The binomial distribution is a commonly used discrete distribution in statistics. The normal distribution as opposed to a binomial distribution is a continuous distribution. 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.

**Binomial Distribution in Statistics: **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 question is asked for a yes/no with a series of n experiments. Then in the binomial probability distribution, the boolean-valued outcome the success/yes/true/one is represented with probability p and the failure/no/false/zero with probability q (q = 1 − p). In a single experiment when n = 1, the binomial distribution is called a Bernoulli distribution.

Mục Lục

## What Is the Binomial Distribution Formula?

The binomial distribution formula is for any random variable X, given by; P(x:n,p) = ^{n}Cxx p^{x }(1-p)^{n-x} **Or** P(x:n,p) = ^{n}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}C_{x} = n!/x!(n-x)!. Hence, P(x:n,p) = n!/[x!(n-x)!].p^{x}.(q)^{n-x}

### KEY TAKEAWAYS

- The binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent values under a given set of parameters or assumptions.
- The underlying assumptions of the binomial distribution are that there is only one outcome for each trial, that each trial has the same probability of success, and that each trial is mutually exclusive or independent of one another.
- The binomial distribution is a common discrete distribution used in statistics, as opposed to a continuous distribution, such as the normal distribution.

## Examples on Binomial Distribution Formula

**Example 1: If a coin is tossed 5 times, using binomial distribution find the probability of:**

**(a) Exactly 2 heads**

**(b) At least 4 heads.**

**Solution:**

**(a) **The repeated tossing of the coin is an example of a Bernoulli trial. According to the problem:

Number of trials: n=5

Probability of head: p= 1/2 and hence the probability of tail, q =1/2

For exactly two heads:

x=2

P(x=2) = ^{5}C2 p^{2} q^{5-2 }= 5! / 2! 3! × (½)^{2}× (½)^{3}

P(x=2) = 5/16

**(b) **For at least four heads,

x ≥ 4, P(x ≥ 4) = P(x = 4) + P(x=5)

Hence,

P(x = 4) = ^{5}C4 p^{4} q^{5-4} = 5!/4! 1! × (½)^{4}× (½)^{1} = 5/32

P(x = 5) = ^{5}C5 p^{5} q^{5-5} = (½)^{5} = 1/32

**Answer: Therefore, P(x ≥ 4) = 5/32 + 1/32 = 6/32 = 3/16**

**Example 2: For the same question given above, find the probability of getting at most 2 heads.**

**Solution:**

Solution: P(at most 2 heads) = P(X ≤ 2) = P (X = 0) + P (X = 1)

P(X = 0) = (½)^{5} = 1/32

P(X=1) = 5C1 (½)^{5}.= 5/32

**Answer:** **Therefore, P(X ≤ 2) = 1/32 + 5/32 = 3/16**

**Example 3: 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.**

**Solution:**

Let’s Identify ‘n’ and ‘X’ from the problem.

The number of sports car owners are randomly selected is n = 10, and

The number to find the probability is X = 7.

Given: p = 60%, or 0.6.

Therefore, the probability of failure is q = 1 – 0.6 = 0.4

Now, using the binomial distribution formula

**Answer: The probability that exactly 7 are men is 0.215 or 21.5%.**

## What is a Binomial Distribution?

A **binomial distribution** can be thought of as simply the probability of a SUCCESS or FAILURE outcome in an experiment or survey that is repeated multiple times. The binomial is a type of distribution that has **two possible outcomes** (the prefix “bi” means two, or twice). For example, a coin toss has only two possible outcomes: heads or tails and taking a test could have two possible outcomes: pass or fail.

- The first variable in the binomial formula, n, stands for the number of times the experiment runs.
- The second variable, p, represents the probability of one specific outcome.

For example, let’s suppose you wanted to know the probability of getting a 1 on a die roll. if you were to roll a die 20 times, the probability of rolling a one on any throw is 1/6. Roll twenty times and you have a binomial distribution of (n=20, p=1/6). SUCCESS would be “roll a one” and FAILURE would be “roll anything else.” If the outcome in question was the probability of the die landing on an even number, the binomial distribution would then become (n=20, p=1/2). That’s because your probability of throwing an even number is one half.

