There are two possible outcomes when using the binomial distribution model (hence the name "binomial").It might be appropriate to use a multinomial probability model if there were more than two distinct outcomes, but here we are looking at the dichotomous case.

Medications might relieve allergies for adults or not, antibiotic therapy for children with a bacterial infection might work or not, adults with a myocardial infarction might survive the heart attack or not, and a medical device, such as a coronary stent, might be successful or not.There are a number of processes in which an outcome is dichotomous (i.e., has two possible values).As a result, the two outcomes are often labeled as "success" and "failure", with success referring to the presence of the desired outcome.However, you should be aware that for many medical and public health questions, the outcome or event of interest is actually the occurrence of disease, which obviously is not a success.Nevertheless, using this terminology is typical when discussing binomial distributions.Consequently, whenever we use the binomial distribution, we must clearly specify what outcome is a "success" and what outcome is a "failure".

A binomial distribution model allows us to compute the probability of experiencing a specified number of "successes" when a specific number of times (e.g., for a group of patients) is repeated and what happens is either a success or a failure.To begin, we will introduce some of the notation necessary for the binomial distribution model.

"N" represents the number of observations or the number of times the process has been repeated, and "x" represents the number of "successes" or events of interest occurring during "n" observations."P" is the probability of success or occurrence of the outcome of interest.

Factorials are also used in the binomial equation.A factorial of a non-negative integer is denoted by k!.k!k! (which is the product of the positive integers and the integer k).

Based on this descriptor, the binomial distribution model is:

This model describes the binomial distribution

Three assumptions are required when using a binomial distribution:

The following KhanAcademy.org video provides a more intuitive explanation of the binomial distribution.

## Examples of Use of the Binomial Model

### 1. Relief of Allergies

Let's say 80% of adults suffering from allergies experience relief from symptoms after taking a specific medication.How likely is it that the medication will work in exactly seven out of 10 new patients with allergies?

We must first determine whether the three assumptions of the binomial distribution model are satisfied.

As we know:

Based on 7 successes, the probability is:

Accordingly, this would be equivalent to:

The numerator and denominator, however, are often contradictory,

This can be simplified as follows:

Exactly 7 of 10 patients are likely to report relief when the probability that any one person will report relief is 80%.

You can also compute binomial probabilities such as this using Excel's =BINOMDIST function.Place the cursor into an empty cell and enter:

The BINOMDIST(x,n,p,false) function

Where x is the number of successes, n is the number of replications or observations, and p is the probability of success on a single observation.

Is it likely that none report relief?

The following is equivalent to

The result simplifies to

There is a very small probability that none of the ten patients will report relief from their symptoms if the probability for each individual patient is 80%.

From a possible 10 patients, how many are most likely to report relief?.If 80% of patients report relief and 10 patients are considered, we would expect that 8 patients will report relief.Can we be sure that exactly eight of 10 patients report relief?

### 2. The Probability of Dying after a Heart Attack

.If we have 5 patients who have a heart attack, what is the possibility for all of them to survive?.We will call a success in this case a fatal attack (p = 0.04).

Again, we need to assess the assumptions.A fatal attack is 4% likely to occur for all patients, and the outcomes for individual patients are independent.We should be cautious of assuming that the probability of success applies to all patients.Many factors affect whether a patient will die from a heart attack, such as their age, the severity of the attack, and other comorbid conditions.It is necessary to assume that all patients are at the same risk of fatal attack in order to apply the 4% probability.We should also thoroughly examine the assumption of independent events.If the events are unrelated, the assumption is usually appropriate.A related or correlated prognosis of disease may exist among members of the same family or among cohabiting individuals.Let us assume, for the purposes of this example, that the five patients being investigated are unrelated, of a similar age, and devoid of any comorbidities.

In an attack where the probability of death for any patient is 4%, there is 81.546% probability that all patients will survive.The possible outcomes in this example are 0, 1, 2, 3, 4 or 5 successes (fatalities).Due to the low likelihood of mortality, the most likely response is 0 (all patients survive).A binomial formula determines the probability of observing exactly x successes out of n.

## Computing the Probability of a Range of Outcomes

If you want to compute the probability of all possible outcomes, you will need to apply the formula more than once.Suppose we wanted to calculate the probability that no more than one person dies from a heart attack in the example.Or, none or one, but not more.Specifically, P(no more than 1 success) = P(0 or 1 success) + P(1 success).This probability is solved twice with the binomial formula.

The previous step was to compute P(0 successes); we now compute P(1 success):

A success is defined as (1) (no more than one) = (1) (no more than one) + (1) (no more than one success).

0.9851 is equal to 0.8154 + 0.1697.

A probability of 98.51% is that fewer than 1 out of 5 individuals will die from the attack.

Does it seem likely that two or more of the five will die in an attack?.In this case, we are interested in calculating P(2 or more successes).It is possible to observe 0, 1, 2, 3, 4 or 5 successes, and the probability of observing any of these outcomes is 1, so we are certain to observe one of those outcomes.As P(0 or 1 success) = 0.9851, P(2, 3, 4 or 5 success) = 1 - P(0 or 1 success) = 0.0149.

## Mean and Standard Deviation of a Binomial Population

The number of successes:

Deviation standard: