1. Sample Space and Probability Part IV: Pascal Triangle and Bernoulli Trials. ECE 302 Spring 2012 Purdue University, School of ECE Prof.
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1 1. Sample Space and Probability Part IV: Pascal Triangle and Bernoulli Trials ECE 302 Spring 2012 Purdue University, School of ECE Prof. Ilya Pollak
2 ConnecGon between Pascal triangle and probability theory: Number of successes in a sequence of independent Bernoulli trials A Bernoulli trial is any probabilisgc experiment with two possible outcomes Ilya Pollak
3 ConnecGon between Pascal triangle and probability theory: Number of successes in a sequence of independent Bernoulli trials A Bernoulli trial is any probabilisgc experiment with two possible outcomes, e.g., Will CiGgroup become insolvent during next 12 months? Democrats or Republicans in the next elecgon? Will Dow Jones go up tomorrow? Will a new drug cure at least 80% of the pagents? Ilya Pollak
4 ConnecGon between Pascal triangle and probability theory: Number of successes in a sequence of independent Bernoulli trials A Bernoulli trial is any probabilisgc experiment with two possible outcomes, e.g., Will CiGgroup become insolvent during next 12 months? Democrats or Republicans in the next elecgon? Will Dow Jones go up tomorrow? Will a new drug cure at least 80% of the pagents? Terminology: somegmes the two outcomes are called success and failure. Suppose the probability of success is p. What is the probability of k successes in n independent trials? Ilya Pollak
5 Probability of k successes in n independent Bernoulli trials n independent coin tosses, P(H) = p Ilya Pollak
6 Probability of k successes in n independent Bernoulli trials n independent coin tosses, P(H) = p E.g., P(HTTHHH) = p(1- p)(1- p)p 3 = p 4 (1- p) 2 Ilya Pollak
7 Probability of k successes in n independent Bernoulli trials n independent coin tosses, P(H) = p E.g., P(HTTHHH) = p(1- p)(1- p)p 3 = p 4 (1- p) 2 P(specific sequence with k H s and (n- k) T s) = p k (1- p) n- k Ilya Pollak
8 Probability of k successes in n independent Bernoulli trials n independent coin tosses, P(H) = p E.g., P(HTTHHH) = p(1- p)(1- p)p 3 = p 4 (1- p) 2 P(specific sequence with k H s and (n- k) T s) = p k (1- p) n- k P(k heads) = (number of k- head sequences) p k (1- p) n- k Ilya Pollak
9 Probability of k successes in n independent Bernoulli trials n independent coin tosses, P(H) = p E.g., P(HTTHHH) = p(1- p)(1- p)p 3 = p 4 (1- p) 2 P(specific sequence with k H s and (n- k) T s) = p k (1- p) n- k P(k heads) = (number of k- head sequences) p k (1- p) n- k Ilya Pollak
10 An interesgng property of binomial coefficients Since P(zero H's) + P(one H) + P(two H's) + + P(n H's) =1, it follows that n k = 0 n p k (1 p) n k =1. k Another way to show the same thing is to realize that n k = 0 n p k (1 p) n k = (p + (1 p)) n =1 n =1. k Ilya Pollak
11 Binomial probabiliges: illustragon Ilya Pollak
12 Binomial probabiliges: illustragon Ilya Pollak
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27 Comments on binomial probabiliges and the bell curve Summing many independent random contribugons usually leads to the bell- shaped distribugon. Ilya Pollak
28 Comments on binomial probabiliges and the bell curve Summing many independent random contribugons usually leads to the bell- shaped distribugon. This is called the central limit theorem (CLT). We have not yet covered the tools to precisely state the CLT, but we will later in the course. Ilya Pollak
29 Comments on binomial probabiliges and the bell curve Summing many independent random contribugons usually leads to the bell- shaped distribugon. This is called the central limit theorem (CLT). We have not yet covered the tools to precisely state the CLT, but we will later in the course. The behavior of the binomial distribugon for large n shown above is a manifestagon of the CLT. Ilya Pollak
30 InteresGngly, we get the bell curve even for asymmetric binomial probabiliges Ilya Pollak
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65 This tells us how to empirically esgmate the probability of an event! To esgmate the probability p based on n flips, divide the observed number of H s by the total number of experiments: k/n. To see the distribugon of k/n for any n, simply rescale the x- axis in the distribugon of k. This distribugon will tell us What we should expect our esgmate to be, on average, and What error we should expect to make, on average Ilya Pollak
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70 Note: o for 50 flips, the most likely outcome is the correct one, 0.8 o it s also close to the average outcome o it s very unlikely to make a mistake of more than 0.2 Ilya Pollak
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73 If p=0.8, when estimating based on 1000 flips, it s extremely unlikely to make a mistake of more than Ilya Pollak
74 If p=0.8, when estimating based on 1000 flips, it s extremely unlikely to make a mistake of more than Hence, when the goal is to forecast a two-way election, and the actual p is reasonably far from 1/2, polling a few hundred people is very likely to give accurate results. Ilya Pollak
75 If p=0.8, when estimating based on 1000 flips, it s extremely unlikely to make a mistake of more than Hence, when the goal is to forecast a two-way election, and the actual p is reasonably far from 1/2, polling a few hundred people is very likely to give accurate results. However, o independence is important; o getting a representative sample is important (for a country with 300M population, this is tricky!) o when the actual p is extremely close to 1/2 (e.g., the 2000 presidential election in Florida or the 2008 senatorial election in Minnesota), pollsters forecasts are about as accurate as a random guess. Ilya Pollak
76 The 2008 Franken- Coleman elecgon Franken 1,212,629 votes Coleman 1,212,317 votes In our analysis, we will disregard third party candidate who got 437,505 votes (he actually makes pre- elecgon polling even more complicated) EffecGvely, p Ilya Pollak
77 ProbabiliGes for fracgons of Franken vote in pre - elecgon polling based on n=2.5m (more than all Franken and Coleman votes combined) Even though we are unlikely to make an error of more than 0.001, this is not enough because p-0.5= ! Note: 42% of the area under the bell curve is to the left of 1/2. When the election is this close, no poll can accurately predict the outcome. In fact, the noise in the voting process itself (voting machine malfunctions, human errors, etc) becomes very important in determining the outcome. Ilya Pollak
78 EsGmaGng the probability of success in a Bernoulli trial: summary As the number n of independent experiments increases, the empirical fracgon of occurrences of success becomes close to the actual probability of success, p. The error goes down proporgonately to n 1/2. I.e., error aler 400 trials is twice as small as aler 100 trials. This is called the law of large numbers. This result will be precisely described later in the course. Ilya Pollak
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