Lecture3_Spring11

Lecture3_Spring11 - Elec 210: Lecture 3 Computing...

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Elec210 Lecture 3 1 Elec 210: Lecture 3 Computing Probabilities using Counting Methods Sample Size Computation and Examples Probabilities and Poker!
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Elec210 Lecture 3 2 Computing Probabilities Using Counting Methods In experiments where the outcomes are equiprobable , we can compute the probability of any event by counting the number of outcomes in the event and dividing by the total number of outcomes in the sample space. Thus, P[A] is a measure of the size of the set A. Here we develop several formulas that are useful for counting problems which are posed as sampling (choosing) problems Balls from an urn Cards from a deck Objects from a population Answers to a multiple-choice question space sample in outcomes of number A in outcomes of number ] [ = A P
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Number of ordered 2-tuples Assume a multiple choice exam with 2 questions. Question 1 has n 1 possible answers: a 1 , a 2 , . .., a n 1 Question 2 has n 2 possible answers: b 1 , b 2 , . .., b n 2 Let x 1 represent the answer to question 1, and similarly for x 2 . Each possible pair is called a 2-tuple, e.g., The number of possible ways to answer the test is n 1 n 2 . Elec210 Lecture 3 3 () 12 38 ,, x xa b =
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Number of ordered k-tuples Assume a multiple choice exam with k questions. Question i has n i possible answers. Each possible way of answering the test is called a k -tuple: where x i represents the chosen answer to question i . The total number of possible ways to answer the test is Elec210 Lecture 3 4 123 1 k ik i nn n n n = = () ,,, , k xx x x The Counting Principle
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Example Suppose the test has 5 questions, and each question has 4 possible answers. One student’s answer can be represented by the 5-tuple: i.e., for the first question the student chose answer 4, for the second question the student chose answer 3, and so on. .. Since each question has 4 possible answers, Thus, the total number of possible ways to answer the test is Elec210 Lecture 3 5 () 12345 ,,,, ( 4 , 3 , 1 , 4 , 2 ) xx x x x = 12 345 4 nn n n n == === 5 4 nnnnn =
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Elec210 Lecture 3 6 Sampling (Replacement and Ordering) We can choose k objects from a set A that has n members in different ways Replacement With replacement : after selecting an object and noting its identity, the object is put back before the next selection.
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This note was uploaded on 03/27/2011 for the course ELEC 202 taught by Professor ? during the Spring '11 term at HKUST.

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Lecture3_Spring11 - Elec 210: Lecture 3 Computing...

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