What is Binomial Distribution – Meaning, Formulas & Examples

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The distribution of a statistical dataset is the spread of the data which shows all possible values or intervals of the data and how they occur. A distribution is simply a collection of data or scores on a variable. Usually, these scores are arranged in order from ascending to descending and then they can be presented graphically.

There are many types of distributions in statistics suited for some particular application or environment. Some of the most common types of distributions in statistics are Normal or Gaussian distribution, Bernoulli Distribution, Binomial distribution, Poisson distribution, Exponential distribution, Gamma distribution, and Weibull distribution.

Let’s understand what is binomial distribution, and the formulas used with examples.

What is Binomial Distribution Definition? 

The binomial distribution is a commonly used discrete distribution in statistics. It represents the probability for $x$ successes of an experiment in $n$ trials, given a success probability $p$ for each trial at the experiment. The binomial distribution forms the base for the famous binomial test of statistical importance. A test that has a single outcome such as success/failure is also called a Bernoulli trial or Bernoulli experiment, and a series of outcomes is called a Bernoulli process. 

Consider an experiment where each time a coin is tossed for a head/tail with a series of $n$ experiments. Then in the binomial probability distribution, the boolean-valued outcome the success(say getting a head) denoted by yes/true/$1$ is represented with probability $p$ and the failure(say getting a tail) no/false/zero with probability $q (q = 1 − p)$. In a single experiment when $n = 1$, the binomial distribution is called a Bernoulli distribution.

Examples of Binomial Distribution

The following are some of the examples of binomial distribution.

• The number of successes (say head) in an experiment of $n$ trials of tossing a coin.
• The number of successes (say six) in an experiment of $n$ trials of rolling a die.
• The number of successes (say defective items) in an experiment of $n$trials of examining $n$ items.
• Taking a survey of positive and negative reviews from the public for any specific product or place.
• Finding the number of male and female employees in an organization.
• The number of votes collected by a candidate in an election is counted based on 0 or 1 probability.

Properties of Binomial Distribution

The following are the distinctive properties of a binomial distribution.

• There are two possible outcomes: true or false, success or failure, yes or no, $1$ or $0$.
• There is $n$ number of independent trials or a fixed number of $n$ times repeated trials.
• The probability of success or failure remains the same for each trial.
• Only the number of successes is calculated out of n independent trials.
• Every trial is an independent trial, which means the outcome of one trial does not affect the outcome of another trial.

When Should We Use Binomial Distribution?

Binomial distributions are useful in a number of settings. It is important to know when this type of distribution should be used.  The basic features that we must have been for a total of $n$ independent trials are conducted and we want to find out the probability of $x$ successes, where each success has probability $p$ of occurring. The following are the conditions that need to be checked before using a binomial distribution.

• Fixed number of trials
• Independent trials
• Two different Outcomes
• The probability of success remains the same for all trials

Fixed Trials

The process being investigated must have a clearly defined number of trials that do not vary. We cannot alter this number midway through our analysis. Each trial must be performed the same way as all of the others, although the outcomes may vary. The number of trials is indicated by an n in the formula.

An example of having fixed trials for a process would involve studying the outcomes from rolling a die ten times. Here each roll of the die is a trial. The total number of times that each trial is conducted is defined from the outset.

Independent Trials

Each of the trials has to be independent. Each trial should have absolutely no effect on any of the others. The classical examples of rolling two dice or flipping several coins illustrate independent events. Since the events are independent we are able to use the multiplication rule to multiply the probabilities together.

In practice, especially due to some sampling techniques, there can be times when trials are not technically independent. A binomial distribution can sometimes be used in these situations as long as the population is larger relative to the sample.

Two Different Outcomes

Each of the trials is grouped into two possible outcomes – successes and failures. Although we typically think of success as a positive thing, we should not read too much into this term. We are indicating that the trial is a success in that it lines up with what we have determined to call a success.

Consider this example. Suppose we are testing the failure rate of light bulbs. If we want to know how many in a batch will not work, we could define success for our trial to be when we have a light bulb that fails to work. A failure of the trial is when the light bulb works. This may sound a bit backward, but there may be some good reasons for defining the successes and failures of our trial as we have done. It may be preferable, for marking purposes, to stress that there is a low probability of a light bulb not working rather than a high probability of a light bulb working.

Same Probabilities

The probabilities of successful trials must remain the same throughout the process we are studying. Flipping coins is one example of this. No matter how many coins are tossed, the probability of flipping a head is $\frac{1}{2}$ each time.

Suppose there are $20$ blue marbles out of $100$ marbles in a bag. The probability of choosing a blue marble at random is $\frac{20}{100} = 0.2$. Now choose again from the remaining marbles. There are $19$ blue marbles out of $99$ marbles. The probability of selecting another blue marble is $\frac{19}{99} = 0.19$. As you can see the probability of success is changing. Therefore in such cases, we cannot use binomial distribution.

