The binomial probability calculator will calculate a probability based on the binomial probability formula. You will also get a step by step solution to follow. Enter the trials, probability, successes, and probability type.

- Trials, n, must be a whole number greater than 0. This is the number of times the event will occur.
- Probability, p, must be a decimal between 0 and 1 and represents the probability of success on a single trial.
- Successes, X, must be a number less than or equal to the number of trials. This number represents the number of desired positive outcomes for the experiment.
- The probability type can either be a single success (“exactly”), or an accumulation of successes (“less than”, “at most”, “more than”, “at least”).

### Binomial Probability Calculator

### Answer:

$ P(2) $ Probability of exactly 2 successes: 0.095102109375

### Solution:

**$P(2)$ 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 (PDF). 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
$$ P(2) = 0.095102109375 $$

### Complete Binomial Distribution Table

If we apply the binomial probability formula, or a calculator's binomial probability distribution (PDF) function, to all possible values of X for 6 trials, we can construct a complete binomial distribution table. The sum of the probabilities in this table will always be 1. The complete binomial distribution table for this problem, with p = 0.65 and 6 trials is:

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