## Criteria

Binomial distributions must also meet the following three criteria:

**The number of observations or trials is fixed.**In other words, you can only figure out the probability of something happening if you do it a certain number of times. This is common sense—if you toss a coin once, your probability of getting a tails is 50%. If you toss a coin a 20 times, your probability of getting a tails is very, very close to 100%.**Each observation or trial is**independent. In other words, none of your trials have an effect on the probability of the next trial.- The
**probability of success**(tails, heads, fail or pass) is**exactly the same**from one trial to another.

Once you know that your distribution is binomial, you can apply the **binomial distribution formula **to calculate the probability.

## What is a Binomial Distribution? The Bernoulli Distribution.

The binomial distribution is closely related to the Bernoulli distribution. According to Washington State University, “If each Bernoulli trial is independent, then the number of successes in Bernoulli trails has a binomial Distribution. On the other hand, the Bernoulli distribution is the Binomial distribution with n=1.”

A Bernoulli distribution is a set of Bernoulli trials. Each Bernoulli trial has one possible outcome, chosen from S, success, or F, failure. In each trial, the probability of success, P(S) = p, is the same. The probability of failure is just 1 minus the probability of success: P(F) = 1 – p. (Remember that “1” is the total probability of an event occurring…probability is always between zero and 1). Finally, all Bernoulli trials are independent from each other and the probability of success doesn’t change from trial to trial, even if you have information about the other trials’ outcomes.

## What is a Binomial Distribution? Real Life Examples

Many instances of binomial distributions can be found in real life. For example, if a new drug is introduced to cure a disease, it either cures the disease (it’s successful) or it doesn’t cure the disease (it’s a failure). If you purchase a lottery ticket, you’re either going to win money, or you aren’t. Basically, anything you can think of that can only be a success or a failure can be represented by a binomial distribution.

## The Binomial Distribution Formula

The binomial distribution formula is:

**b(x; n, P) = _{n}C_{x} * P^{x} * (1 – P)^{n – x}**

Where:

b = binomial probability

x = total number of “successes” (pass or fail, heads or tails etc.)

P = probability of a success on an individual trial

n = number of trials

**Note:** The binomial distribution formula can also be written in a slightly different way, because _{n}C_{x} = n! / x!(n – x)! (this binomial distribution formula uses factorials (What is a factorial?). “q” in this formula is just the probability of failure (subtract your probability of success from 1).

## Using the First Binomial Distribution Formula

The binomial distribution formula can calculate the probability of success for binomial distributions. Often you’ll be told to “plug in” the numbers to the **formula and calculate**. This is easy to say, but not so easy to do—unless you are *very* careful with order of operations, you won’t get the right answer. If you have a Ti-83 or Ti-89, the calculator can do much of the work for you. If not, here’s how to break down the problem into simple steps so you get the answer right—every time.

**Example 1**

**Q. A coin is tossed 10 times. What is the probability of getting exactly 6 heads?**

I’m going to use this formula: b(x; n, P) – _{n}C_{x} * P^{x} * (1 – P)^{n – x}

The number of trials (n) is 10

The odds of success (“tossing a heads”) is 0.5 (So 1-p = 0.5)

x = 6

P(x=6) = _{10}C_{6} * 0.5^6 * 0.5^4 = 210 * 0.015625 * 0.0625 = 0.205078125

**Tip:** You can use the **combinations calculator** to figure out the value for _{n}C_{x}.

## How to Work a Binomial Distribution Formula:

**Example 2**

**80% of people who purchase pet insurance are women. If 9 pet insurance owners are randomly selected, find the probability that exactly 6 are women.**

**Step 1:** Identify ‘n’ from the problem. Using our example question, n (the number of randomly selected items) is 9.

**Step 2:** Identify ‘X’ from the problem. X (the number you are asked to find the probability for) is 6.

**Step 3:** Work the first part of the formula. The first part of the formula is

n! / (n – X)! X!

Substitute your variables:

9! / ((9 – 6)! × 6!)

Which equals 84. Set this number aside for a moment.

**Step 4:** Find p and q. p is the probability of success and q is the probability of failure. We are given p = 80%, or .8. So the probability of failure is 1 – .8 = .2 (20%).

**Step 5:** Work the second part of the formula.

p^{X}

= .8^{6}

= .262144

Set this number aside for a moment.

