**Definition. **A discrete random variable *X* is a **binomial random variable** if:

- An experiment, or trial, is performed in exactly the same way
*n* times.
- Each of the
*n* trials has only two possible outcomes. One of the outcomes is called a "success," while the other is called a "failure." Such a trial is called a **Bernoulli trial**.
- The
*n* trials are independent.
- The probability of success, denoted
*p*, is the same for each trial. The probability of failure is *q* = *1 − p*.
- The random variable
*X* = the number of successes in the *n *trials.

Example

A coin is weighted in such a way so that there is a 70% chance of getting a head on any particular toss. Toss the coin, in exactly the same way, 100 times. Let *X* equal the number of heads tossed. Is *X* a binomial random variable?

**Solution.** Yes, *X* is a binomial random variable, because:

- The coin is tossed in exactly the same way 100 times.
- Each toss results in either a head (success) or a tail (failure).
- One toss doesn't affect the outcome of another toss. The trials are independent.
- The probability of getting a head is 0.70 for each toss of the coin.
*X* equals the number of heads (successes).

Example

A college administrator randomly samples students until he finds four that have volunteered to work for a local organization. Let *X* equal the number of students sampled. Is *X* a binomial random variable?

**Solution.** No, *X* is not a binomial random variable, because the number of trials *n* was not fixed in advance, and *X* does not equal the number of volunteers in the sample.

Example

A Quality Control Inspector (QCI) investigates a lot containing 15 skeins of yarn. The QCI randomly samples (without replacement) 5 skeins of yarn from the lot. Let *X* equal the number of skeins with acceptable color. Is *X* a binomial random variable?

**Solution.** No, *X* is not a binomial random variable, because *p*, the probability that a randomly selected skein has acceptable color changes from trial to trial. For example, suppose, unknown to the QCI, that 9 of the 15 skeins of yarn in the lot are acceptable. For the first trial, *p* equals 9/15. However, for the second trial, *p* equals either 9/14 or 8/14 depending on whether an acceptable or unacceptable skein was selected in the first trial. Rather than being a binomial random variable, *X* is a hypergeometric random variable. If we continue to assume that 9 of the 15 skeins of yarn in the lot are acceptable, then *X* has the following probability mass function:

$$f(x) = P(X=x) = \frac{\binom{9}{x} \binom{6}{5-x} }{\binom{15}{5}} \text{for x} =0,1,2,3,4,5 $$

Example

A Gallup Poll of *n* = 1000 random adult Americans is conducted. Let *X* equal the number in the sample who own a sport utility vehicle (SUV). Is *X* a binomial random variable?

**Solution.** No, *X* is technically a hypergeometric random variable, not a binomial random variable, because, just as in the previous example, sampling takes place without replacement. Therefore, *p*, the probability of selecting an SUV owner, has the potential to change from trial to trial. To make this point concrete, suppose that Americans own a total of *N* = 270,000,000 cars. Suppose too that half (135,000,000) of the cars are SUVs, while the other half (135,000,000) are not. Then, on the first trial, *p* equals ½ (from 135,000,000 divided by 270,000,000). Suppose an SUV owner was selected on the first trial. Then, on the second trial, p equals 134,999,999 divided by 269,999,999, which equals.... punching into a calculator... 0.499999... Hmmmmm! Isn't that 0.499999... close enough to ½ to just call it ½? Yes...that's what we do!

In general, when the sample size *n* is small in relation to the population size *N*, we assume a random variable *X*, whose value is determined by sampling without replacement, follows (approximately) a binomial distribution. On the other hand, if the sample size *n* is close to the population size *N*, then we assume the random variable *X* follows a hypergeometric distribution.