Binomial Distribution Calculator

Enter the number of trials n, the probability of success on each trial p, and the number of successes k. You get the probability of exactly k successes, the cumulative probabilities of at most k and at least k, the mean, variance and standard deviation, and a bar chart of the whole distribution.

Number of trials n

Probability of success p

Number of successes k

Enter the probability p as a decimal from 0 to 1 (for 25%, enter 0.25) or as a fraction such as 1/4. n is a whole number from 1 to 50, and k is a whole number from 0 to n.

Result and graph

Enter the number of trials and the probability in the fields on the left and press "Calculate". The result and a bar chart of the distribution will appear here.

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What you can do on this page
What is this calculation used for?
How to Use
Formulas and graphs
Symbols and terms
- Symbols
- Terms
Good to know before you start
How to calculate it in Excel
How to calculate it in Google Sheets
How to calculate it in Python
How to write it in LaTeX and other math languages (copy and paste)
How to have ChatGPT do the calculation
DataChef Features
Related Features
NumberChef Calculators List

What you can do on this page

Enter just 3 numbers to find "the probability that something with probability \(p\) happens exactly \(k\) times in \(n\) tries", \(P(X = k) = {}_{n}\mathrm{C}_{k}\, p^{k} (1-p)^{n-k}\), on the spot
The answer is shown both as a fraction in lowest terms such as \(\dfrac{3}{8}\) (the exact value) and as a decimal and percent. You can also see the steps for the number of combinations \({}_{n}\mathrm{C}_{k}\)
It also finds the cumulative probabilities, "at most \(k\) times" \(P(X \le k)\) and "at least \(k\) times" \(P(X \ge k)\), at the same time
It shows the expected value \(E(X) = np\), the variance \(V(X) = np(1-p)\) and the standard deviation \(\sigma\), and checks the rule of thumb for using the normal approximation
The whole distribution, from 0 up to \(n\) successes, is drawn as a bar chart, so you can see at a glance which counts are most likely

This works for repeated trials where the probability is the same every time and the results do not affect each other (they are independent). It does not work when the probability changes from one try to the next, such as drawing tickets without putting them back.

What is this calculation used for?

Estimating "k wins out of n" for loot boxes and prize draws (games)

If you open 100 loot boxes (or gacha pulls) with a 1% drop rate, the number of rare drops follows the binomial distribution \(B(100,\ 0.01)\) (assuming, as published, that each try has the same chance and is independent). The chance of getting nothing at all is \(0.99^{100} \approx 36.6\%\), so the distribution shows that the gut feeling "100 tries at 1% is almost a sure thing" is risky.
Estimating "the expected value \(np\) plus the spread" when you plan how much to spend is exactly the calculation on this page.

Designing acceptance rules for sampling inspection (manufacturing and quality control)

Instead of checking every item, factories use acceptance sampling: "take \(n\) items from a lot, and accept the lot if there are \(c\) or fewer defects". The chance that a lot with a defect rate \(p\) is accepted is exactly the cumulative binomial probability \(P(X \le c)\). For example, if you take 20 items from a lot with a 1% defect rate and accept it only with 0 defects, the chance of acceptance is \(0.99^{20} \approx 81.8\%\).
This calculation is used to design inspection rules that accept good lots with high probability and reject bad ones.

Estimating lucky guesses on multiple-choice tests (education and exams)

If you guess on all 20 questions of a test with 4 choices each, the number of correct answers follows the binomial distribution \(B(20,\ 0.25)\), and the expected value is \(20 \times 0.25 = 5\) questions. The chance of getting 8 or more right by luck is only about 10%, which confirms with probability that guessing alone rarely reaches a passing score.
Test makers also keep this distribution in mind when they set the number of questions and the passing score, so that it is hard to pass by luck.