**Step 6:** Work the third part of the formula.

q^{(n – X)}

= .2^{(9-6)}

= .2^{3}

= .008

**Step 7:** Multiply your answer from step 3, 5, and 6 together.

84 × .262144 × .008 = 0.176.

**Example 3**

**60% of people who purchase sports cars are men. If 10 sports car owners are randomly selected, find the probability that exactly 7 are men.**

**Step 1:**: Identify ‘n’ and ‘X’ from the problem. Using our sample question, n (the number of randomly selected items—in this case, sports car owners are randomly selected) is 10, and X (the number you are asked to “find the probability” for) is 7.

**Step 2:** Figure out the first part of the formula, which is:

n! / (n – X)! X!

Substituting the variables:

10! / ((10 – 7)! × 7!)

Which equals 120. Set this number aside for a moment.

**Step 3:** Find “p” the probability of success and “q” the probability of failure. We are given p = 60%, or .6. therefore, the probability of failure is 1 – .6 = .4 (40%).

**Step 4:** Work the next part of the formula.

p^{X}

= .6^{7}

= .0.0279936

Set this number aside while you work the third part of the formula.

**Step 5:** Work the third part of the formula.

q^{(.4 – 7)}

= .4^{(10-7)}

= .4^{3}

= .0.064

**Step 6:** Multiply the three answers from steps 2, 4 and 5 together.

120 × 0.0279936 × 0.064 = 0.215.

That’s it!

## Binomial Probability Distribution

To understand binomial distributions and binomial probability, it helps to understand binomial experiments and some associated notation; so we cover those topics first.

## Binomial Experiment

A **binomial experiment** is a statistical experiment that has the following properties:

- The experiment consists of
*n*repeated trials. - Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure.
- The probability of success, denoted by
*P*, is the same on every trial. - The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.

Consider the following statistical experiment. You flip a coin 2 times and count the number of times the coin lands on heads. This is a binomial experiment because:

- The experiment consists of repeated trials. We flip a coin 2 times.
- Each trial can result in just two possible outcomes – heads or tails.
- The probability of success is constant – 0.5 on every trial.
- The trials are independent; that is, getting heads on one trial does not affect whether we get heads on other trials.

## Notation

The following notation is helpful, when we talk about binomial probability.

*x*: The number of successes that result from the binomial experiment.*n*: The number of trials in the binomial experiment.*P*: The probability of success on an individual trial.*Q*: The probability of failure on an individual trial. (This is equal to 1 –*P*.)*n!*: The factorial of n (also known as n factorial).- b(
*x*;*n, P*): Binomial probability – the probability that an*n*-trial binomial experiment results in exactly*x*successes, when the probability of success on an individual trial is*P*. _{n}C_{r}: The number of combinations of*n*things, taken*r*at a time.

## Binomial Distribution

A **binomial random variable** is the number of successes *x* in *n* repeated trials of a binomial experiment. The probability distribution of a binomial random variable is called a **binomial distribution**.

Suppose we flip a coin two times and count the number of heads (successes). The binomial random variable is the number of heads, which can take on values of 0, 1, or 2. The binomial distribution is presented below.

Number of heads | Probability |
---|---|

0 | 0.25 |

1 | 0.50 |

2 | 0.25 |

The binomial distribution has the following properties:

- The mean of the distribution (μ
_{x}) is equal to*n***P*. - The variance (σ
^{2}_{x}) is*n***P** ( 1 –*P*). - The standard deviation (σ
_{x}) is sqrt[*n***P** ( 1 –*P*) ].

## Binomial Formula and Binomial Probability

The **binomial probability** refers to the probability that a binomial experiment results in exactly *x* successes. For example, in the above table, we see that the binomial probability of getting exactly one head in two coin flips is 0.50.

Given *x*, *n*, and *P*, we can compute the binomial probability based on the binomial formula:

**Binomial Formula.** Suppose a binomial experiment consists of *n* trials and results in *x* successes. If the probability of success on an individual trial is *P*, then the binomial probability is:

b(*x*; *n, P*) = _{n}C_{x} * P^{x} * (1 – P)^{n – x}

or

b(*x*; *n, P*) = { n! / [ x! (n – x)! ] } * P^{x} * (1 – P)^{n – x}

**Example 1**

Suppose a die is tossed 5 times. What is the probability of getting exactly 2 fours?

