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 yields has exactly two possible outcomes then we have BT B c 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 trial to trial 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 first tool for calculation of course is the multiplication rule that we learned in Chapter 1 For 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 px q n x for x 0 1 n x n x 17 2 1 To use this formula recall that n is read n factorial 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 definition 0 1 Note to the extremely interested reader If you want to see why Equation 2 1 is correct go to the link to the Revised Student Study Guide on my webpage and read the More Mathematics sections of Chapters 2 3 and 5 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 60 0 0 40 5 0 0102 0 5 5 0 60 1 0 40 4 0 0768 1 4 5 P X 2 0 60 2 0 40 3 0 2304 2 3 5 0 60 3 0 40 2 0 3456 P X 3 3 2 5 P X 4 0 60 4 0 40 1 0 2592 4 1 5 P X 5 0 60 5 0 40 0 0 0778 5 0 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 For n 9 I recommend using a statistical software package on a computer or the website I describe later For example the probability distribution for X for n 25 and p 0 50 is presented in Table 2 1 Equation 2 1 is called the binomial probability distribution with parameters n and p it is denoted by Bin n p With this notation we see that my earlier by hand effort was the Bin 5 0 60 and Table 2 1 is the Bin 25 0 50 Sadly life is a bit more complicated than the above In particular a statistical software package should not be considered a panacea for the binomial For example if I direct my computer to calculate the Bin n 0 50 for any n 1023 I get an error message the computer program is smart enough to realize that it has messed up and its answer is wrong hence it does not give me an answer Why does this happen Well consider the computation of P X 1000 for the Bin 2000 0 50 This involves a really huge number 2000 divided by the square of a really huge number 1000 and them multiplied by a really really small positive number 0 50 2000 Unless the computer programmer exhibits incredible care in writing the code the result will be an overflow or an underflow or both P X 1 18 Table 2 1 The Binomial Distribution for n 25 and p 0 50 x P X x P X x P X x 0 0 0000 0 0000 1 0000 1 0 0000 0 0000 1 0000 2 0 0000 0 0000 1 0000 3 0 0001 0 0001 1 0000 4 0 0004 0 0005 0 9999 5 0 0016 0 0020 0 9995 6 0 0053 0 0073 0 9980 7 0 0143 0 0216 0 9927 8 0 0322 0 0539 0 9784 9 0 0609 0 1148 0 9461 10 0 0974 0 2122 0 8852 11 0 1328 0 3450 0 7878 12 0 1550 0 5000 0 6550 13 0 1550 0 6550 0 5000 14 0 1328 0 7878 0 3450 15 0 0974 0 8852 0 2122 16 0 0609 0 9461 0 1148 17 0 0322 0 9784 0 0539 18 0 0143 0 9927 0 0216 19 0 0053 0 9980 0 0073 20 0 0016 0 9995 0 0020 21 0 0004 0 9999 0 0005 22 0 0001 1 0000 0 0001 23 0 0000 1 0000 0 0000 24 0 0000 1 0000 0 0000 25 0 0000 1 0000 0 0000 Total 1 0000 19 Figure 2 1 The Bin 100 0 5 Distribution 0 08 0 06 0 04 0 02 35 40 45 50 55 60 Figure 2 2 The Bin 100 0 2 Distribution 0 10 0 08 0 06 0 04 0 02 10 15 20 25 Figure 2 3 The Bin 25 0 5 Distribution 0 16 0 14 0 12 0 10 0 08 0 06 0 04 0 02 8 11 20 14 17 30 65 Figure 2 4 The Bin 50 0 1 Distribution 0 20 0 18 0 16 0 14 0 12 0 10 0 08 0 06 0 04 0 02 0 3 6 9 12 Before we condemn the programmer for carelessness or bemoan the limits of the human mind note the following We do not need to evaluate Equation 2 1 for large n s b c there is a very easy way to obtain a good approximation to the exact answer Figures 2 1 2 4 probability histograms for several binomial probability distributions Here is how they are drawn The method I am going to give you works only if the possible values of the random variable are equally spaced on the number line This definition can be modified for other situations but we won t need to do this 1 On a horizontal number line mark all possible values of …
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