Reading the ups and downs of sports stats (basketball and more)

When a 70% free-throw shooter takes 10 shots, the chance of making exactly 7 is only \({}_{10}\mathrm{C}_{7} \times 0.7^{7} \times 0.3^{3} \approx 26.7\%\) (treating each shot as independent with a fixed success rate). A 70% shooter making exactly 7 of 10 is not as ordinary as it sounds. Making 6 or 8 happens all the time.
The binomial distribution helps you tell how much of a "hot streak" or "slump" is just normal random variation.

Where the margin of error in polls and TV ratings comes from (surveys)

In approval rating polls and TV ratings, the number of people out of the \(n\) surveyed who give a certain answer is treated as following a binomial distribution. The "plus or minus a few points" margin of error is calculated from the standard deviation \(\sqrt{np(1-p)}\) and the normal approximation.
Deciding how many people to survey to get the margin of error down to a target size is based directly on the properties of the binomial distribution.

Formulas and graphs

Binomial probability (probability of exactly \(k\) successes)

Graph

Standard notation (the usual math form)

\(P(X = k)\) \(=\) \({}_{n}\mathrm{C}_{k}\) \(\times\) \(p\) \(k\) \(\times\) \((1 - p)\) \(n - k\)

In words (symbols replaced with words)

⑥ \(P(X = k)\): probability of exactly \(k\) successes \(=\) ⑤ \({}_{n}\mathrm{C}_{k}\): ways to choose which trials succeed \(\times\) ① \(p\): probability of success ② \(k\): number of successes \(\times\) ③ \(1 - p\): probability of failure ④ \(n - k\): number of failures

The formula in words

① Take the \(p\): probability of success

② and multiply it by itself as many times as the \(k\): number of successes

③ then take the \(1 - p\): probability of failure

④ and multiply it by itself as many times as the \(n - k\): number of failures

⑤ multiply the results by the \({}_{n}\mathrm{C}_{k}\): ways to choose which trials succeed

⑥ and you get the \(P(X = k)\): probability of exactly \(k\) successes

Quick example

If you flip a coin (probability of heads \(\dfrac{1}{2}\)) 4 times, the probability of exactly 2 heads is

probability of exactly 2 \(P(X = 2)\) \(=\) number of orders \({}_{4}\mathrm{C}_{2}\) \(\times\) heads \(\dfrac{1}{2}\), twice \(\times\) tails \(\dfrac{1}{2}\), twice

\({}_{4}\mathrm{C}_{2} = \dfrac{4 \times 3}{2 \times 1} = 6\)

\(P(X = 2) = 6 \times \left(\dfrac{1}{2}\right)^{2} \times \left(\dfrac{1}{2}\right)^{2} = \dfrac{6}{16} = \dfrac{3}{8}\ \ (37.5\%)\)

Key idea

The probability of one particular order with \(k\) successes and \(n-k\) failures is \(p^{k} (1-p)^{n-k}\). There are \({}_{n}\mathrm{C}_{k}\) such orders (ways to choose which trials are the successes), so multiplying by that number gives the answer. The distribution that collects the probability of every possible count \(X\) (0, 1, …, \(n\) successes) is called the binomial distribution and is written \(B(n,\ p)\). The calculator on this page draws the whole distribution as a bar chart.

Cumulative probability (at most \(k\) and at least \(k\) successes)

Standard notation (the usual math form)

\(P(X \le k)\) \(=\) \(P(X=0) + P(X=1) + \cdots + P(X=k)\)

\(P(X \ge k)\) \(=\) \(1\) \(-\) \(P(X \le k - 1)\)

In words (symbols replaced with words)

② \(P(X \le k)\): probability of at most \(k\) \(=\) ① sum of the probabilities of exactly 0 through exactly \(k\)

⑤ \(P(X \ge k)\): probability of at least \(k\) \(=\) ③ \(1\): total probability \(-\) ④ probability of at most \(k-1\)

The formula in words

① Add up all the probabilities of exactly 0, exactly 1, …, exactly \(k\)

② and you get the \(P(X \le k)\): probability of at most \(k\)