*Solution:* This is a binomial experiment in which the number of trials is equal to 5, the number of successes is equal to 2, and the probability of success on a single trial is 1/6 or about 0.167. Therefore, the binomial probability is:

b(2; 5, 0.167) = _{5}C_{2} * (0.167)^{2} * (0.833)^{3}

b(2; 5, 0.167) = 0.161

## Cumulative Binomial Probability

A **cumulative binomial probability** refers to the probability that the binomial random variable falls within a specified range (e.g., is greater than or equal to a stated lower limit and less than or equal to a stated upper limit).

For example, we might be interested in the cumulative binomial probability of obtaining 45 or fewer heads in 100 tosses of a coin (see Example 1 below). This would be the sum of all these individual binomial probabilities.

b(x < 45; 100, 0.5) =

b(x = 0; 100, 0.5) + b(x = 1; 100, 0.5) + … + b(x = 44; 100, 0.5) + b(x = 45; 100, 0.5)

**Example 2**

What is the probability of obtaining 45 or fewer heads in 100 tosses of a coin?

*Solution:* To solve this problem, we compute 46 individual probabilities, using the binomial formula. The sum of all these probabilities is the answer we seek. Thus,

b(x < 45; 100, 0.5) = b(x = 0; 100, 0.5) + b(x = 1; 100, 0.5) + . . . + b(x = 45; 100, 0.5)

b(x < 45; 100, 0.5) = 0.184

**Example 3**

The probability that a student is accepted to a prestigious college is 0.3. If 5 students from the same school apply, what is the probability that at most 2 are accepted?

*Solution:* To solve this problem, we compute 3 individual probabilities, using the binomial formula. The sum of all these probabilities is the answer we seek. Thus,

b(x < 2; 5, 0.3) = b(x = 0; 5, 0.3) + b(x = 1; 5, 0.3) + b(x = 2; 5, 0.3)

b(x < 2; 5, 0.3) = 0.1681 + 0.3601 + 0.3087

b(x < 2; 5, 0.3) = 0.8369

**Example 4**

What is the probability that the world series will last 4 games? 5 games? 6 games? 7 games? Assume that the teams are evenly matched.

*Solution:* This is a very tricky application of the binomial distribution. If you can follow the logic of this solution, you have a good understanding of the material covered in the tutorial, to this point.

In the world series, there are two baseball teams. The series ends when the winning team wins 4 games. Therefore, we define a success as a win by the team that ultimately becomes the world series champion.

For the purpose of this analysis, we assume that the teams are evenly matched. Therefore, the probability that a particular team wins a particular game is 0.5.

Let’s look first at the simplest case. What is the probability that the series lasts only 4 games. This can occur if one team wins the first 4 games. The probability of the National League team winning 4 games in a row is:

b(4; 4, 0.5) = _{4}C_{4} * (0.5)^{4} * (0.5)^{0} = 0.0625

Similarly, when we compute the probability of the American League team winning 4 games in a row, we find that it is also 0.0625. Therefore, probability that the series ends in four games would be 0.0625 + 0.0625 = 0.125; since the series would end if either the American or National League team won 4 games in a row.

Now let’s tackle the question of finding probability that the world series ends in 5 games. The trick in finding this solution is to recognize that the series can only end in 5 games, if one team has won 3 out of the first 4 games. So let’s first find the probability that the American League team wins exactly 3 of the first 4 games.

b(3; 4, 0.5) = _{4}C_{3} * (0.5)^{3} * (0.5)^{1} = 0.25

Okay, here comes some more tricky stuff, so listen up. Given that the American League team has won 3 of the first 4 games, the American League team has a 50/50 chance of winning the fifth game to end the series. Therefore, the probability of the American League team winning the series in 5 games is 0.25 * 0.50 = 0.125. Since the National League team could also win the series in 5 games, the probability that the series ends in 5 games would be 0.125 + 0.125 = 0.25.

The rest of the problem would be solved in the same way. You should find that the probability of the series ending in 6 games is 0.3125; and the probability of the series ending in 7 games is also 0.3125.

