# 5.4.1: Binomial Distribution Formula

- Page ID
- 10936

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Example \(\PageIndex{1}\): Shock Study

Suppose we randomly selected four individuals to participate in the "shock" study. What is the chance exactly one of them will be a success? Let's call the four people Allen (A), Brittany (B), Caroline (C), and Damian (D) for convenience. Also, suppose 35% of people are successes as in the previous version of this example.

**Solution**

Let's consider a scenario where one person refuses:

\[ \begin{align*} P(A = \text{refuse}; B = \text{shock}; C = \text{shock}; D = \text{shock}) &= P(A = \text{refuse}) P(B = \text{shock}) P(C = \text{shock}) P(D = \text{shock}) \\[5pt] &= (0.35)(0.65)(0.65)(0.65) \\[5pt] &= (0.35)^1(0.65)^3 \\[5pt] &= 0.096 \end{align*}\]

But there are three other scenarios: Brittany, Caroline, or Damian could have been the one to refuse. In each of these cases, the probability is again

\[P=(0.35)^1(0.65)^3. \nonumber\]

These four scenarios exhaust all the possible ways that exactly one of these four people could refuse to administer the most severe shock, so the total probability is

\[4 \times (0.35)^1(0.65)^3 = 0.38. \nonumber\]

Exercise \(\PageIndex{1}\)

Verify that the scenario where Brittany is the only one to refuse to give the most severe shock has probability \((0.35)^1(0.65)^3.\)

**Answer**-
\[ \begin{align*} P(A = shock; B = refuse; C = shock; D = shock) &= (0.65)(0.35)(0.65)(0.65) \\[5pt] &= (0.35)^1(0.65)^3.\end{align*}\]

## The Binomial Distribution

The scenario outlined in Example \(\PageIndex{1}\) is a special case of what is called the binomial distribution. The binomial distribution** **describes the probability of having exactly k successes in n independent Bernoulli trials with probability of a success p (in Example \(\PageIndex{1}\), n = 4, k = 1, p = 0.35). We would like to determine the probabilities associated with the binomial distribution more generally, i.e. we want a formula where we can use n, k, and p to obtain the probability. To do this, we reexamine each part of the example.

There were four individuals who could have been the one to refuse, and each of these four scenarios had the same probability. Thus, we could identify the nal probability as

\[ \text {[# of scenarios]} \times \text {P(single scenario)} \label {3.39}\]

The first component of this equation is the number of ways to arrange the k = 1 successes among the n = 4 trials. The second component is the probability of any of the four (equally probable) scenarios.

Consider P(single scenario) under the general case of k successes and n-k failures in the n trials. In any such scenario, we apply the Multiplication Rule for independent events:

\[ p^k (1- p)^{n-k}\]

This is our general formula for P(single scenario).

Secondly, we introduce a general formula for the number of ways to choose k successes in n trials, i.e. arrange k successes and n - k failures:

\[ \binom {n}{k} = \dfrac {n!}{k!(n - k)!} \label{3.4.Y}\]

The quantity \( \binom {n}{k} \) is read **n choose k**.^{30} The exclamation point notation (e.g. k!) denotes a **factorial **expression.

\[ \begin{align} 0! &= 1 \nonumber \\[5pt] 1! &= 1 \nonumber \\[5pt] 2! &= 2 \times 1 = 2 \nonumber \\[5pt] 3! &= 3 \times 2 \times 1 = 6 \nonumber \\[5pt] 4! &= 4 \times 3 \times 2 \times 1 = 24 \nonumber \\[5pt] & \vdots \nonumber \\[5pt] n! &= n \times (n - 1) \times \dots \times 3 \times 2 \times 1 \label{eq3.4.X} \end{align} \]

Substituting Equation \ref{eq3.4.X} into Equation \ref{3.4.Y}, we can compute the number of ways to choose \(k = 1\) successes in \(n = 4\) trials:

\[ \begin{align} \binom {4}{1} &= \dfrac {4!}{1! (4 - 1)!} \\[5pt] &= \dfrac {4!}{1! 3!} \\[5pt]&= \dfrac {4 \times 3 \times 2 \times 1}{(1)(3 \times 2 \times 1)} \\[5pt]&= 4 \end{align} \]

This result is exactly what we found by carefully thinking of each possible scenario in Example \(\PageIndex{1}\).

Other notations

Other notation for **n choose k** includes \(nC_k\), \(C^k_n\), and \(C(n, k)\).

Substituting \(n\) choose \(k\) for the number of scenarios and \(p^k(1 - p)^{n-k}\) for the single scenario probability in Equation \ref{3.39} yields the general binomial formula (Equation \ref{3.40}).

Definition: Binomial distribution

Suppose the probability of a single trial being a success is p. Then the probability of observing exactly k successes in n independent trials is given by

\[ \binom {n}{k} p^k (1 - p)^{n - k} = \dfrac {n!}{k!(n - k)!} p^k (1 - p)^{n - k} \label {3.40}\]

Additionally, the mean, variance, and standard deviation of the number of observed successes are

\[\mu = np \sigma^2 = np(1 - p) \sigma = \sqrt {np(1- p)} \label{3.41}\]

TIP: Four conditions to check if it is binomial?

- The trials independent.
- The number of trials, \(n\), is fixed.
- Each trial outcome can be classified as a success or failure.
- The probability of a success, \(p\), is the same for each trial.

Example \(\PageIndex{2}\)

What is the probability that 3 of 8 randomly selected students will refuse to administer the worst shock, i.e. 5 of 8 will?

**Solution**

We would like to apply the binomial model, so we check our conditions. The number of trials is fixed (n = 8) (condition 2) and each trial outcome can be classi ed as a success or failure (condition 3). Because the sample is random, the trials are independent (condition 1) and the probability of a success is the same for each trial (condition 4).