③ For "at least \(k\)", the complement gives a shorter calculation. Start from the \(1\): total probability

④ subtract the \(P(X \le k-1)\): probability of at most \(k-1\)

⑤ and you get the \(P(X \ge k)\): probability of at least \(k\)

Quick example

If you flip a coin 4 times, the probability of at most 2 heads and the probability of at least 2 heads are

probability of at most 2 \(P(X \le 2)\) \(=\) sum for 0, 1 and 2

\(P(X \le 2) = \dfrac{1}{16} + \dfrac{4}{16} + \dfrac{6}{16} = \dfrac{11}{16}\ \ (68.75\%)\)

\(P(X \ge 2) = 1 - P(X \le 1) = 1 - \dfrac{5}{16} = \dfrac{11}{16}\ \ (68.75\%)\)

Key idea

"At most" and "at least" include the number itself (at most 2 means 0, 1 or 2). The standard approach is to simply add up the exact probabilities for "at most \(k\)", and to subtract "at most \(k-1\)" from the total of 1 for "at least \(k\)" (using the complement). "The probability that it happens at least once" is \(P(X \ge 1) = 1 - P(X = 0)\). This is the most common use of the complement.

Mean, variance and standard deviation of the binomial distribution

Standard notation (the usual math form)

\(E(X)\) \(=\) \(n\) \(\times\) \(p\)

\(V(X)\) \(=\) \(n\) \(\times\) \(p\) \(\times\) \((1 - p)\)

\(\sigma\) \(=\) \(\sqrt{V(X)}\)

In words (symbols replaced with words)

③ \(E(X)\): expected value (average number of successes) \(=\) ① \(n\): number of trials \(\times\) ② \(p\): probability of success

⑤ \(V(X)\): variance (how spread out) \(=\) \(n\): number of trials \(\times\) \(p\): probability of success \(\times\) ④ \(1 - p\): probability of failure

⑦ \(\sigma\): standard deviation (typical spread) \(=\) ⑥ square root of the variance \(V(X)\)

The formula in words

① Multiply the \(n\): number of trials

② by the \(p\): probability of success

③ and you get the \(E(X)\): expected value (average number of successes)

④ Multiply this further by the \(1 - p\): probability of failure

⑤ and you get the \(V(X)\): variance (how spread out)

⑥ The square root of the variance \(V(X)\) is the

⑦ \(\sigma\): standard deviation (typical spread)

Quick example

For the number of 1s when you roll a die 30 times (\(n = 30\), \(p = \dfrac{1}{6}\)):

expected value \(E(X)\) \(=\) number of trials \(30\) \(\times\) probability of a 1 \(\dfrac{1}{6}\)

\(E(X) = 30 \times \dfrac{1}{6} = 5\)

\(V(X) = 30 \times \dfrac{1}{6} \times \dfrac{5}{6} = \dfrac{25}{6}\)

\(\sigma = \sqrt{\dfrac{25}{6}} = \dfrac{5}{\sqrt{6}} = \dfrac{5\sqrt{6}}{6} \approx 2.04\)

Key idea

The expected value \(np\) matches your intuition exactly: "Do something with probability \(\dfrac{1}{6}\) 30 times, and on average it happens about \(30 \times \dfrac{1}{6} = 5\) times." The variance and standard deviation tell you how far the actual count tends to be from the expected value. With a standard deviation of about 2, you can read the result as "usually somewhere around 3 to 7 times" rather than "exactly 5 times". For the same \(n\), the variance \(np(1-p)\) is largest when \(p = \dfrac{1}{2}\) (when success and failure are equally likely, the result is hardest to predict).

