**Binomial Distribution**

## Binomial DistributionIf |

Seldom are we satisfied with performing one Bernoulli trial. Instead, we want to conduct multiple Bernoulli trials, observing the outcome of each one. Suppose we have *n independent* Bernoulli trials, each with the probability *p* of success. Let *X* be the number of successes observed in the *n* trials. Then *X* is a binomial random variable. The probability of *x* successes in *n* trials may be written as:

In the above equation, is the number of ways to choose *x* items from *n* and is called the number of combinations of *n* things taken *x* at a time. (Recall that *n*! = *n*(*n* – 1)(*n* – 2) . . . (1) so that 5! = 5(4)(3)(2)(1) =120.) We will consider some examples of the binomial distribution and then provide an explanation of why the probabilities are computed as stated here.

Suppose we randomly select 25 apples from the orchard and count how many are damaged. If one apple being damaged has no effect on whether or not the next selected apple is damaged, then the number of damaged apples would have a binomial distribution with *n* = 25 and *p* equal to the proportion of damaged apples in the orchard.

In a quality control study, we could randomly choose 50 television sets from the production line and carefully test each to determine how many are defective. If we assume that whether or not one television is defective is independent of the next television being defective, then the number of defective televisions would have a binomial distribution with *n* = 50 and *p* equal to the proportion of defective sets produced during the period of the study.

To understand why the probabilities associated with the binomial random variable are as given in this way, first suppose we flip a coin twice and observe the number of heads. Each flip may be considered to be a Bernoulli trial. Because the outcome on one flip does not affect the outcome on the next flip, the two trials are independent. Let *X* be the random variable denoting the number of heads observed on the two flips. Then *X* = 0, 1, or 2. In Figure 10.2, we have a tree diagram representing this experiment. Notice that each flip results in a set of two branches, one representing heads and the other tails. The tree has four terminal branches, representing the outcomes *HH*,*HT*,*TH*, and *TT*, where *H* represents heads and *T* represents tails. The random variable *X* assigns 2 to *HH*, 1 to *HT*, 1 to *TH*, and 0 to *TT*. Because all four outcomes are equally likely, we have the following probability distribution for *X*.

Notice that because the two flips are independent, the probability of two heads is *p*(2) = *p*^{2}. Here, so Other outcomes can be computed in the same manner. Also note that the values of *X* are not equally likely. Because there are two ways to obtain one head (*HT* and *TH*) and only one way to obtain either zero heads (*TT*) or two heads (*HH*), the probability that *X*=1 is twice that of *X* = 0 or *X* = 2. Thus, we have 2*p*(1 – *p*) as the probability of one head.

When experiments increase in size, as would be the case if we flipped the coin 40 times, it becomes unreasonable to construct a tree diagram or to list all possible outcomes. We need a general way of counting the number of ways to get *x* successes in *n* trials. The number of ways to choose *x* items from *n* is and is called the number of combinations of *n* things taken *r* at a time.

For the coin flipping example, we have *n* = 2. For *X* = 0, we have . When *X* = 1, . When *X* = 2, this function is again 1. Thus, we have accurately counted the number of ways to get 0, 1, or 2 heads. The probability of any particular sequence of heads and tails is *p*^{x}(1 –*p*)^{n – x} because we have *x* successes, each with probability *p*, and (*n* – *x*) failures, each with probability 1 – *p*. Thus, the probability of *X* successes in *n* trials may be written as which is the probability function given earlier.

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