# Binomial Distribution Calculator with Step by Step Solution

## Binomial distribution Calculator with Step by Step

Binomial distribution is one of the most important discrete distribution in statistics. In this tutorial we will discuss about how to solve numerical examples based on binomial distribution. Upon successful completion of this tutorial, you will be able to understand how to calculate binomial probabilities.

For the theoretical proof of various properties of binomial distribution, read Binomial Distribution

## Binomial Distribution Calculator

Binomial distribution calculator is used to find the probability and cumulative probabilities for binomial random variable given the number of trials ($n$) and probability of success ($p$).

## Calculator

Binomial Distribution Calculator
Number of trials ($n$):
Probability of success ($p$):
Number of success (x):
Binomial Distribution Calculator Result
Mean : $np$
Variance : $np(1-p)$
Probability : P(X = x)
Cumulative Probability : P(X ≤ x)
Cumulative Probability : P(X < x)
Cumulative Probability : P(X ≥ x)
Cumulative Probability : P(X > x)

### How to use Binomial Distribution Calculator with step by step?

Step 1 - Enter the number of trials $n$

Step 2 - Enter the probability of success $p$

Step 3 - Enter the number of successes $x$

Step 4 - Click on "Calculate" button to get Binomial probabilities

Step 5 - Gives output for mean of binomial distribution

Step 6 - Gives the output for variance of binomial distribution

Step 7 - Calculate probability of $X=x$ and the various cumulative probabilities for binomial distribution

## Binomial Distribution Theory and Formula

A discrete random variable $X$ is said to have Binomial distribution with parameter $n$ and $p$ if its probability mass function is

 \begin{aligned} P(X=x)&= \begin{cases} & \binom{n}{x} p^x q^{n-x}, & x=0,1,2,\cdots, n \\ & & 0 < p < 1, q=1-p \\ & 0, & Otherwise. \end{cases} \end{aligned}

## Mean and variance of Binomial distribution

The mean or expected value of binomial random variable $X$ is $\mu=E(X) = np$.

The variance of binomial random variable $X$ is $\sigma^2=V(X) = np(1-p)$. Thus the standard deviation of binomial random variable is $\sigma = \sqrt{np(1-p)}$.

Below are the few numerical problems solved using binomial distribution calculator with step by step solution.

## Example -1 Binomial Distribution Calculator

Suppose that a short quiz consists of 6 multiple choice questions.Each question has four possible answers of which ony one in correct. A student guesses on every question. Find the probability that a student will answer

a. five or more quetions correctly,
b. all questions correctly,
c. at most 1 questions correctly,
d. between 4 and 5 (inclusive) questions correctly.

### Solution

Let $X$ be the number of questions guessed correctly out of $6$ questions. Let $p$ be the probability of correct guess.

The random variable $X$ follows a Binomial distribution with parameter $n=6$ and $p=0.25$. That is, $X\sim B(6, 0.25)$.

The probability mass function (pmf) of binomial distribution $X$ is

 \begin{aligned} P(X=x) &= \binom{6}{x} (0.25)^x (1-0.25)^{6-x}, \; x=0,1,\cdots, 6\\ &=\binom{6}{x} (0.25)^x (0.75)^{6-x}, \; x=0,1,\cdots,6 \end{aligned}

a. The probability that a student will answer $5$ or more questions correctly is

 \begin{aligned} P(X\geq 5) &= P(X=5)+P(X=6)\\ &= \binom{6}{5} (0.25)^{5} (0.75)^{6-5}+\binom{6}{6} (0.25)^{6} (0.75)^{6-6}\\ &= 0.0044+0.0002\\ & = 0.0046 \end{aligned}

b. The probability that a student will answer all questions correctly is

 \begin{aligned} P(X= 6) & =P(6)\\ &= \binom{6}{6} (0.25)^{6} (0.75)^{6-6}\\ & = 0.0002 \end{aligned}

c. The probability that a student will answer at most $1$ questions correctly is

 \begin{aligned} P(X\leq 1) &= P(X=0)+P(X=1)\\ &= \binom{6}{0} (0.25)^{0} (0.75)^{6-0}+\binom{6}{1} (0.25)^{1} (0.75)^{6-1}\\ &=0.178+0.356\\ &= 0.5339 \end{aligned}

d. The probability that a student will answer between $4$ and $5$ questions correctly is

 \begin{aligned} P(4\leq X\leq 5) & =P(X=4)+P(X=5)\\ &= \binom{6}{4} (0.25)^{4} (0.75)^{6-4}+\binom{6}{5} (0.25)^{5} (0.75)^{6-5}\\ & = 0.033+0.0044\\ &= 0.0374 \end{aligned}

## Example -2 Binomial Distribution Calculator

35% of the adults says cashews are their favorite kind of nuts. If you randomly select 10 adults and ask each adult to name his or her favorite nut, compute the probability that the number of adults who say cashews are their favorite nut is

a. exactly 4,
b. less than 3,
c. at least 3,
d. at the most 2.

### Solution

Let $X$ be the number of adults out of $10$ who say cashew is their favorite nut.

Let $p$ be the probability that an adults favorite nut is cashew.

Given that $p=0.35$ and $n =10$. Here $X$ follows a Binomial distribution. That is, $X\sim B(10, 0.35)$.