Normal approximation (rule of thumb: \(np \ge 5\) and \(n(1-p) \ge 5\))

Graph

Standard notation (the usual math form)

In words (symbols replaced with words)

\(B(n,\ p)\) \(\approx\) \(N(np,\ np(1 - p))\)

① \(B(n,\ p)\): binomial distribution \(\approx\) ② normal distribution with mean \(np\) and variance \(np(1-p)\)

The formula in words

① The \(B(n,\ p)\): binomial distribution can be roughly replaced, when \(np \ge 5\) and \(n(1-p) \ge 5\) (the rule of thumb), by the

② \(N(np,\ np(1-p))\): normal distribution with mean \(np\) and variance \(np(1-p)\)

Quick example

The number of heads when you flip a coin 20 times (\(n = 20\), \(p = \dfrac{1}{2}\)) is

binomial distribution \(B(20,\ \dfrac{1}{2})\) \(\approx\) normal distribution \(N(10,\ 5)\)

\(np = 20 \times \dfrac{1}{2} = 10 \ge 5,\quad n(1 - p) = 20 \times \dfrac{1}{2} = 10 \ge 5\)

\(B\left(20,\ \dfrac{1}{2}\right) \approx N(10,\ 5)\)

Key idea

As \(n\) grows, the bar chart of the binomial distribution gets closer to a symmetric bell shape, almost the same as the normal distribution. So when \(n\) is large, you can replace the long sums of the binomial distribution with the normal distribution table (or z-scores). This is the normal approximation, and \(np \ge 5\) and \(n(1-p) \ge 5\) is the common rule of thumb for when to use it. Inside \(N(np,\ np(1-p))\), the order is "mean, variance". After approximating, you can find things like "the probability of at least a certain count" on our Normal Distribution Probability Calculator page.

The binomial distribution is the distribution of "how many times something with probability \(p\) happens in \(n\) tries". The probability of exactly \(k\) successes is \(P(X = k) = {}_{n}\mathrm{C}_{k}\, p^{k} (1-p)^{n-k}\), the expected value is \(np\), and the variance is \(np(1-p)\). When \(n\) is large, it can be approximated by the normal distribution \(N(np,\ np(1-p))\).

Symbols and terms

Symbols

\(X\)	X	The random variable for the number of successes. By convention, numbers decided by chance are written with capital letters from the end of the alphabet, such as \(X,\ Y,\ Z\).
\(P(X = k)\)	P of X equals k	The probability that the number of successes \(X\) is exactly \(k\). \(P\) stands for "probability".
\({}_{n}\mathrm{C}_{k}\)	n choose k	The number of combinations of \(k\) items chosen from \(n\). \(C\) stands for "combination". Many textbooks write it as \(\dbinom{n}{k}\) instead.
\(n\)	en	The number of trials (how many times you repeat it). The letter \(n\), for "number", is often used for counts.
\(p\)	pee	The probability of success on one trial. It comes from "probability", and the lowercase letter is often used for the probability of a single try.
\(1 - p\)	one minus p	The probability of failure on one trial (not succeeding). Either success or failure always happens, so it is what is left after subtracting \(p\) from the total of \(1\). It is also written \(q\).
\(k\)	kay	The number of successes, a whole number from \(0\) to \(n\). The letter \(k\) is often used to say "how many times".
\(P(X \le k)\)	P of X less than or equal to k	The probability of at most \(k\) successes (any of 0 through \(k\)). It is a cumulative probability, the exact probabilities added up through \(k\).
\(P(X \ge k)\)	P of X greater than or equal to k	The probability of at least \(k\) successes (any of \(k\) through \(n\)). Subtracting \(P(X \le k-1)\) from the total of \(1\) (the complement) shortens the sum.
\(E(X)\)	E of X	The expected value (the average number of successes). \(E\) stands for "expectation". For the binomial distribution, \(E(X) = np\). It is also called the mean and written \(\mu\).
\(V(X)\)	V of X	The variance (how spread out the results are). \(V\) stands for "variance". For the binomial distribution, \(V(X) = np(1-p)\). It is also written \(\mathrm{Var}(X)\) or \(\sigma^2\).
\(\sigma\)	sigma	The standard deviation. It is the lowercase Greek letter sigma, the Greek "s" as in "standard deviation". It is the square root of the variance, \(\sigma = \sqrt{V(X)}\), and shows the typical distance from the expected value.
\(B(n,\ p)\)	B of n, p	The binomial distribution with \(n\) trials and probability of success \(p\). \(B\) stands for "binomial". "\(X\) follows a binomial distribution" is written \(X \sim B(n,\ p)\).
\(N(\mu,\ \sigma^2)\)	N of mu, sigma squared	The normal distribution with mean \(\mu\) and variance \(\sigma^2\). \(N\) stands for "normal". By convention, the order inside the parentheses is "mean, variance".
\(\approx\)	approximately equal to	The symbol for "approximately equal". On this page it shows that the binomial distribution can be roughly replaced by the normal distribution.

