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Unformatted text preview: Page 1 of 2 ECO220Y: Homework, Lectures 13 & 14 – SOLUTIONS (1) The number of trials and the probability of success. Go over how this relates to the example and provide the intuition. (2) The first example would be Binomial but the second case would not. Drawing cards without replacement would lead to a violation of the independence requirement for a Binomial Experiment. (3) Answer: n = 2. With two tosses you have a 50% chance of getting 50% heads (HT or TH). Each possible outcome has a 25% chance: HH, HT, TH, TT. For n = to any odd number (1, 3, 5, …) there is a 0% chance of getting 50% heads. For n = 4 or a larger even number the probability of getting 50% heads goes down. For example: n = 4 has 2 4 = 16 possible outcomes: (HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH THTT, TTHH, TTHT, TTTH, TTTT). Of these 16 possible outcomes only 37.5% (=6/16) have 50% heads. [You could, of course, show this using the combinatorial formula.] The reason is that as the number of tosses goes could, of course, show this using the combinatorial formula....
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 Spring '11
 tanaka
 Probability distribution, Probability theory, Discrete probability distribution, discrete random variable, Continuous probability distribution

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