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# lecture2 - ISYE2028A and B Probability Lecture 2 Dr Kobi...

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Unformatted text preview: ISYE2028A and B: Probability Lecture 2 Dr. Kobi Abayomi February 10, 2009 1 The basic principles of counting It is fitting that a probability class begins with counting, for a probability of an event may be seen as an enumeration of the ratio: number of possible ways for the event to happen to number of possible ways for anything to happen. The counting principle is: If an event can happen in any of n ways, and when the event has occurred, another event can happen in any of m ways, then the number of ways in which both events can happen is nm . Example : If there are 3 candidates for governor and 5 for mayor, the two offices can be filled in 3 5 = 15 ways Example : How many functions defined on n points are possible if each functional value is either 0 or 1: 2 n Example : How many different 7 place license plates are possible if the first 3 places are letters and the final 4 numbers: 26 3 * 10 4 . 1.1 Factorial n Factorial n , denoted by n !, is defined as: n ! = n ( n- 1)( n- 2) ··· 1 Thus 5! = 5 · 4 · 3 · 2 · 1 = 120.The factorial notation is very useful: 1 Example : How many different ‘words’ can be made with using each of the letters a,b,c,d,e only once: Pick the first letter - 5 ways to do it; pick the second letter - 4 ways,...pick the last letter 1 choice left → 5 · 4 · 3 · 2 · 1. Which brings us to... 1.2 Permutations A permutation of n different objects taken i at a time is an arrangement of i out of the n objects, with attention given to the order of arrangement . We denote the number of ar- rangements of n objects taken i at a time as: P n r = n ( n- 1) ··· ( n- i + 1) = n ! ( n- i )! Permutations arise naturally in settings where order matters. If you wanted to know the number of possible “words” of length four in the English alphabet you would: P 26 4 = 26! (26- 4)! = 26! 22! = 26 · 25 · 24 · 23 = 358800 Wow! A way to think about the permutation equation: If we were to count the total of number of ways to arrange n distinct objects, we would have n !. When we only look at i out of the n objects, we (basically) count the total number - n ! - and then divide the overcount - ( n- i )! out. The number of permutations of n objects consisting of k groups of which n 1 are alike, n 2 are alike, etc. is n ! n 1 ! n 2 ! ··· n k . For example, the number of permutations of letters in the word statistics: 10! 3!3!1!2!1! = 50400, since there are 3ss, 3ts, 1a, 2is, and 1c. Example : How many different letter arrangements can be formed using the letters in the word PEPPER : 6! 3!2! = 60. 2 Combinations, Binomial Theorem, Multinomials A combination of n different objects taken i at a time is a selection of i out of the n objects, with no attention given to the order of arrangement . Our notation here is: 2 C n i = n ( n- 1) ··· ( n- i + 1) i !...
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lecture2 - ISYE2028A and B Probability Lecture 2 Dr Kobi...

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