Binomial distribution

A binomial distribution is a type of discrete probability distribution that results from a trial in which there are only two mutually exclusive outcomes. These outcomes are labeled as a success or a failure.

The binomial distribution is commonly used to determine the probability of a certain number of successes in n trials, where the probability of success on a single trial does not change. One such example is the flip of a coin. The two possible outcomes of a coin flip are heads or tails, and the probability of heads or tails occurring is the same for each trial (50% for a fair coin). Thus, the flip of a coin meets the conditions of a binomial distribution, and probabilities such as the probability of 3 heads occurring in 5 flips of a coin can be determined.

In summary, for an event to exhibit a binomial distribution, the following conditions must be met:

The probability mass function (pmf) of a binomial distribution is:

, where

p is the probability of success
q = 1 - p is the probability of failure
n is the number of trials
x is the number of successes from the n trials
, referred to as the binomial coefficient

Thus, if X is a discrete random variable that exhibits a binomial distribution, the probability, P(X = x) = f(x), where f(x) is defined as above.


A fair 6-sided die is rolled 4 times. Using a binomial distribution,

  1. find the probability of rolling a 6 three times.
  2. find the probability of rolling a 6 at least two times.

i. Rolling a 6 is a success in this case. Rolling anything else is a failure. Thus, p = 1/6 and q = 5/6. There are 4 trials, and we want to determine the probability of 3 successes, or x = 3. Thus:


Thus, there is about a 1.5% chance of rolling a 6 three times in 4 trials.

ii. Rolling a 6 at least two times means that we can roll a 6 twice, three times, or four times. The probability of rolling a 6 at least two times is the sum of the probabilities of rolling a 6 two, three, or four times.


Thus, there is a 13% probability of rolling a 6 at least 2 times in 4 rolls.

Mean and variance

The mean (or expected value), μ, and variance, σ2, of a binomially distributed random variable can be found using the following formulas:


200 ping pong balls of various colors are placed in a bin. 25 of the ping pong balls are red.

  1. Find the probability of selecting a red ball on any single draw given that the selected ball is replaced prior to each subsequent selection.
  2. Find the expected value and variance given that 50 balls are randomly selected (with replacement) from the bin.
  3. Find the probability of selecting 6 red balls from the 50 balls that were randomly selected from the bin.

i. There are 25 red balls and a total of 200 balls, so:

p = 25/200 = 1/8

ii. n = 50 and p = 1/8, so:

μ = np = 50 × 1/8 = 6.25

Thus, we would expect around 6 of the selected ping pong balls to be red. The variance is computed as:

σ2 = 50 × 1/8 × 7/8 = 5.47

iii. The experiment meets the criteria of a binomial distribution, so:

Thus, while we would expect to choose 6 red balls from the 50, the probability of doing so is only 17%. The binomial distribution for this experiment is shown in the figure below: