6.2 The Binomial Probability Model

Many applications of probability involve considering situations where an experiment has only two possible outcomes. A flip of a coin produces either heads or tails. A child is either male or female. A household either owns a dog or it does not. A voter selects a certain candidate or not.

These situations occur so frequently that mathematicians have studied them in great detail. We call each repetition of the experiment a trial; and we call these situations binomial experiments if four conditions are true:

1. Each trial has only two outcomes; to simplify discussion, we call one outcome success and the other outcome failure.
2. The probability of success remains the same from trial to trial. We denote the probability of success as p. (Because there are only two outcomes, this means that the probability of failure is 1 – p.)
3. The trials are independent; what happens on one trial does not influence what happens on any other trial.
4. The experiment is performed a specific number of times; we use n to denote the number of trials.

In the previous section, we looked at the sex of children in a three-child family. This is a binomial experiment with each child constituting at “trial,” because it satisfies the required conditions.

1. Each trial has only two outcomes; each child is either a boy or a girl. We will call the child being a boy success, and being a girl failure (although things work out just the same if "girl" is success).
2. The probability of success remains the same from trial to trial; the probability of being a boy is 0.5 for each child in the family.
3. The trials are independent; whether the first child is a boy does not influence the sex of the later children.
4. The experiment is performed a specific number of times; there are three children and thus, three trials.

When we perform a binomial experiment and record the number X of successes that occur, then X is a discrete random variable, and we say that X has a binomial probability model. The pattern of probability values that occurs is determined by two numbers, the number of trials we perform (we call this number n) and the probability of success (p). If X represents the number of boys in a three-child family, then n = 3 and p = 0.5. The binomial experiment of tossing a coin 3 times and recording the number of heads has the same probability model. In Section 6.1, we constructed this probability model by using the tree diagram for the sex of children in a 3-child family. We give this model again here, using X to indicate the number of either boys or heads, and P(X) to indicate the corresponding probability.

The probability of success (boy or head) is the same as the probability of failure (girl or tail) in this case. Both p and 1 – p are 0.5. Therefore, the probability model is the same if we count boys or girls, heads or tails.

6.2.1 Finding Binomial Probabilities By Hand

While making a tree diagram was a reasonable tool to determine the probabilities associated with the number of boys in a three-child family, this approach becomes more cumbersome as we increase the number of trials. If we added two more children to our theoretical family, we would have to add many more branches to our tree. Surely, there is a better way.

The “better way” involves a fairly ugly looking mathematical formula. It may not seem better at first glance, but the advantage is that statistical software can easily do the necessary calculations.

Let’s deconstruct the formula to make it a bit more understandable:

• \( \binom{n}{k} \) is called the binomial coefficient and is shorthand that tells us how many different ways the k successes can be arranged in n trials. (If you have one boy in a three child family, there are 3 different ways for this to occur. The boy can be the oldest child, the middle child, or the youngest child.)
• \( \binom{n}{k} = \frac{n!}{k!(n - k)!} \), where the exclamation point indicates a factorial. n! is the product of all the whole numbers starting at 1 and ending at n. When n=0, n! is defined to be 1.
• \( p^{k} (1-p)^{n-k} \) is the probability of k successes and n - k failures occurring in some particular way. (For example, the first child is a boy, the second child is a girl, the third child is a girl.)

If we have k successes, \( p^{k} (1-p)^{n-k} \) is the same number regardless of the order in which the successes occur. So our formula finds the probability for one such way \( (p^{k} (1-p)^{n-k}) \) and then multiplies that result by the total number of ways those successes can occur \( \binom{n}{k} \).

Now we can try this out on our example of boys in a three-child family. To find the probability of 2 boys in a three-child family, we use

And so,

Happily, this is the same value that we obtained from the tree diagram and showed in the probability model above.

6.2.2 Generating a Binomial Probability Model

Finding a single binomial probability using the formula may be manageable and even somewhat novel. But none of us is likely to enjoy repeating this procedure over and over to create a table of probabilities for a binomial experiment with even a reasonably small number of trials. Before the wide availability of computer and calculator software that easily finds these numbers, reference works gave tables of binomial probabilities for various values of n and p. These tables can still be found on the web; if you search for “table of binomial probabilities,” you are likely to find many.

The table below shows a portion of such a table for a binomial experiment with 4 trials. The lefthand column shows the possible number of successes (0, 1, 2, 3, or 4). Each of the remaining columns shows the corresponding probabilities for a different value of p.

If we pull out the column representing the tetrahedral dice example above, where p = 0.25, we obtain the binomial probability model below. Notice that the value for P(X = 2) is the same (rounded to 4 decimal places) as the value we calculated using the formula.

Even using these tables has its drawbacks, however. These tables typically use a limited number of trials (n), and they generally show only certain values for p. If p were, for example, 0.22, and n were 8, how would we proceed?

Fortunately, most statistical software, either on a computer or a calculator, can easily compute binomial probability values. The process for doing this varies from software to software. The table below was generated using CrunchIt!, finding each (3 decimal place) probability separately by entering n and p, and then changing the value of X for each number of successes (0 through 6).

Some calculators can generate a whole table at once. You should find the technology that works best for you, and learn to use it well.

6.2.3 Finding Specific Binomial Probabilities

We now return to a question we posed at the beginning of Chapter 6--what is the probability that you get at least 7 out of 10 correct on a multiple-choice pop quiz if you guess at each answer? Many multiple-choice questions have four answers. If you guess at an answer, your probability of success, p, is ¼ or 0.25. Each question constitutes a trial, so n = 10. This situation is a binomial experiment, with n = 10 and p = 0.25 as shown in the accompanying table.

We can see from this table that the probability of exactly 7 successes (answers correct) is 0.00039.

(We have used 5 decimal places here to limit the number of “essentially zero” probabilities that appear. In this case, “.00000” gives the probability of 10 successes as a value rounded to 5 decimal places. It definitely does not mean that the probability is exactly 0 nor that 10 successes cannot occur.)

But what about the probability of getting at least 7 correct? “At least 7” correct means 7 or more, that is, 7 or 8 or 9 or 10 correct. Because these numbers of successes represent disjoint events, the probability of the number of successes being 7 or 8 or 9 or 10 is the sum of their individual probabilities. That is, P(X ≥ 7) = 0.00309 + 0.00039 + 0.00003 + 0.00000 = 0.00351. This probability suggests that guessing is very unlikely to produce a passing score on the quiz, assuming that 70% is required to pass.

What is the probability that a student gets at most 5 correct on the quiz by guessing? “At most 5” means 5 or fewer correct, that is, 0, 1, 2, 3, 4, or 5. Once again, the desired probability is the sum of the individual probabilities. So P(X ≤ 5) = 0.05631 + 0.18771 + 0.28157 + 0.25028 + 0.14600 + 0.05840 = 0.98027.

6.2.4 Binomial Mean and Standard Deviation

Binomial random variables are a particular type of discrete random variable. A binomial random variable, X, is the number of successes in the n trials of the binomial experiment. The formulas for mean and standard deviation presented in Section 6.1 apply here as well.

The mean or expected value of any discrete random variable X is

\( \mu_{x} = \sum x \cdot P(x) \),

and its standard deviation is

\( \sigma _{x} = \sqrt{\sum (x - \mu_x)^2 P(x)} \).

Let’s look for the last time at the three-child family experiment, where X represents the number of boys in the family. The probability model for this experiment is shown below.

To find the mean or expected value of X, we calculate

\( \mu_x = 0 \times 0.125 + 1 \times 0.375 + 2 \times 0.375 + 3 \times 0.125 = 0.375 + 0.750 + 0.375 = 1.500 \).

Perhaps you notice something interesting about this value. For this experiment, n is 3 and p is 0.5, and the product of those numbers is 1.5. Is this a happy coincidence? As it turns out, it is not. The mean (expected value) of the binomial probability model is always \( \mu_x = n \times p \), where n is the number of trials and p is the probability of success. This greatly simplifies our calculation, particularly when n is large (and we have many possible values of X) or the values of P(X) are reported to many decimal places.

Recall that the mean represents what the average number of boys per family should be if we randomly select many, many three-child families and record how many boys each contains.

To calculate the standard deviation for a binomial probability model when n = 3 and p = 0.5, we use

\( \sigma_{x} = \sqrt{(0 - 1.5)^2 (0.125) + (1- 1.5)^2 (0.375) + (2-1.5)^2 (0.375) + (3 - 1.5) (0.125)} \)

\( = \sqrt{(2.25) (0.125) + (0.25) (0.375) + (0.25) (0.375) + (2.25) (0.125)} = \sqrt{0.750} = 0.866 \)

rounded to three decimal places.

While it is probably not obvious from the above calculation, the value under the square sign is actually \( n \times p \times (1-p) \). So this makes the formula for the standard deviation of the binomial random variable X look much simpler:

\( \sigma_x = \sqrt{n \times p \times (1-p)} \).

As we saw in the last section, if the binomial probability model is bell-shaped (as this one is), then the Empirical Rule applies. So about 68% of the time, we expect the number of boys in a three-child family to lie within one standard deviation of the mean, that is, between 0.634 and 2.366. Effectively, about 68% of the time, there will be one or two boys in a three-child family.

Because we are so often interested in counting the number of successes that occur in a series of trials, we are frequently considering binomial experiments. Agreeing or disagreeing with a survey question, having an IQ higher than 120 or not, selecting an M&M’s candy that is red (or a different specific color) are all situations that might lead to a binomial model. Before you proceed to use binomial probability techniques, however, it is important for you to verify that your situation satisfies all four conditions for a binomial experiment.

Not all discrete probability models are binomial ones. And not all probability models are discrete. In the next chapter, we will consider continuous probability models in general, and a particular model that is probably the most famous (or infamous, depending of your perspective) probability model of all.