The probability mass function (pmf) of binomial distribution $X$ is

 \begin{aligned} P(X=x) &= \binom{10}{x} (0.35)^x (1-0.35)^{10-x},\\ &\quad \; x=0,1,\cdots, 10. \end{aligned}

The mean of $X$ is

 \begin{aligned} \mu =E(X) &= n*p\\ &= 10*0.35\\ &= 3.5 \end{aligned}

and the standard deviation of $X$ is

 \begin{aligned} \sigma=\sqrt{V(X)} &=\sqrt{n*p*(1-p)}\\ &=\sqrt{10*0.35* (1- 0.35)}\\ &= 1.5083 \end{aligned}

a. The probability that exactly 4 adults say cashews are their favorite nut is

 \begin{aligned} P(X= 4) & =P(4)\\ &= \binom{10}{4} (0.35)^{4} (1-0.35)^{10-4}\\ &= 210\times 0.015\times 0.0754\\ & = 0.2377 \end{aligned}

b. The probability that less than 3 adults say cashews are their favorite nut is

 \begin{aligned} P(X< 3) & =P(X\leq 2)\\ &= P(0)+P(1)\\ & = 0.0135+0.0725\\ & = 0.2616 \end{aligned}

c. The probability that at least 3 adults say cashews are their favorite nut is

 \begin{aligned} P(X\geq 3) & =1-P(X\leq 2)\\ &= 1-(P(0)+P(1))\\ & = 1-\big(0.0135+0.0725 \big)\\ & = 0.7384 \end{aligned}

d. The probability that at most 2 adults say cashews are their favorite nut is

 \begin{aligned} P(X\leq 2) & =\sum_{x=0}^{2} P(x)\\ & =P(0) + P(1) + P(2)\\ & = 0.0135+0.0725+0.1757\\ &= 0.2616 \end{aligned}

## Example -3 Binomial Distribution Calculator

Working Women and Computer Use : It is reported that 72% of working women use computers at work.
Choose 5 working women at random. Find

a. The probability that at least 1 does not use a
computer at work.
b. The probability that all 5 use a computer in their jobs.

### Solution

Let $X :$ the number of womens who use computer at work out of $5$.

Here $n = 5$ and $p=0.72$. The probability distribution of $X$ is Binomial distribution. That is $X\sim B(5,0.72)$.

The probability mass function (pmf) of binomial distribution $X$ is

 \begin{aligned} P(X=x) &= \binom{5}{x} (0.72)^x (1-0.72)^{5-x},\\ &\quad \; x=0,1,\cdots, 5 \end{aligned}

a. Probability that at least 1 doesn't use a computer at work is same as the probability that none use computer at work.

 \begin{aligned} P(X= 0) & =\binom{5}{0} (0.72)^{0} (1-0.72)^{5-0}\\ & = 0.0017\\ \end{aligned}

b. Probability that all 5 use computer at work is

 \begin{aligned} P(X= 5) & =\binom{5}{5} (0.72)^{5} (1-0.72)^{5-5}\\ & = 0.1935 \end{aligned}

## Example -4 Binomial Distribution

A new surgical procedure is said to be successful 75% of the time. If 4 patients undergo the surgical procedure, what is the probability that

a. all four operations are successful?
b. Exactly three operations are successful?
c. Less than two are successful?

### Solution

Here $X$ denote the number successful operations out of 4.

Let $p$ be the probability that operation using new surgical procedure is successful.

Given that $p=0.75$ and $n =4$. Thus$X\sim B(4, 0.75)$.

The probability mass function (pmf) of binomial distribution $X$ is
 \begin{aligned} P(X=x) &= \binom{4}{x} (0.75)^x (1-0.75)^{4-x}, \\ &\quad \; x=0,1,\cdots, 4 \end{aligned}

a. The probability that all four operations are successful is

 \begin{aligned} P(X=4) & =\binom{4}{4} (0.75)^{4} (1-0.75)^{4-4}\\ &=0.3164 \end{aligned}

b. The probability that exactly three operations are successful is

 \begin{aligned} P(X=3) & =\binom{4}{3} (0.75)^{3} (1-0.75)^{4-3}\\ &=0.4219 \end{aligned}

c. The probability that less than two operations are successful is

 \begin{aligned} P(X< 2) &=P(X\leq 1)\\ &=P(X=0)+P(X=1)\\ &= 0.0039+0.0469\\ &= 0.0508 \end{aligned}

## Example -5 Binomial Distribution Calculator

According to a recent survey, outside of their own family members, 26% of adult Americans have no close friend to confide in. If this is the prevailing probability today, find the probability that in a random sample of n = 5 adults

a. two or more have no close friend.
b. at most two have no close friend.
c. Find the expected number of persons who have no close friend.

### Solution

Here $X$ denote the number of adult Americans who have no close friend to confide.

$p$ be the probability that adult American who have no close friend to confide.

Given that $p=0.26$ and $n =5$. Thus$X\sim B(5, 0.26)$.

The probability mass function (pmf) of binomial distribution $X$ is

 \begin{aligned} P(X=x) &= \binom{5}{x} (0.26)^x (1-0.26)^{5-x},\\ &\quad\; x=0,1,\cdots, 5 \end{aligned}

a. The probability that 2 or more have no close friend is

 \begin{aligned} P(X\geq 2) & =1-P(X\leq 1)\\ &= 1-\sum_{x=0}^{1} P(x)\\ & = 1-0.6117 \\ &= 1- (0.2219+0.3898)\\ & = 0.3883 \\ \end{aligned}

b. The probability that at most two have no close friend is

 \begin{aligned} P(X\leq 2) & =\sum_{x=0}^{2} P(x)\\ & =\sum_{x=0}^{2}\binom{5}{x}(0.26)^x(1-0.26)^{5-x}\\ & = 0.2219+0.3898+0.2739\\ &=0.8857. \end{aligned}

c. The expected number of persons who have no close friend is $np = 5 0.26 = 1.3$.

For the theoretical proof of various properties of binomial distribution, read Binomial Distribution

Let me know in the comments if you have any questions on Binomial Distribution calculator with step by step solution on solved examples and your thought on this article.

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.