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(3) $ Probability of exactly 3 successes: 0.336415625
Solution:
$P(3)$ Probability of exactly 3 successes
If using a calculator, you can enter $ \text{trials} = 5 $, $ p = 0.65 $, and $ X = 3 $ 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 = 5 $, $ p = 0.65 $, and $ X = 3 $.
$$ P(3) = \frac{5!}{3!(5-3)!} \cdot 0.65^3 \cdot (1-0.65)^{5-3} $$
Evaluating the expression, we have
$$ P(3) = 0.336415625 $$
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 5 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 5 trials is:
P(0) = 0.0052521875
P(1) = 0.0487703125
P(2) = 0.181146875
P(3) = 0.336415625
P(4) = 0.3123859375
P(5) = 0.1160290625