What is Binomial Distribution Formula? 

The formula used to compute the probability of $x$ successes in $n$ trials for any random variate $\text{X}$ is as follows.

$\text{P}(x:n,p) = ^n \text{C}_x  p^x  (1-p)^{n-x}$

Or $\text{P}(x:n,p) = ^n \text{C}_x p^x (q)^{n-x}$

where,

$n$ = the number of experiments

$x = 0, 1, 2, 3, 4, …$

$p$ = Probability of success in a single experiment

$q$ = Probability of failure in a single experiment (= $1 – p$)

The binomial distribution formula is also written in the form of $n$-Bernoulli trials, where $^n \text{C}_x = \frac{n!}{x!(n-x)!}$. Hence, $\text{P}(x:n,p) = \frac{n!}{x!(n-x)!}.p^x.(q)^{n-x}$

Examples on Binomial Distribution Formula

Example 1: If a coin is tossed $5$ times, find the probability of getting exactly $2$ heads.

Number of trials $n = 5$

Probability of success $p = \frac{1}{2}$

Probability of failure $q = 1 – p = 1 – \frac{1}{2} = \frac{1}{2}$

Number of successes $x = 2$

$\text{P}(x:n,p) = ^n \text{C}_x p^x (q)^{n-x}$

Therefore $\text{P}\left(2:5,\frac{1}{2} \right) = ^5 \text{C}_2 \left(\frac{1}{2} \right)^2 \left(\frac{1}{2} \right)^{5-2}$

$= 10 \times \frac{1}{4} \times \frac{1}{8}$

$= \frac{10}{32} = \frac{5}{16}$

Example 2: If a die is tossed $4$ times, find the probability of getting exactly $3$ sixes.

Number of trials $n = 4$

Probability of success $p = \frac{1}{6}$

Probability of failure $q = 1 – p = 1 – \frac{1}{6} = \frac{5}{2}$

Number of successes $x = 3$

$\text{P}(x:n,p) = ^n \text{C}_x p^x (q)^{n-x}$

Therefore $\text{P}\left(3:4,\frac{1}{6} \right) = ^4 \text{C}_3 \left(\frac{1}{6} \right)^3 \left(\frac{1}{6} \right)^{4-3}$

$= 4 \times \frac{1}{216} \times \frac{1}{6}$

$= 4 \times \frac{1}{1296} = \frac{1}{324}$

Example 3: If a die is tossed $6$ times, find the probability of getting prime number $4$ times.

Number of trials $n = 6$

Number of total possible outcomes in a dice = $6$

Prime numbers in dice are $2, 3, 5$

Number of prime numbers in a dice = $3$

Therefore probability of getting a prime number = $\frac{3}{6} = \frac{1}{2}$

Probability of success $p = \frac{1}{2}$

Probability of failure $q = 1 – p = 1 – \frac{1}{2} = \frac{1}{2}$

Number of successes $x = 4$

$\text{P}(x:n,p) = ^n \text{C}_x p^x (q)^{n-x}$

$= ^6 \text{C}_4 \times \left(\frac{1}{2} \right)^4 \times \left(\frac{1}{2} \right)^{6-4}$

$= 15 \times \frac{1}{16} \times \frac{1}{4}$

$= 15 \times \frac{1}{64} $

$= \frac{15}{64} $

Example 4: If a coin is tossed $5$ times, find the probability of getting at least $3$ heads.

Number of trials $n = 5$

Probability of success $p = \frac{1}{2}$

Probability of failure $q = 1 – p = 1 – \frac{1}{2} = \frac{1}{2}$

At least $3$ heads means number of heads = $3$, $4$, or $5$.

Number of successes $x = 3, 4, 5$

$\text{P}(x:n,p) = ^n \text{C}_x p^x (q)^{n-x}$

$\text{P}(3) = ^5 \text{C}_3 \left( \frac{1}{2} \right)^3 \times \left(\frac{1}{2} \right)^{5-3}$

$= 10 \times \frac{1}{8} \times \frac{1}{4}$

$= 10 \times \frac{1}{32} = \frac{5}{16}$

$\text{P}(4) = ^5 \text{C}_4 \left( \frac{1}{2} \right)^4 \times \left(\frac{1}{2} \right)^{5-4}$

$= 5 \times \frac{1}{16} \times \frac{1}{2}$

$= 5 \times \frac{1}{32} =\frac{5}{32}$

$\text{P}(5) = ^5 \text{C}_5 \left( \frac{1}{2} \right)^5 \times \left(\frac{1}{2} \right)^{5-5}$

$= 1 \times \frac{1}{32} \times 1 = \frac{1}{32}$

Therefore required probability = $\frac{5}{16} + \frac{5}{32} + \frac{1}{32}$

$= \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2}$

Example 5: $60\%$ of people who purchase sports cars are men. Find the probability that exactly $7$ are men if $10$ sports car owners are randomly selected.

