▷ Binomial Distribution | Formula, properties and example

Here you will find the binomial distribution formula, the properties of the binomial distribution, how to identify a binomial distribution and an example of the binomial distribution explained step by step!

Many figures are written, but the most practical is to truncate up to 2 significant figures.

Binomial distribution formula

P\{ X = k \} = \left( \begin{array}{c} n \\ k \end{array} \right)p^{k}(1 - p)^{n - k}

$n$ means the number of tests performed.
$p$ means the probability of success.
$k$ the variable of times that we want to calculate that success occurs, it is any value that we give between $0$ and $n$ , since a value less than zero will result in an error and a value greater than $n$ will result in a probability of zero.
The $\left( \begin{array}{c} n \\ k\end{array}\right)$ means that we are making $n$ combinations of $k$ .

Properties of the binomial distribution

If you want to calculate the mean of the binomial distribution, you have to apply the following formula:

$\mu = np$

If you want to calculate the variance of the binomial distribution, you have to apply the following formula:

$\sigma^{2} = np (1 - p)$

If you want to calculate the standard distribution of the binomial distribution, you have to apply the following formula:

$\sigma=\sqrt{np(1 - p)}$

How to identify a binomial distribution

The experiment consists of a fixed number $n$ of trials, resulting in success or failure.
The trials are identical and independent, so the probability of success remains constant from trial to trial.
The random variable $X$ (which is equal to $k$ ), denotes the number of successes obtained from $n$ trials.

Binomial distribution example

A study determined that 40% of the students of a university eat in one of the cafeterias of the campus. If 8 students are randomly chosen from that campus one afternoon, determine the probability that they ate at one of the cafeterias on your campus …:

Remember the formula:

P\{ X = k \} = \left( \begin{array}{c} n \\ k \end{array} \right)p^{k}(1 - p)^{n - k}

Remember also that the probability $p$ that students will eat in one of the cafeterias on campus is 40%, or 0.4.

… exactly two of them

As they ask us for exactly two of them, the value of $k$ is replaced by a $2$ :

P\{ X = 2\} = \left( \begin{array}{c} 8 \\ 2 \end{array} \right) 0.4^{2} (1-0.4)^{8-2}

P\{ X = 2\} = \left( \begin{array}{c} 8 \\ 2 \end{array} \right) 0.4^{2}(0.6)^{6}

The probability that exactly 2 people ate at a cafeteria on your campus is $0.20$

… at least two of them

In this case, the only thing that was done was to subtract the probability from the unit when one and none of the students eat breakfast in one of the cafeterias. If we take the probability when two, three, four, five, six, seven and eight students eat in one of the cafeterias, it would be more work, so it is easier to subtract from the unit the amount of one and zero people who eat in the cafeterias, see:

$1 - \text{a student has breakfast} - \text{no student eats breakfast} =$

1 - \left( \begin{array}{c}8 \\ 1 \end{array}\right)0.4^{1}(0.6)^{7} - \left( \begin{array}{c} 8\\ 0 \end{array}\right)0.4^{0}(0.6)^{8}=

$1 - 0.8957952-0.01679616 = 0.89362432$

Therefore, the probability that at least two students ate breakfast in a cafeteria on campus is $0.89$

… none of them

It is requested that none of the students have had breakfast in some of the cafeterias, a calculation is easily made:

P\{ X = 0\} = \left( \begin{array}{c} 8 \\ 0 \end{array}\right)0.4^{0}(0.6)^{8} = 0.01679616

The probability that none of the students had breakfast in any of the cafeterias is $0.016$

… no more than three of them

For this subsection, the only thing that was done was to sum the probabilities when $X = 0,1,2,3$ , observe:

\left( \begin{array}{c} 8 \\ 0 \end{array}\right)0.4^{0}(0.6)^{8}+\left( \begin{array}{c} 8 \\ 1 \end{array}\right)0.4^{1}(0.6)^{7}+\left( \begin{array}{c} 8 \\ 2 \end{array}\right)0.4^{2}(0.6)^{6}+\left( \begin{array}{c} 8 \\ 3 \end{array}\right)0.4^{3}(0.6)^{5}=

$0.01679616+0.08957952+0.20901888+0.27869184=0.5940864$

The probability that no more than three students ate breakfast in one of the cafeterias on campus is $0.59$

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