The Binomial Distribution Calculator will construct a complete binomial distribution and find the mean and standard deviation. A binomial probability is the chance of an event occurring given a number of trials and number of successes. Enter p, probability, and the number of trials, then the calculator will find all the binomial probabilities from 0 to # trials. If you want to find the binomial probability for a specific success only, then have a look at the Binomial Probability Calculator.

Enter a value for p and trials. Probability, p, is the probability of an event occurring on a single trial. Trials is the number of times you’ll conduct the experiment. The calculator will display the binomial distribution, its mean and its standard deviation. Then, it will also give you a step by step solution for how to find all the binomial probabilities.

### Answer:

Mean of the Binomial Distribution, $\mu$: 3.9

Standard Deviation of the Binomial Distribution, $\sigma$: 1.1683321445548

P(0) = 0.001838265625

P(1) = 0.02048353125

P(2) = 0.095102109375

P(3) = 0.2354909375

P(4) = 0.328005234375

P(5) = 0.24366103125

P(6) = 0.075418890625

### Solution:

The mean of the binomial distribution is interpreted as the mean number of successes for the distribution. To find the mean, use the formula $$ \mu = n \cdot p $$ where n is the number of trials and p is the probability of success on a single trial. Substituting values for this problem, we have $$ \mu = 6 \cdot 0.65 $$ Multiplying the expression we have $$ \mu = 3.9 $$

The standard deviation of the binomial distribution is interpreted as the standard deviation of the number of successes for the distribution. To find the standard deviation, use the formula $$ \sigma = \sqrt{n \cdot p \cdot (1 - p)} $$ where n is the umber of trials and p is the probability of success on a single trial. Substituting values fo this problem, we have $$ \sigma = \sqrt{6 \cdot 0.65 \cdot (1 - 0.65)} $$ Evaluating the expression on the right, we have $$ \sigma = \sqrt{1.365)} $$
$$ \sigma = 1.1683321445548 $$

To complete a binomial distribution table, first identify all of the possible values of X. Since there are 6 trials, the values of X range from X = 0 to X = 6.

Next, find each individual binomial probability for each value of X. In this problem, we will be finding 7 probabilities. The sum of all these probabilities will be 1.

### $P(0)$ Probability of exactly 0 successes

If using a calculator, you can enter $ \text{trials} = 6 $, $ p = 0.65 $, and $ X = 0 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 6 $, $ p = 0.65 $, and $ X = 0 $.
$$ P(0) = \frac{6!}{0!(6-0)!} \cdot 0.65^0 \cdot (1-0.65)^{6-0} $$
Remember, 0! is 1. Evaluating the expression, we have
$$ P(0) = 0.001838265625 $$

### $P(1)$ Probability of exactly 1 successes

If using a calculator, you can enter $ \text{trials} = 6 $, $ p = 0.65 $, and $ X = 1 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 6 $, $ p = 0.65 $, and $ X = 1 $.
$$ P(1) = \frac{6!}{1!(6-1)!} \cdot 0.65^1 \cdot (1-0.65)^{6-1} $$
Evaluating the expression, we have
$$ P(1) = 0.02048353125 $$

### $P(2)$ Probability of exactly 2 successes

If using a calculator, you can enter $ \text{trials} = 6 $, $ p = 0.65 $, and $ X = 2 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 6 $, $ p = 0.65 $, and $ X = 2 $.
$$ P(2) = \frac{6!}{2!(6-2)!} \cdot 0.65^2 \cdot (1-0.65)^{6-2} $$
Evaluating the expression, we have
$$ P(2) = 0.095102109375 $$

### $P(3)$ Probability of exactly 3 successes

If using a calculator, you can enter $ \text{trials} = 6 $, $ p = 0.65 $, and $ X = 3 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 6 $, $ p = 0.65 $, and $ X = 3 $.
$$ P(3) = \frac{6!}{3!(6-3)!} \cdot 0.65^3 \cdot (1-0.65)^{6-3} $$
Evaluating the expression, we have
$$ P(3) = 0.2354909375 $$

### $P(4)$ Probability of exactly 4 successes

If using a calculator, you can enter $ \text{trials} = 6 $, $ p = 0.65 $, and $ X = 4 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 6 $, $ p = 0.65 $, and $ X = 4 $.
$$ P(4) = \frac{6!}{4!(6-4)!} \cdot 0.65^4 \cdot (1-0.65)^{6-4} $$
Evaluating the expression, we have
$$ P(4) = 0.328005234375 $$

### $P(5)$ Probability of exactly 5 successes

If using a calculator, you can enter $ \text{trials} = 6 $, $ p = 0.65 $, and $ X = 5 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 6 $, $ p = 0.65 $, and $ X = 5 $.
$$ P(5) = \frac{6!}{5!(6-5)!} \cdot 0.65^5 \cdot (1-0.65)^{6-5} $$
Evaluating the expression, we have
$$ P(5) = 0.24366103125 $$

### $P(6)$ Probability of exactly 6 successes

If using a calculator, you can enter $ \text{trials} = 6 $, $ p = 0.65 $, and $ X = 6 $ into a binomial probability distribution function (binomPDF). If doing this by hand, apply the binomial probability formula:
$$ P(X) = \binom{n}{X} \cdot p^X \cdot (1-p)^{n-X} $$
The binomial coefficient, $ \binom{n}{X} $ is defined by
$$ \binom{n}{X} = \frac{n!}{X!(n-X)!} $$
The full binomial probability formula with the binomial coefficient is
$$ P(X) = \frac{n!}{X!(n-X)!} \cdot p^X \cdot (1-p)^{n-X} $$
where $n$ is the number of trials, $p$ is the probability of success on a single trial, and $X$ is the number of successes. Substituting in values for this problem, $ n = 6 $, $ p = 0.65 $, and $ X = 6 $.
$$ P(6) = \frac{6!}{6!(6-6)!} \cdot 0.65^6 \cdot (1-0.65)^{6-6} $$
Evaluating the expression, we have
$$ P(6) = 0.075418890625 $$