Science, Tech, Math › Math Expected Value of a Binomial Distribution Share Flipboard Email Print A histogram of a binomial distribution. C.K.Taylor Math Statistics Probability & Games Statistics Tutorials Formulas Descriptive Statistics Inferential Statistics Applications Of Statistics Math Tutorials Geometry Arithmetic Pre Algebra & Algebra Exponential Decay Functions Worksheets By Grade Resources View More By Courtney Taylor Professor of Mathematics Ph.D., Mathematics, Purdue University M.S., Mathematics, Purdue University B.A., Mathematics, Physics, and Chemistry, Anderson University Courtney K. Taylor, Ph.D., is a professor of mathematics at Anderson University and the author of "An Introduction to Abstract Algebra." our editorial process Courtney Taylor Updated January 29, 2019 Binomial distributions are an important class of discrete probability distributions. These types of distributions are a series of n independent Bernoulli trials, each of which has a constant probability p of success. As with any probability distribution we would like to know what its mean or center is. For this we are really asking, “What is the expected value of the binomial distribution?” Intuition vs. Proof If we carefully think about a binomial distribution, it is not difficult to determine that the expected value of this type of probability distribution is np. For a few quick examples of this, consider the following: If we toss 100 coins, and X is the number of heads, the expected value of X is 50 = (1/2)100.If we are taking a multiple choice test with 20 questions and each question has four choices (only one of which is correct), then guessing randomly would mean that we would only expect to get (1/4)20 = 5 questions correct. In both of these examples we see that E[ X ] = n p. Two cases is hardly enough to reach a conclusion. Although intuition is a good tool to guide us, it is not enough to form a mathematical argument and to prove that something is true. How do we prove definitively that the expected value of this distribution is indeed np? From the definition of expected value and the probability mass function for the binomial distribution of n trials of probability of success p, we can demonstrate that our intuition matches with the fruits of mathematical rigor. We need to be somewhat careful in our work and nimble in our manipulations of the binomial coefficient that is given by the formula for combinations. We begin by using the formula: E[ X ] = Σ x=0n x C(n, x)px(1-p)n – x. Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0, and so we can actually write: E[ X ] = Σ x = 1n x C(n , x) p x (1 – p) n – x . By manipulating the factorials involved in the expression for C(n, x) we can rewrite x C(n, x) = n C(n – 1, x – 1). This is true because: x C(n, x) = x n!/(x!(n – x)!) = n!/((x – 1)!(n – x)!) = n(n – 1)!/((x – 1)!((n – 1) – (x – 1))!) = n C(n – 1, x – 1). It follows that: E[ X ] = Σ x = 1n n C(n – 1, x – 1) p x (1 – p) n – x . We factor out the n and one p from the above expression: E[ X ] = np Σ x = 1n C(n – 1, x – 1) p x – 1 (1 – p) (n – 1) - (x – 1) . A change of variables r = x – 1 gives us: E[ X ] = np Σ r = 0n – 1 C(n – 1, r) p r (1 – p) (n – 1) - r . By the binomial formula, (x + y)k = Σ r = 0 kC( k, r)xr yk – r the summation above can be rewritten: E[ X ] = (np) (p +(1 – p))n – 1 = np. The above argument has taken us a long way. From beginning only with the definition of expected value and probability mass function for a binomial distribution, we have proved that what our intuition told us. The expected value of the binomial distribution B( n, p) is n p.