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## The Big Picture:

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**The Big Picture:**• Counting events in a sample space allows us to calculate probabilities • The key to calculating the probabilities of events is to count the occurrences of events • Our goal is to calculate probabilities of ALL possible events • We can do this by counting the total number of events possible in the sample space, using combinatorial mathematics to do this**Permutations**• Permutation is a sequence or ordering of events. • Basic Question: if I have N objects, how many different orderings of them are there? • N! • Formula: N(N-1)(N-2)…(1) • Example: 5(5-1)(5-2)(5-3)(5-4) • 5*4*3*2*1 = 120**Permutations**• General formula for finding the number of permutations of size k taken from n objects**Example**e.g. 10 compact discs and a 6 disc carousel- What is the number 6 disc orderings that we can make from the 10 cds?: 10! = 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 (10 - 6)! 4! 4 x 3 x 2 x 1 = 10 x 9 x 8 x 7 x 6 x 5 = 151200**Combinations**• General formula for finding the number of unique combinations of k objects you can choose from a set of n objects**Example**e.g. How many sets of 6 discs can we make from 10 cds (without repeating the same combinations)? 10! = 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 6!(10 - 6)! 6!(4!) (6 x 5 x 4 x 3 x 2 x 1)( 4 x 3 x 2 x 1) = 10 x 9 x 8 x 7 = 5040 = 210 (4 x 3 x 2 x 1) 24**Note the distinction**• Before we were counting our cd groupings without regard to the same groupings of a different order • But when looking for combinations, similar groupings are not counted as different • Pavement, The New Year, Built to Spill, Pixies, Grifters, Sonic Youth • Built to Spill, Grifters, Pavement, Sonic Youth, The New Year, Pixies • 2 permutations, but 1 combination (same set of 6 are involved)**Bernoulli Trials**• Bernoulli trials = 2 mutually exclusive outcomes • Distribution of outcomes • e.g. the probability of various outcomes in terms of numbers of heads and tails • Order of items does not matter • N = # trials = 3**Coin toss**• How many possible outcomes of the 3 coin tosses are there? • List them out: • HHH HHT HTT TTT TTH THH THT HTH • Now condense them ignoring order • i.e. HTT = THT = TTH • Only one occurrence of heads on 3 trials • Probability of 0 heads, only 1 heads, 2 heads, all 3 heads?**Now how about 10 coin flips? That’d be a lot of work**writing out all the possibilities. What’s another way to find the probability of coin flips?**Binomial distribution**• Find a probability for an event using: • N = number of trials • r = number of ‘successes’ • p = probability of ‘success’ on any trial • q = 1-p (probability of ‘failure’) • CNr=The number of combinations of N things taken r at a time**So if I want to know the odds of getting 9 heads out of 10**coin flips or p(H,H, H,H, H,H, H,H, H,T): • p(9) = • 10(.001953)(.5)=.0098 = .01**Now if we did this for all possible hits (heads) on 10**flips:**Using these probabilities**• What is the probability of getting 4 or fewer heads in 10 coin tosses? • Addition • p(4 or less) = p(4) + p(3) + p(2) + p(1) + p(0) = .205 + .117 + .044 +.010 + .001 = p = .377 About 38% of the time**Now take it out a step**• Suppose you were giving some sort of treatment to depressed individuals and assumed the treatment in general would have a 50/50 chance of doing so if it wasn’t anything special (i.e. just a placebo). Then it worked an average of 9 times out of 10 administrations. • Would you think there was something special going on or just chance?**Not just 50/50**• Not every 2 outcome situation has equal probabilities associated with each option • There are two parameters we are concerned with when considering a binomial distribution. • p = the probability of a success. • n = the number of Bernoulli trials p = .8 N = 10 coins**Also...**• More info about binomial distribution m = Np s2=Nqp In Excel =BINOMDIST(success,total N, prob., FALSE)**Binomial distribution shape**Approximately “normal” curve when: • p is close to 0.5 • If not then “skewed” distribution • N large • If not then not a representative distribution**Binomial in Action**• Say someone claims ESP and we give them a test. The card can only be a star or a circle on the one side and we let them guess as to what it is. We test them 20 times and they guess right on 13, which chance alone would dictate only 10 or .5 of the time. • 2 Hypes: • H0 = their performance is really no different from chance • H1 = their performance is above chance. • The probability of guessing 13 is ~.07. Pretty unlikely! • However our question isn’t really about guessing 13 is it? It is about guessing 13 or more. Guessing at least 13 right has a chance of (or p value) ~.13 Still not that likely but not unreasonable. Think if there was a 13% chance of rain. We’d be surprised but not completely amazed.**The problem with probabilities**• Can be very hard to grasp • e.g. Monty Hall problem • TV show “Let’s make a deal” • 3 closed doors, behind 1 is a prize • Select a door • Monty Hall opens one of the remaining doors that does NOT contain a prize • Now allowed to keep your original door or switch to the other one • Does it make a difference if you switch?