# The Binomial Distribution

“Bi” means “two” (like a bicycle has two wheels) …… so this is about things with two results. |

## Let’s Toss a Coin!

Toss a fair coin** three times** … what is the chance of getting **two Heads**?

Tossing a coin three times (**H** is for heads, **T** for Tails) can get any of these 8 **outcomes**:

### Which outcomes do we want?

“Two Heads” could be in any order: “HHT”, “THH” and “HTH” all have two Heads (and one Tail).

So **3 of the outcomes** produce “Two Heads”.

### What is the probability of each outcome?

Each outcome is equally likely, and there are 8 of them, so each outcome has a probability of 1/8

So the probability of **event** “Two Heads” is:

## 3 Heads, 2 Heads, 1 Head, None

The calculations are (P means “Probability of”):

- P(Three Heads) = P(
**HHH**) =**1/8** - P(Two Heads) = P(
**HHT**) + P(**HTH**) + P(**THH**) = 1/8 + 1/8 + 1/8 =**3/8** - P(One Head) = P(
**HTT**) + P(**THT**) + P(**TTH**) = 1/8 + 1/8 + 1/8 =**3/8** - P(Zero Heads) = P(
**TTT**) =**1/8**

We can write this in terms of a Random Variable, X, = “The number of Heads from 3 tosses of a coin”:

- P(X = 3) = 1/8
- P(X = 2) = 3/8
- P(X = 1) = 3/8
- P(X = 0) = 1/8

And this is what it looks like as a graph:

## Making a Formula

Now imagine we want the chances of **5 heads in 9 tosses**: to list all 512 outcomes will take a long time!

So let’s make a formula.

In our previous example, how can we get the values 1, 3, 3 and 1 ?

### Example: with 3 tosses, what are the chances of 2 Heads?

We have **n=3** and **k=2**:

So there are 3 outcomes that have “2 Heads”

(We knew that already, but now we have a formula for it.)

Let’s use it for a harder question:

### Example: with 9 tosses, what are the chances of 5 Heads?

We have **n=9** and **k=5**:

About a **25% chance**.

(Easier than listing them all.)

## Bias!

So far the chances of success or failure have been **equally likely**.

But what if the coins are biased (land more on one side than another) or choices are not 50/50.

### Example: You sell sandwiches. 70% of people choose chicken, the rest choose something else.

### What is the probability of selling 2 chicken sandwiches to the next 3 customers?

This is just like the heads and tails example, but with 70/30 instead of 50/50.

**The “Two Chicken” cases are highlighted.**

The probabilities for “two chickens” all work out to be **0.147**, because we are multiplying two 0.7s and one 0.3 in each case. In other words

0.147 = 0.7 × 0.7 × 0.3

Or, using exponents:

= 0.7^{2} × 0.3^{1}

The **0.7** is the probability of each choice we want, call it **p**

The **2** is the number of choices we want, call it **k**

And we have (so far):

= p^{k} × 0.3^{1}

The **0.3** is the probability of the opposite choice, so it is: **1−p**

The **1** is the number of opposite choices, so it is: **n−k**

Which gives us:

= **p ^{k}(1-p)^{(n-k)}**

Where

**p**is the probability of each choice we want**k**is the the number of choices we want**n**is the total number of choices

### Example: (continued)

- p = 0.7 (chance of chicken)
- k = 2 (chicken choices)
- n = 3 (total choices)

So we get:p^{k}(1-p)^{(n-k)} =0.7^{2}(1-0.7)^{(3-2)}=0.7^{2}(0.3)^{(1)}=0.7 × 0.7 × 0.3=0.147

which is what we got before, but now using a formula

Now we know the probability of each outcome is 0.147

But we need to include that there are **three** such ways it can happen: (chicken, chicken, other) or (chicken, other, chicken) or (other, chicken, chicken)

### Example: (continued)

The total number of “two chicken” outcomes is:

OK. That was a lot of work for something we knew already, but now we have a formula we can use for harder questions.

### Example: Sam says “70% choose chicken, so 7 of the next 10 customers should choose chicken” … what are the chances Sam is right?

So we have:

- p = 0.7
- n = 10
- k = 7

And we get:p^{k}(1-p)^{(n-k)} =0.7^{7}(1-0.7)^{(10-7)}=0.7^{7}(0.3)^{(3)}=**0.0022235661**

That is the probability of each outcome.