Probability of success $p = 60\% = 0.6$

Therefore probability of failure $q = 1 – 0.6 = 0.4$

Number of trials $n = 10$

Number of success $x = 7$

$\text{P}(x:n,p) = ^n \text{C}_x p^x (q)^{n-x}$

$=> \text{P}(7:10,0.6) = ^{10} \text{C}_7 \times 0.6^7 \times 0.4^{10-7}$

$= 120 \times 0.0279936 \times 0.064 = 0.2150$ (Rounded off to $4$ decimal places)

Binomial Distribution Mean and Variance

For a binomial distribution, the mean, variance, and standard deviation for the given number of successes are represented using the formulas

Mean, $\mu = np$

Variance, $\sigma ^2 = npq$

Standard Deviation $\sigma= \sqrt{npq}$

where 

$p$ is the probability of success

$q$ is the probability of failure ($q = 1-p$)

Examples on Binomial Distribution Mean and Variance

Example 1: What is the mean of a binomial random variable with $n = 18$ and $p = 0.4$?

$n = 18$ 

$p = 0.4$

Mean, $\mu = np = 18 \times 0.4 = 7.2$

Example 2: What is the standard deviation of a binomial distribution with $n = 18$ and $p = 0.4$? Round your answer to two decimal places.

$n = 18$ 

$p = 0.4$

$q = 1 – p = 1 – 0.4 = 0.6$

Standard Deviation $\sigma= \sqrt{npq}$

$=\sqrt{18 \times 0.4 \times 0.6} = \sqrt{4.32} = 2.08$ (Rounded off to two decimal places)

Example 3: Find the variance of the binomial distribution having $12$ trials and a probability of success as $0.5$.

Number of trials $n = 12$ 

Probability of success $p = 0.5$

Probability of failure $q = 1 – p = 1 – 0.5 = 0.5$

Variance, $\sigma ^2 = npq = 12 \times 0.5 \times 0.5 = 3$

Example 4: Find the mean, variance, and standard deviation of the binomial distribution having $16$ trials, and a probability of success as $0.8$.

Number of trials $n = 16$ 

Probability of success $p = 0.8$

Probability of failure $q = 1 – p = 1 – 0.8 = 0.2$

Mean, $\mu = np = 16 \times 0.8 = 12.8$

Variance, $\sigma ^2 = npq = 16 \times 0.8 \times 0.2 = 2.56$

Standard Deviation $\sigma= \sqrt{npq} = \sqrt{16 \times 0.8 \times 0.2} = \sqrt{2.56} = 1.6$

Practice Problems

1. A coin is tossed four times. Calculate the probability of obtaining more heads than tails.
2. An agent sells life insurance policies to five equally aged, healthy people. According to recent data, the probability of a person living in these conditions for $30$ years or more is $\frac{2}{3}$. Calculate the probability that after $30$ years:
   • All five people are still living.
   • At least three people are still living.
   • Exactly two people are still living.
3. If from six to seven in the evening one telephone line in every five is engaged in a conversation: what is the probability that when $10$ telephone numbers are chosen at random, only two are in use?
4. The probability of a man hitting the target at a shooting range is $\frac{1}{4}$. If he shoots $10$ times, what is the probability that he hits the target exactly three times? What is the probability that he hits the target at least once?
5. There are $10$ red and $20$ blue balls in a box. A ball is chosen at random and it is noted whether it is red. The process repeats, returning the ball $10$ times. Calculate the expected value and the standard deviation of this game.
6. It has been determined that $5\%$ of drivers checked at a road stop show traces of alcohol and $10\%$ of drivers checked do not wear seat belts. In addition, it has been observed that the two infractions are independent of one another. If an officer stops five drivers at random:
   • Calculate the probability that exactly three of the drivers have committed any one of the two offenses.
   • Calculate the probability that at least one of the drivers checked has committed at least one of the two offenses.

FAQs

Conclusion

The binomial distribution represents the probability for $x$ successes of an experiment in $n$ trials, given a success probability $p$ for each trial at the experiment. The binomial distribution forms the base for the famous binomial test of statistical importance. The formula used in the binomial distribution is $\text{P}(x:n,p) = ^n \text{C}_x p^x (q)^{n-x}$ where, $n$ = the number of experiments, $x = 0, 1, 2, 3, 4, …$, $p$ = Probability of success in a single experiment and $q$ = Probability of failure in a single experiment (= $1 – p$).

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