Terms

binomial distribution	The probability distribution of "how many times something with probability \(p\) happens in \(n\) tries". The name comes from the fact that the \({}_{n}\mathrm{C}_{k}\) in the formula is the same as the binomial coefficient (the coefficients in the binomial theorem). In the US it is usually taught in high school Statistics or AP Statistics.
random variable	A variable whose value is decided by chance. On this page, the number of successes is the random variable. It takes the values 0 up to n, each with its own probability.
trial	One run of an experiment or observation whose result is decided by chance, such as one coin flip or one draw of a ticket.
repeated trials	Repeating the same trial with the same probability each time, where the results do not affect each other (independently). They are also called Bernoulli trials. The binomial distribution is the distribution of the number of successes in repeated trials.
independent	When the result of one trial does not affect the result of another. The first and second rolls of a die are independent, but drawing tickets without putting them back is not (the earlier results change how many winning tickets are left).
combination	A way to choose \(k\) of \(n\) items where order does not matter. The total number of them is \({}_{n}\mathrm{C}_{k}\). In the binomial distribution, it counts the ways to choose which of the \(n\) trials are successes.
cumulative probability	The exact probabilities added up to a certain count. The probability of at most \(k\), \(P(X \le k)\), is the typical example.
complement	The event that something does not happen. The probabilities always add up to 1, so a probability like "at least \(k\)" is easy to find by subtracting the opposite side (at most \(k-1\)) from 1.
expected value	The average value when you repeat the trials many times. For the binomial distribution it is the average number of successes, given by the simple formula \(np\).
variance	A measure of how far the actual values tend to be from the expected value (the spread). It is defined as the average of the squared differences.
standard deviation	The square root of the variance. The variance is in squared units, so taking the square root brings it back to the original unit (here, the number of successes) as a measure of spread.
normal distribution	A symmetric, bell-shaped distribution where values near the mean are most likely and values farther away are less likely. Many real-world quantities, such as heights and measurement errors, come close to this shape, and the binomial distribution also gets close to it when \(n\) is large.
normal approximation	Replacing the binomial distribution with the normal distribution \(N(np,\ np(1-p))\) when \(n\) is large. The common rule of thumb is \(np \ge 5\) and \(n(1-p) \ge 5\).

Good to know before you start

Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.
If you get stuck, going back over the topics in this table is the fastest way forward.

Basic probability (Grade 7)	Knowing that a probability shows how likely something is as a number from 0 to 1 (including reading 0.25 as 25%) Knowing that the probabilities of all possible outcomes always add up to 1
Working with fractions (Grades 5–6)	Being able to multiply fractions (\(\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\)), simplify them and find common denominators Being comfortable keeping answers as fractions (and turning \(\dfrac{3}{8}\) into 0.375)
Powers and exponents (Grades 6–8, Algebra 1)	Knowing that the small number at the upper right (the exponent) tells how many times to multiply, as in \(p^{3} = p \times p \times p\) Being able to raise a fraction to a power, as in \(\left(\dfrac{1}{2}\right)^{4} = \dfrac{1}{16}\)
Counting and combinations (high school, Algebra 2 or Statistics)	Knowing that \({}_{n}\mathrm{C}_{k}\) is "the number of ways to choose \(k\) of \(n\) items when order does not matter" Being able to calculate it, as in \({}_{4}\mathrm{C}_{2} = \dfrac{4 \times 3}{2 \times 1} = 6\)
Probability of repeated trials and the complement (high school, Statistics)	Understanding the setup of repeated trials, "repeated independently with the same probability each time" Being able to use the complement to find "at least once" as \(1 - P(X = 0)\)
Random variables, expected value and variance (high school, AP Statistics)	Knowing what a random variable (a variable whose value is decided by chance) and a probability distribution are Knowing that the expected value is the average, and the variance and standard deviation show the spread
The normal distribution (high school, AP Statistics)	Knowing that the normal distribution is a symmetric, bell-shaped distribution where values near the mean are most likely (you need this for the normal approximation card) Knowing that the order inside \(N(\mu,\ \sigma^2)\) is "mean, variance"

