This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 2 Bernoulli Trials 2.1 The Binomial Distribution In Chapter 1 we learned about i.i.d. trials. In this chapter, we study a very important special case of these, namely Bernoulli trials (BT). If each trial has exactly two possible outcomes, then we have BT. Because this is so important, I will be a bit redundant and explicitly present the assumptions of BT. The Assumptions of Bernoulli Trials. There are three: 1. Each trial results in one of two possible outcomes, denoted success ( S ) or failure ( F ). 2. The probability of S remains constant from trialtotrial and is denoted by p . Write q = 1 p for the constant probability of F . 3. The trials are independent. When we are doing arithmetic, it will be convenient to represent S by the number 1 and F by the number 0. One reason that BT are so important, is that if we have BT, we can calculate probabilities of a great many events. Our Frst tool for calculation, of course, is the multiplication rule that we learned in Chapter 1. or example, suppose that we have n = 5 BT with p = 0 . 70 . The probability that the BT yield four successes followed by a failure is: P ( SSSSF ) = ppppq = (0 . 70) 4 (0 . 30) = 0 . 0720 . Our next tool is extremely powerful and very useful in science. It is the binomial probability distribution . Suppose that we plan to perform/observe n BT. Let X denote the total number of successes in the n trials. The probability distribution of X is given by the following equation. P ( X = x ) = n ! x !( n x )! p x q n x , for x = 0 , 1 , . . ., n. (2.1) 19 To use this formula, recall that n ! is read nfactorial and is computed as follows. 1! = 1; 2! = 2(1) = 2; 3! = 3(2)(1) = 6 , 4! = 4(3)(2)(1) = 24; and so on. By special deFnition, 0! = 1 . I will do an extended example to illustrate the use of Equation 2.1. Suppose that n = 5 and p = 0 . 60 . I will obtain the probability distribution for X . P ( X = 0) = 5! 0!5! (0 . 60) (0 . 40) 5 = 1(1)(0 . 0102) = 0 . 0102 . P ( X = 1) = 5! 1!4! (0 . 60) 1 (0 . 40) 4 = 5(0 . 60)(0 . 0256) = 0 . 0768 . P ( X = 2) = 5! 2!3! (0 . 60) 2 (0 . 40) 3 = 10(0 . 36)(0 . 064) = 0 . 2304 . P ( X = 3) = 5! 3!2! (0 . 60) 3 (0 . 40) 2 = 10(0 . 216)(0 . 16) = 0 . 3456 . P ( X = 4) = 5! 4!1! (0 . 60) 4 (0 . 40) 1 = 5(0 . 1296)(0 . 40) = 0 . 2592 . P ( X = 5) = 5! 5!0! (0 . 60) 5 (0 . 40) = 1(0 . 0778)(1) = 0 . 0778 . You should check the above computations to make sure you are comfortable using Equation 2.1. Here are some guidelines for this class. If n 8 , you should be able to evaluate Equation 2.1 by hand as I have done above for n = 5 . or n 9 , I recommend using a statistical software package on a computer or the website I describe later. or example, the probability distribution for X for n = 25 and p = 0 . 50 is presented in Table 2.1....
View Full
Document
 Fall '11
 hanlon
 Bernoulli, Binomial

Click to edit the document details