 Binomial Random Variable

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 \(q=1p\).
 The random variable \(X=\)the number of successes in the \(n\) trials.
Example 102 Section
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?
Answer
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 103 Section
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?
Answer
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 104 Section
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?
Answer
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 \(\frac{9}{15}\). However, for the second trial, \(p\)equals either \(\frac{9}{14}\) or \(\frac{8}{14}\)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:
\(f(x)=P(X=x)=\dfrac{\dbinom{9}{x} \dbinom{6}{5x}}{\dbinom{15}{5}}\) for \(x=0, 1, \ldots, 5\)
Example 105 Section
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?
Answer
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 \(\frac{1}{2}\) (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 \(\frac{1}{2}\) to just call it \(\frac{1}{2}\)?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.