How to calculate it in Excel

Copy the whole table below and paste it into cell A1 in Excel. It works as is.

Table to find the probability of exactly k successes

Number of trials n	4
Probability of success p	0.5
Number of successes k	2
Probability of exactly k P(X=k)	=BINOM.DIST(B3,B1,B2,FALSE)

Table to find the cumulative probabilities (at most k, at least k)

Number of trials n	4
Probability of success p	0.5
Number of successes k	2
Probability of at most k P(X≤k)	=BINOM.DIST(B3,B1,B2,TRUE)
Probability of at least k P(X≥k)	=1-BINOM.DIST(B3-1,B1,B2,TRUE)

Table to find the mean, variance and standard deviation

Number of trials n	30
Probability of success p	=1/6
Mean E(X) = np	=B1*B2
Variance V(X) = np(1-p)	=B1B2(1-B2)
Standard deviation σ	=SQRT(B4)

Table to check the rule of thumb for the normal approximation

Number of trials n	20
Probability of success p	0.5
np	=B1*B2
n(1-p)	=B1*(1-B2)
Rule of thumb met?	=AND(B3>=5,B4>=5)

After pasting, the upper rows (n, p and k) are your inputs and the lower rows are calculated automatically.
BINOM.DIST is the binomial distribution function. Its arguments are, in order, "number of successes, number of trials, probability of success, cumulative or not". FALSE at the end gives the exact probability, and TRUE gives the cumulative probability of that count or fewer.
The first table is the example of exactly 2 heads in 4 coin flips, and the answer is 0.375 (= 3/8).
Both answers in the second table are 0.6875. "At least k" is found by subtracting "at most k−1" from 1 (this formula gives an error if k is 0, but the probability of 0 or more is always 1).
The third table is the number of 1s in 30 rolls of a die: mean 5, variance 4.1667 and standard deviation 2.0412.
The fourth table shows TRUE when both np and n(1−p) are 5 or more (the rule of thumb for the normal approximation is met), and FALSE otherwise.

How to calculate it in Google Sheets

Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.

Table to find the probability of exactly k successes

Number of trials n	4
Probability of success p	0.5
Number of successes k	2
Probability of exactly k P(X=k)	=BINOMDIST(B3,B1,B2,FALSE)

Table to find the cumulative probabilities (at most k, at least k)

Number of trials n	4
Probability of success p	0.5
Number of successes k	2
Probability of at most k P(X≤k)	=BINOMDIST(B3,B1,B2,TRUE)
Probability of at least k P(X≥k)	=1-BINOMDIST(B3-1,B1,B2,TRUE)

Table to find the mean, variance and standard deviation

Number of trials n	30
Probability of success p	=1/6
Mean E(X) = np	=B1*B2
Variance V(X) = np(1-p)	=B1B2(1-B2)
Standard deviation σ	=SQRT(B4)

Table to check the rule of thumb for the normal approximation

Number of trials n	20
Probability of success p	0.5
np	=B1*B2
n(1-p)	=B1*(1-B2)
Rule of thumb met?	=AND(B3>=5,B4>=5)

In Google Sheets, the function is called BINOMDIST (the arguments are in the same order as Excel's BINOM.DIST: "number of successes, number of trials, probability of success, cumulative or not"). Copy the whole table, paste it into cell A1, and replace n, p and k with your own numbers.
As in Excel, the "at least k" formula gives an error if k is 0 (the probability of 0 or more is always 1).

