pixels Background Face Pixels Other Example A manager of Athletics store has to decide, which brands to keep in the new season. 200 runners were asked to indicate their favorite type of running shoe. Type of shoe # of runners % of total Nike 92 46.0 Adidas 49 24.5 Reebok 37 18.5 Asics 13 6.5 Other 9 4.5 Example: Pie chart for running shoes 18.50% 6.50% 4.50% Nike Adidas Reebok Asics Other 24.50% 46% We can express this in words by saying the probability of Nike is 46% and the probability of Reebok is 18.5% The probability of an event is the proportion of times the event is expected to occur in repeated experiments Probability Properties The probability of an event, say event A, is denoted P(A). All probabilities are between 0 and 1. (i.e. 0 < P(A) < 1) Sample Space – set of all possible events. In previous example Set = {Nike, Adidas, Reebok, Asic, Other} The sum of the probabilities of all possible outcomes (sample space) must be 1. NOTE: it is possible to us a scale of 100% instead of 1 but, in statistics we use the scale of 1. What are the Probabilities 18.50% 6.50% 4.50% Nike Adidas Reebok Asics Other 24.50% 46% P(Nike) = 46/100 = .46 P(Adidas) = 24.5/100 = .245 P(Reebok) = 18.5/100 = .185 P(Asics) = 6.5/100 = .065 P(Other) = 4.5/100 = .045 Assigning Probabilities Guess based on prior knowledge alone Guess based on knowledge of probability distribution (to be discussed later) Assume equally likely outcomes Use relative frequencies Guess based on prior knowledge alone a priori Knowledge Event B = {It rains Tomorrow} Weth R. Guy says “There is a 30% chance of rain tomorrow.” P(B) = .30 What do to when no prior knowledge and no training data …..Assume equally likely outcomes Use Relative Frequencies Gather training data to estimate probabilities….Flip a coin how many times get head versus tails. i.e. Take a bunch of images of the data and see what it means to be yellow for a banana? Additional material…. Beyond the very beginning Complement* The complement of an event A, denoted by A, is the set of outcomes that are not in A A means A does not occur * Some texts use Ac to denote the complement of A Law of Complement P(A) = Probability of any event except A occurring = P(all Events) - P(A) = Sum(all events i P(i)) – P(A) = 1 – P(A) Union The union of two events A and B, denoted by A U B, is the set of outcomes that are in A, or B, or both If A U B occurs, then either A or B or both occur Intersection The intersection of two events A and B, denoted by AB, is the set of outcomes that are in both A and B. If AB occurs, then both A and B occur Addition Law P(A U B) = P(A) + P(B) - P(AB) (The probability of the union of A and B is the probability of A plus the probability of B minus the probability of the intersection of A and B) Mutually Exclusive Events* Two events are mutually exclusive if their intersection is empty. Two events, A and B, are mutually exclusive if and only if P(AB) = 0 Addition Law for Mutually Exclusive Events P(A U B) = P(A) + P(B) Conditional Probability The probability of event A occurring, given that event B has occurred, is called the conditional probability of event A given event B, denoted P(A|B) Conditional Probability P(AB) P(A|B) = -------P(B) or P(AB) = P(B)P(A|B) Independence If P(A|B) = P(A) or P(B|A) = P(B) or P(AB) = P(A)P(B) then A and B are independent. Independence Two events A and B are independent if P(A|B) = P(A) or P(B|A) = P(B) or P(AB) = P(A)P(B) NOTE: this is an assumption sometimes researchers make about their systems when they have no a priori knowledge to tell them differently. They do it as it makes math simpler. BE CAREFUL, it may be a WRONG Assumption!!! i.e. in motion tracking – person 1 leaves means nothing about person 2 leaving. They are independent….. But, is this true in practice?