Chapter 2Bernoulli Trials2.1 The Binomial DistributionIn Chapter 1 we learned about i.i.d. trials. In this chapter, we study a very important special case ofthese, namely Bernoulli trials (BT). If each trial yields has exactly two possible outcomes, then wehave BT. B/c this is so important, I will be a bit redundant and explicitly present the assumptionsof 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−pfor 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 thenumber 0.One reason that BT are so important, is that if we have BT, we can calculate probabilities of agreat 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 yieldfour 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 probabilitydistribution.Suppose that we plan to perform/observe n BT. Let X denote the total number of successes inthe n trials. The probability distribution of X is given by the following equation.P (X = x) =n!x!(n − x)!pxqn−x, for x = 0, 1, . . . , n. (2.1)17To 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, goto the link to the Revised Student Study Guide on my webpage and read the More Mathematicssections 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!5!(0.60)0(0.40)5= 0.0102.P (X = 1) =5!1!4!(0.60)1(0.40)4= 0.0768.P (X = 2) =5!2!3!(0.60)2(0.40)3= 0.2304.P (X = 3) =5!3!2!(0.60)3(0.40)2= 0.3456.P (X = 4) =5!4!1!(0.60)4(0.40)1= 0.2592.P (X = 5) =5!5!0!(0.60)5(0.40)0= 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.For n ≥ 9, I recommend using a statistical software package on a computer or the website Idescribe later. For example, the probability distribution for X for n = 25 and p = 0.50 is presentedin Table 2.1.Equation 2.1 is called the binomial probability distribution with parameters n and p; it isdenoted 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 packageshould not be considered a panacea for the binomial. For example, if I direct my computer tocalculate the Bin(n,0.50) for any n ≥ 1023 I get an error message; the computer program is smartenough to realize that it has messed up and its answer is wrong; hence, it does not give me ananswer. Why does this happen?Well, consider the computation of P (X = 1000) for the Bin(2000,0.50). This involves a reallyhuge number (2000!) divided by the square of a really huge number (1000!), and them multipliedby a really really small positive number ((0.50)2000). Unless the computer programmer exhibitsincredible care in writing the code, the result will be an overflow or an underflow or both.18Table 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 1.00001 0.0000 0.0000 1.00002 0.0000 0.0000 1.00003 0.0001 0.0001 1.00004 0.0004 0.0005 0.99995 0.0016 0.0020 0.99956 0.0053 0.0073 0.99807 0.0143 0.0216 0.99278 0.0322 0.0539 0.97849 0.0609 0.1148 0.946110 0.0974 0.2122 0.885211 0.1328 0.3450 0.787812 0.1550 0.5000 0.655013 0.1550 0.6550 0.500014 0.1328 0.7878 0.345015 0.0974 0.8852 0.212216 0.0609 0.9461 0.114817 0.0322 0.9784 0.053918 0.0143 0.9927 0.021619 0.0053 0.9980 0.007320 0.0016 0.9995 0.002021 0.0004 0.9999 0.000522 0.0001 1.0000 0.000123 0.0000 1.0000 0.000024 0.0000 1.0000 0.000025 0.0000 1.0000 0.0000Total 1.000019Figure 2.1: The Bin(100, 0.5) Distribution.0.020.040.060.0835 40 45 50 55 60 65Figure 2.2: The Bin(100, 0.2) Distribution.0.020.040.060.080.1010 15 20 25 30Figure 2.3: The Bin(25, 0.5) Distribution.0.020.040.060.080.100.120.140.16811 14 1720Figure 2.4: The Bin(50, 0.1) Distribution.0.020.040.060.080.100.120.140.160.180.200 3 6912Before 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 easyway to obtain a good approximation to the exact answer.Figures 2.1–2.4 probability histograms for several binomial probability distributions. Here ishow they are drawn. The method I am going to give you works only if the possible values of therandom variable are equally spaced on the number line. This definition can be modified for othersituations, but we won’t need to do this.1. On a horizontal number line, mark all possible values of X. For the binomial, these are 0, 1,2, ...n.2. Determine the value of δ (Greek lower case delta) for the random variable of interest. Thenumber δ is the distance between any two consecutive values of the random variable. Forthe binomial, δ = 1, which makes life much simpler.3. Above each x draw a rectangle, with its center at x, its base equal to δ and its height equalto P (X = x)/δ. Of course, in the current case, δ = 1, so the height of each rectangle equalsthe probability of its center value.For a probability histogram (PH) the area of a rectangle equals the probability of its center value,b/cArea = Base × Height = δ ×P (X = x)δ= P (X = x).A PH allows us to ‘see’ a probability distribution. For example, for our pictures we can see thatthe binomial is symmetric for p = 0.50 and not symmetric for p 6= 0.50. This is not an accident of21the four pictures I chose to present; indeed, the binomial is symmetric if, and only if, p = 0.50. Bythe way, I am making the tacit assumption that 0 < p < 1; that is, that p is neither 1 nor 0. Trialsfor which
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