How to calculate it in Python

from fractions import Fraction
from math import comb, sqrt

trial_count = 4                  # number of trials n
success_prob = Fraction(1, 2)    # probability of success p (a decimal such as 0.3 also works)
success_count = 2                # number of successes k

failure_prob = 1 - success_prob

# probability of exactly k successes P(X=k) = nCk × p^k × (1−p)^(n−k)
exact_prob = (comb(trial_count, success_count)
              * success_prob ** success_count
              * failure_prob ** (trial_count - success_count))
print(f"Probability of exactly {success_count}: {exact_prob} = {float(exact_prob)}")

# cumulative probabilities (at most k, at least k)
prob_le = sum(comb(trial_count, i) * success_prob ** i * failure_prob ** (trial_count - i)
              for i in range(success_count + 1))
prob_ge = 1 - prob_le + exact_prob
print(f"Probability of at most {success_count}: {prob_le} = {float(prob_le)}")
print(f"Probability of at least {success_count}: {prob_ge} = {float(prob_ge)}")

# mean, variance and standard deviation
mean = trial_count * success_prob
variance = mean * failure_prob
print(f"Mean: {mean}  Variance: {variance}  Standard deviation: {sqrt(variance)}")

Runs with the standard library only. The fractions module keeps the calculation as exact fractions, with no decimal rounding errors (comb is the number of combinations nCk). This example is exactly 2 heads in 4 coin flips. Running it shows the probability 3/8 = 0.375, 11/16 = 0.6875 for both at most 2 and at least 2, and mean 2, variance 1 and standard deviation 1.0. Change the three numbers n, p and k and run it.

How to write it in LaTeX and other math languages (copy and paste)

Binomial probability (probability of exactly \(k\) successes)

P(X = k) = nCk pᵏ (1 − p)ⁿ⁻ᵏ

P(X = k) = \binom{n}{k} p^{k} (1 - p)^{n - k}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>P</mi><mo>(</mo><mi>X</mi><mo>=</mo><mi>k</mi><mo>)</mo>
    <mo>=</mo>
    <mrow>
      <mo>(</mo>
      <mfrac linethickness="0"><mi>n</mi><mi>k</mi></mfrac>
      <mo>)</mo>
    </mrow>
    <msup><mi>p</mi><mi>k</mi></msup>
    <msup>
      <mrow><mo>(</mo><mn>1</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow>
      <mrow><mi>n</mi><mo>&#x2212;</mo><mi>k</mi></mrow>
    </msup>
  </mrow>
</math>

P(X = k) = C(n,k) p^k (1-p)^(n-k)

PDF[BinomialDistribution[n, p], k]

binomial(n, k)*p^k*(1-p)^(n-k);

binopdf(k, n, p)

P(X=k) = C(n,k) p^k (1-p)^(n-k)

Cumulative probability (at most \(k\) and at least \(k\) successes)

P(X ≤ k) = Σ[i=0→k] nCi pⁱ (1 − p)ⁿ⁻ⁱ

P(X \le k) = \sum_{i=0}^{k} \binom{n}{i} p^{i} (1 - p)^{n - i}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>P</mi><mo>(</mo><mi>X</mi><mo>&#x2264;</mo><mi>k</mi><mo>)</mo>
    <mo>=</mo>
    <munderover>
      <mo>&#x2211;</mo>
      <mrow><mi>i</mi><mo>=</mo><mn>0</mn></mrow>
      <mi>k</mi>
    </munderover>
    <mrow>
      <mo>(</mo>
      <mfrac linethickness="0"><mi>n</mi><mi>i</mi></mfrac>
      <mo>)</mo>
    </mrow>
    <msup><mi>p</mi><mi>i</mi></msup>
    <msup>
      <mrow><mo>(</mo><mn>1</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow>
      <mrow><mi>n</mi><mo>&#x2212;</mo><mi>i</mi></mrow>
    </msup>
  </mrow>
</math>

