4Equiprobable_Outcomes

4Equiprobable_Outcomes - Chapter 4 Equiprobable Outcomes...

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Unformatted text preview: Chapter 4 Equiprobable Outcomes and Combinatorics The simplest probabilistic scenario is perhaps one where the sample space Ω is finite and all the possible outcomes are equally likely. In such cases, computing the probability of an event amounts to counting the number of outcomes comprising this event and dividing the sum by the total number of elements contained in the sample space. This situation was first explored in Section 2.2.1, with an explicit formula for computing probabilities appearing in (2.2). While counting outcomes may appear intuitively straightforward, it is in many circumstances a daunting task. For instance, consider the number of distinct subsets of the integers { 1 , 2 , . . . , n } that do not contain two consecutive integers. This number is equal to φ n +2- (1- φ ) n +2 √ 5 , where φ = (1 + √ 5) / 2 is the golden ratio . This somewhat surprising result can be obtained iteratively through the Fibonacci recurrence relation, and it attests of the fact that counting may at times be challenging. Calculating the number of ways that certain patterns can be formed is part of the field of combinatorics . In this chapter, we introduce useful counting techniques that can be applied to situations pertinent to probability. 33 34 CHAPTER 4. EQUIPROBABLE OUTCOMES 4.1 The Counting Principle The counting principle is a guiding rule for computing the number of elements in a cartesian product. Suppose that S and T are finite sets with m and n elements, respectively. The cartesian product of S and T is given by S × T = { ( x, y ) | x ∈ S and y ∈ T } . The number of elements in the cartesian product S × T is equal to mn . This is illustrated in Figure 4.1. 1 1 1 1 2 2 2 2 3 3 3 3 a a a a b b b b Figure 4.1: This figure provides a graphical interpretation of the cartesian product of S = { 1 , 2 , 3 } and T = { a, b } . In general, if S has m elements and T contains n elements, then the cartesian product S × T consists of mn elements. Example 16. Consider an experiment consisting of flipping a coin and rolling a die. There are two possibilities for the coin, heads or tales, and the die has six faces. The total number of outcomes for this experiment is 2 × 6 = 12 . That is, there are twelve different ways to flip a coin and roll a die. The counting principle can be extended to computing the number of el- ements in the cartesian product of multiple sets. Consider the finite sets S 1 , S 2 , . . . , S r and their cartesian product S 1 × S 2 ×···× S r = { ( s 1 , s 2 , . . . , s r ) | s p ∈ S p } . 4.2. PERMUTATIONS 35 If we denote the cardinality of S p by n p = | S p | , then the number of distinct ordered r-tuples of the form ( s 1 , s 2 , . . . , s r ) is n = n 1 n 2 · · · n r ....
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This note was uploaded on 03/30/2010 for the course ECEN 303 taught by Professor Chamberlain during the Fall '07 term at Texas A&M.

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4Equiprobable_Outcomes - Chapter 4 Equiprobable Outcomes...

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