In the outcome of interest, there are k = 3 successes in n = 8 trials, and the probability of a success is p = 0.35. So the probability that 3 of 8 will refuse is given by

\[ \begin{align*} \binom {8}{3} {(0.35)}^k (1 - 0.35)^{8 - 3} &= \dfrac {8!}{3!(8 - 3)!} {(0.35)}^k (1 - 0.35)^{8 - 3} \\[5pt] &= \dfrac {8!}{3! 5!} {(0.35)}^3 {(0.65)}^5 \end{align*}\]

Dealing with the factorial part:

\[ \begin{align*} \dfrac {8!}{3!5!} &= \dfrac {8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)( 5 \times 4 \times 3 \times 2 \times 1 )} \\[5pt] &= \dfrac {8 \times 7 \times 6 }{ 3 \times 2 \times 1 } \\[5pt] & = 56 \end{align*}\]

Using \((0.35)^3(0.65)^5 \approx 0.005\), the final probability is about 56 * 0.005 = 0.28.

TIP: computing binomial probabilities

The rst step in using the binomial model is to check that the model is appropriate. The second step is to identify n, p, and k. The final step is to apply the formulas and interpret the results.

TIP: computing n choose k

In general, it is useful to do some cancelation in the factorials immediately. Alternatively, many computer programs and calculators have built in functions to compute n choose k, factorials, and even entire binomial probabilities.

Exercise \(\PageIndex{2A}\)

If you ran a study and randomly sampled 40 students, how many would you expect to refuse to administer the worst shock? What is the standard deviation of the number of people who would refuse? Equation \ref{3.41} may be useful.

**Answer**-
We are asked to determine the expected number (the mean) and the standard deviation, both of which can be directly computed from the formulas in Equation \ref{3.41}:

\[\mu = np = 40 \times 0.35 = 14 \nonumber\]

and

\[\sigma = \sqrt{np(1 - p)} = \sqrt { 40 \times 0.35 \times 0.65} = 0.02. \nonumber\]

Because very roughly 95% of observations fall within 2 standard deviations of the mean (see Section 1.6.4), we would probably observe at least 8 but less than 20 individuals in our sample who would refuse to administer the shock.

Exercise \(\PageIndex{2B}\)

The probability that a random smoker will develop a severe lung condition in his or her lifetime is about 0:3. If you have 4 friends who smoke, are the conditions for the binomial model satisfied?

**Answer**-
One possible answer: if the friends know each other, then the independence assumption is probably not satis ed. For example, acquaintances may have similar smoking habits.

Example \(\PageIndex{3}\)

Suppose these four friends do not know each other and we can treat them as if they were a random sample from the population. Is the binomial model appropriate? What is the probability that

- none of them will develop a severe lung condition?
- One will develop a severe lung condition?
- That no more than one will develop a severe lung condition?

**Solution**

To check if the binomial model is appropriate, we must verify the conditions. (i) Since we are supposing we can treat the friends as a random sample, they are independent. (ii) We have a fixed number of trials (n = 4). (iii) Each outcome is a success or failure. (iv) The probability of a success is the same for each trials since the individuals are like a random sample (p = 0.3 if we say a "success" is someone getting a lung condition, a morbid choice). Compute parts (a) and (b) from the binomial formula in Equation \ref{3.40}:

\[P(0) = \binom {4}{0}(0.3)^0(0.7)^4 = 1 \times 1 \times 0.7^4 = 0.2401 \nonumber\]

\[P(1) = \binom {4}{1}(0.3)^1(0.7)^3 = 0.4116. \nonumber\]

Note: 0! = 1.

Part (c) can be computed as the sum of parts (a) and (b):

\[P(0)+P(1) = 0.2401+0.4116 = 0.6517.\]

That is, there is about a 65% chance that no more than one of your four smoking friends will develop a severe lung condition.

Exercise \(\PageIndex{3A}\)

What is the probability that at least 2 of your 4 smoking friends will develop a severe lung condition in their lifetimes?

**Answer**-
The complement (no more than one will develop a severe lung condition) as computed in Example \(\PageIndex{3}\) as 0.6517, so we compute one minus this value: 0.3483.

Exercise \(\PageIndex{3B}\)

Suppose you have 7 friends who are smokers and they can be treated as a random sample of smokers.

- How many would you expect to develop a severe lung condition, i.e. what is the mean?
- What is the probability that at most 2 of your 7 friends will develop a severe lung condition.

**Answer a**-
\(\mu\) = 0.3 \times 7 = 2.1.

**Answer b**-
P(0, 1, or 2 develop severe lung condition) = P(k = 0)+P(k = 1)+P(k = 2) = 0:6471.

Below we consider the first term in the binomial probability, **n choose k** under some special scenarios.

Exercise \(\PageIndex{3C}\)

Why is it true that \( \binom {n}{0} = 1\) and \( \binom {n}{n} = 1 \) for any number n?

**Solution**

Frame these expressions into words. How many different ways are there to arrange 0 successes and n failures in n trials? (1 way.) How many different ways are there to arrange n successes and 0 failures in n trials? (1 way.)

Exercise \(\PageIndex{3D}\)

How many ways can you arrange one success and n -1 failures in n trials? How many ways can you arrange n -1 successes and one failure in n trials?

**Solution**

One success and n - 1 failures: there are exactly n unique places we can put the success, so there are n ways to arrange one success and n - 1 failures. A similar argument is used for the second question. Mathematically, we show these results by verifying the following two equations:

\[ \binom {n}{1} = n, \binom {n}{n - 1} = n \]

## Contributors and Attributions

David M Diez (Google/YouTube), Christopher D Barr (Harvard School of Public Health), Mine Çetinkaya-Rundel (Duke University)