P(X <= k) = sum_(i=0)^k C(n,i) p^i (1-p)^(n-i)

CDF[BinomialDistribution[n, p], k]

add(binomial(n, i)*p^i*(1-p)^(n-i), i = 0..k);

binocdf(k, n, p)

P(X≤k) = Σ_(i=0)^k C(n,i) p^i (1-p)^(n-i)

Mean, variance and standard deviation of the binomial distribution

E(X) = np,  V(X) = np(1 − p),  σ = √(np(1 − p))

E(X) = np, \quad V(X) = np(1 - p), \quad \sigma = \sqrt{np(1 - p)}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>E</mi><mo>(</mo><mi>X</mi><mo>)</mo><mo>=</mo><mi>n</mi><mi>p</mi>
    <mo>,</mo>
    <mi>V</mi><mo>(</mo><mi>X</mi><mo>)</mo><mo>=</mo>
    <mi>n</mi><mi>p</mi>
    <mrow><mo>(</mo><mn>1</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow>
    <mo>,</mo>
    <mi>&#x3C3;</mi><mo>=</mo>
    <msqrt><mi>n</mi><mi>p</mi><mrow><mo>(</mo><mn>1</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow></msqrt>
  </mrow>
</math>

E(X) = np, V(X) = np(1-p), sigma = sqrt(np(1-p))

{Mean[BinomialDistribution[n, p]], Variance[BinomialDistribution[n, p]], StandardDeviation[BinomialDistribution[n, p]]}

n*p; n*p*(1-p); sqrt(n*p*(1-p));

[m, v] = binostat(n, p); s = sqrt(v);

E(X) = np, V(X) = np(1-p), σ = √(np(1-p))

Normal approximation (rule of thumb: \(np \ge 5\) and \(n(1-p) \ge 5\))

B(n, p) ≈ N(np, np(1 − p))

B(n,\ p) \approx N(np,\ np(1 - p))

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>B</mi><mo>(</mo><mi>n</mi><mo>,</mo><mi>p</mi><mo>)</mo>
    <mo>&#x2248;</mo>
    <mi>N</mi><mo>(</mo><mi>n</mi><mi>p</mi><mo>,</mo>
    <mi>n</mi><mi>p</mi>
    <mrow><mo>(</mo><mn>1</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow>
    <mo>)</mo>
  </mrow>
</math>

B(n, p) ~~ N(np, np(1-p))

NormalDistribution[n p, Sqrt[n p (1 - p)]]

Normal(n*p, sqrt(n*p*(1-p)));

pd = makedist('Normal', 'mu', n*p, 'sigma', sqrt(n*p*(1-p)));

B(n, p) ≈ N(np, np(1-p))

How to have ChatGPT do the calculation

You are a calculation assistant for math (probability and statistics). Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing).

You enter a prize draw 10 times, and each entry has a 1/6 chance of winning. For the binomial distribution of the number of wins, find:
1. The probability of winning exactly 2 times, P(X=2) (as a fraction in lowest terms and as a decimal)
2. The probability of at most 2 wins, P(X≤2), and of at least 2 wins, P(X≥2)
3. The mean E(X), variance V(X) and standard deviation σ
4. Whether the rule of thumb for the normal approximation, np ≥ 5 and n(1−p) ≥ 5, is met

In Python, calculate exactly with fractions and math.comb from the standard library, and show the formulas you used and the numbers from the execution result.

How to Use

1

Enter your numbers

Type the numbers you want to calculate with into the input fields
2

Calculate

Press the "Calculate" button
3

Check the result

The result appears on the spot. The same page also explains the idea behind the calculation and the formula