Probability

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## Probability

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**Properties of probabilities**• 0 ≤ p(A) ≤ 1 • 0 = never happens • 1 = always happens • A priori definition • p(A) = number of events classifiable as A total number of classifiable events • A posteriori definition • p(A) = number of times A occurred total number of occurrences**So:**p(A)= nA/N = number of events belonging to subset A vs. the total possible (which includes A). If 6 movies are playing at the theater and 5 are crappy but 1 is not so crappy what is the probability that I will be disappointed? 5/6 or p = .8333**Probability in perspective**• Analytic (classical) view • The common approach: if there are 5 bad movies and one good one I have an 83% chance in selecting a bad one. • Fisher • Relative Frequency view • Refers to the long run of events: the probability is the limit of chance i.e. in a hypothetical infinite number of movie weekends I will select a bad movie about 83% of the time • Neyman-Pearson • Subjective view • Probability is akin to a statement of belief and subjective e.g. I always seem to pick a good one. • Bayesian**Experiment**• Experiment -- a Process that produces outcomes • More than one possible outcome • Only one outcome per trial • Trial -- one repetition of the process • Event -- an outcome of an experiment • may be an elementary event, usually represented by an uppercase letter, e.g., A, E1**Generally we can calculate the probability of one of a set**of equally likely events by counting the sample space • Many problems in probability can be solved in this way • probability very often makes use of combinatorics (permutation and combination – we’ll talk about this later)**Sample Space -- Set Notation for Random Sample of Two**Families • S = {(x,y) | x } • x is the family selected on the first draw • y is the family selected on the second draw • Concise description of large sample spaces**Sample Space**• The set of all elementary events for an experiment • Methods for describing a sample space • Listing • Venn diagram**Family**Children in Household Number of Automobiles Listing of Sample Space (A,B), (A,C), (A,D), (B,A), (B,C), (B,D), (C,A), (C,B), (C,D), (D,A), (D,B), (D,C) A B C D Yes Yes No Yes 3 2 1 2 Sample Space -- Listing Example • Experiment: randomly select, without replacement, two families from the residents of Denton • Each ordered pair in the sample space is an elementary event, for example -- (D,C)**Y**X Venn Diagram Venn Diagrams and the union of sets • Venn diagrams helps us pictorially represent many of the algebraic rules of probability • The union of two sets contains an instance of each element of the two sets.**Y**X Venn Diagram Intersection of Sets • The intersection of two sets contains only those elements common to the two sets.**Definitions**• Mutually exclusive events • both events cannot occur simultaneously. • Can’t be a junior and senior • Complementary events • Two mutually exclusive events that are all inclusive • Independent events: • occurrence of one event has no effect on the probability of occurrence of the other**Y**X Mutually Exclusive Events • Events with no common outcomes • Occurrence of one event precludes the occurrence of the other event Venn Diagram**Complementary Events**• All elementary events not in the event ‘A’ are in its complementary event. Sample Space A**Exhaustive sets**• set includes all possible events • the sum of probabilities of all the events in the set = 1 • Equal likelihood • roll a fair die each time the likelihood of 1-6 is the same whichever one we get, we could have just as easily have gotten another • Counter example- put the numbers 1-7 in a hat. What’s the probability of even vs. odd?**AND vs. OR**• How do we find the probability of one event or another occurring? • How do we find the probability of one event and another occurring?**Addition**• p(A or B) = p(A) + p(B) • Probability of getting a grape orlemon skittle in a bag of 60 pieces where there are 15 strawberry, 13 grape, 12 orange, 8 lemon, 12 lime? • p(G) = 13/60 p(L) = 8/60 • 13/60 + 8/60 = 21/60 = .35 or a 35% chance we’ll get one of those two flavors when we open the bag and pick one out**Y**X General Law of Addition (not necessarily mutually exclusive)**Married (Yes/No)**Yes No Total Children (Yes/No) .70 .14 .56 Yes .19 .11 No .30 .33 1.00 .67 Total Example: Marriage and Children**S**N .56 .70 .67 General Law of Addition**Married (Yes/No)**Yes No Total Children (Yes/No) .70 .14 .56 Yes .19 .11 No .30 .33 1.00 .67 Total Contingency Table**Multiplication**If A & B are independent • p(A and B) = p(A)p(B) • p(A and B and C) = p(A)p(B)p(C) • Probability of getting a grape and a lemon after two draws (with replacement) from the bag • p(Grape)*p(Lemon) = 13/60*8/60 = ~.0288**Conditional probabilities**If events X and Y are not independent then: • p(X|Y) = probability that X happens given that Y happens • The probability of X “conditional on” or “given” Y occurs • It’s our ‘and’ type of question from before so we are going to use multiplication, however we don’t have independent events so it will be a little different • p(A and B) = p(A)*p(B|A)**Example: once we grab one skittle we aren’t going to put**it back (sampling without replacement) so: • p(A and B and C) = p(A)*p(B|A)*p(C|A,B) • Probability of getting grape and lemon on successive turns = p(G)*p(L|G) • (13/60)(8/59) = .0293**Conditional and Joint Probabilities**• A conditional probability is one where you are looking for the probability of some event with some sort of information in hand, e.g. the odds of having a boy given that you had a girl already. • A joint event or probability would be the probability of a combination of events e.g. that you have a boy and a girl for children • Aside: In this case the conditional would be higher b/c if we knew there was already a girl that means they’re of child-rearing age, probably interested in having kids etc. We have some additional info that would help us if we were just drawing out people randomly from some population.**Conditional Probability**Venn Diagram N S .56 .70 *note before with our previous conditional probability we were dealing with mutually exclusive events i.e. can’t be grape and lemon at same time**Let’s do a conditional probability: If I have a male,**what is the probability of him being in the ‘Other’ category? Formally: • p(A|B) = p(A and B)/p(B) = • p(O|M) = p(O and M)/p(M) = = (.412*.714)/.588= .5**Easier way by looking at table- there are 10 males and of**those 10 (i.e. given that we are dealing with males) how many are “Other”? • p(O|M) = 5/10 or 50%.**Joint Probability Example**• What is the probability of obtaining a Female Independent from this sample? • In this case we’re looking for the joint probability of someone who is Female and Independent out of all possible outcomes: 2/17 = 11.8**Practice**• a. What is the probability a person is over 25? • b. What is the probability that people under 25 spend at least 10 hours on the internet? • c. What is the probability that someone who does not spend 10 hours on the internet each week is over 25? • d. What is the probability of picking someone who spends less than 10 hours/wk on the internet and is under 25?**a. 45/110 = .41**• b. 50/65 = .77 • c. 20/35 = .57 • d. 15/110 = .14