And the total number of those outcomes is:

So the probability of 7 out of 10 choosing chicken is only about **27%**

Moral of the story: even though the long-run average is 70%, don’t expect 7 out of the next 10.

## Putting it Together

Now we know how to calculate **how many**:

### The General Binomial Probability Formula

Important Notes:

- The trials are independent,
- There are only two possible outcomes at each trial,
- The probability of “success” at each trial is constant.

In this case **n=4**, **p = P(Two) = 1/6**

X is the Random Variable ‘Number of Twos from four throws’.

Substitute x = 0 to 4 into the formula:

Summary: “for the 4 throws, there is a 48% chance of no twos, 39% chance of 1 two, 12% chance of 2 twos, 1.5% chance of 3 twos, and a tiny 0.08% chance of all throws being a two (but it still could happen!)”

This time the graph is not symmetrical:

Summary: “for the 4 next bikes, there is a tiny 0.01% chance of no passes, 0.36% chance of 1 pass, 5% chance of 2 passes, 29% chance of 3 passes, and a whopping 66% chance they all pass the inspection.”

## Mean, Variance and Standard Deviation

Let’s calculate the Mean, Variance and Standard Deviation for the Sports Bike inspections.

There are (relatively) simple formulas for them. They are a little hard to prove, but they do work!

The mean, or “expected value”, is:

μ = np

For the sports bikes:

μ = 4 × 0.9 = 3.6

So we can expect 3.6 bikes (out of 4) to pass the inspection.

Makes sense really … 0.9 chance for each bike times 4 bikes equals 3.6

The formula for Variance is:

Variance: σ^{2} = np(1-p)

And Standard Deviation is the square root of variance:

σ = √(np(1-p))

For the sports bikes:

Variance: σ^{2} = 4 × 0.9 × 0.1 = 0.36

Standard Deviation is:

σ = √(0.36) = 0.6

Note: we could also calculate them manually, by making a table like this:

X | P(X) | X × P(X) | X^{2} × P(X) |

0 | 0.0001 | 0 | 0 |

1 | 0.0036 | 0.0036 | 0.0036 |

2 | 0.0486 | 0.0972 | 0.1944 |

3 | 0.2916 | 0.8748 | 2.6244 |

4 | 0.6561 | 2.6244 | 10.4976 |

SUM: | 3.6 | 13.32 |

The mean is the **Sum of (X × P(X))**:

μ = 3.6

The variance is the **Sum of (X ^{2} × P(X))** minus

**Mean**:

^{2}Variance: σ^{2} = 13.32 − 3.6^{2} = 0.36Standard Deviation is:

σ = √(0.36) = 0.6And we got the same results as before (yay!)

## Summary

- The General Binomial Probability Formula:

## FAQs on Binomial Distribution Formula

### What Is Binomial Distribution and Binomial Distribution Formula in Statistics?

The binomial distribution is a common discrete distribution used in statistics, as opposed to a continuous distribution, such as the normal distribution. The binomial distribution, therefore, represents the probability for x successes in n trials, given a success probability p for each trial. The binomial distribution formula is for any random variable X, given by; P(x:n,p) = ^{n}Cxx p^{x }(1-p)^{n-x} **Or** P(x:n,p) = ^{n}Cxx p^{x} (q)^{n-x}, where, n is the number of experiments, p is probability of success in a single experiment, q is probability of failure in a single experiment (= 1 – p) and takes values as 0, 1, 2, 3, 4, …, n.

### What Is the Purpose of the Binomial Distribution Formula?

The binomial distribution formula allows us to compute the probability of observing a specified number of “successes” when the process is repeated a specific number of times (e.g., in a set of patients) and the outcome for a given patient is either a success or a failure.

### What Is the Formula for Binomial Distribution?

The formula for binomial distribution is:

P(x: n,p) = ^{n}Cx p^{x} (q)^{n-x}

Where p is the probability of success, q is the probability of failure, n = number of trials.

### What Is the Binomial Distribution Formula for the Mean and Variance?

The mean and variance of the binomial distribution are:

Mean = np

Variance = npq

where p is the probability of success, q is the probability of failure, n = number of trials.

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