# Bernoulli Distribution – Explanation & Examples

The definition of the Bernoulli distribution is:

“The Bernoulli distribution is a discrete probability distribution that describes the probability of a random variable with only two outcomes.”

In this topic, we will discuss the Bernoulli distribution from the following aspects:

1. What is a Bernoulli distribution?
2. When to use Bernoulli distribution?
3. Bernoulli distribution formula.
4. Practice questions.

## 1. What is a Bernoulli distribution?

The Bernoulli distribution is a discrete probability distribution that describes the probability of a random variable with only two outcomes.

In the random process called a Bernoulli trial, the random variable can take one outcome, called a success, with a probability p, or take another outcome, called failure, with a probability q = 1-p.

The success outcome is denoted as 1 and the failure outcome is denoted as 0.

The Bernoulli distribution is a special case of the binomial distribution where a single trial is conducted and the binomial distribution is the sum of repeated Bernoulli trials.

The Bernoulli distribution was named after the Swiss mathematician Jacob Bernoulli.

### – Example 1

Tossing a coin can result in only two possible outcomes (head or tail). We call one of these outcomes (head) a success and the other (tail), a failure.

The probability of success (p) or head is 0.5 for a fair coin. The probability of failure (q) or tail = 1-p = 1-0.5 = 0.5.

If we denote head as 1 and tail as 0, we can plot this Bernoulli distribution as follows:

We have two outcomes:

• Tail or 0 with a probability of 0.5.
• Head or 1 with a probability of 0.5 also.

This is an example of a probability mass function where we have the probability for each outcome.

### – Example 2

We have an unfair coin where the probability of success (p) or head is 0.8 and the probability of failure (q) or tail = 1-p = 1-0.8 = 0.2.

If we denote head as 1 and tail as 0, we can plot this Bernoulli distribution as follows:

We have two outcomes:

• Tail or 0 with a probability of 0.2.
• Head or 1 with a probability of 0.8.

### – Example 3

The prevalence of a certain disease in the general population is 10%.

If we randomly select a person from this population, we can have only two possible outcomes (diseased or healthy person). We call one of these outcomes (diseased person) success and the other (healthy person), a failure.

The probability of success (p) or diseased person is 10% or 0.1. So, the probability of failure (q) or healthy person = 1-p = 1-0.1 = 0.9.

If we denote diseased person as 1 and healthy person as 0, we can plot this Bernoulli distribution as follows:

We have two outcomes:

• A healthy person or 0 with a probability of 0.9.
• A diseased person or 1 with a probability of 0.1.

### – Example 4

In the above example of disease prevalence of 10%, if We are interested in healthy persons and call the healthy person a success and the diseased person, a failure.

The probability of success (p) or healthy person is 90% or 0.9. So, the probability of failure (q) or diseased person = 1-p = 1-0.9 = 0.1.

If we denote a healthy person as 1 and diseased person as 0, we can plot this Bernoulli distribution as follows:

We have two outcomes:

• A healthy person or 1 with a probability of 0.9.
• A diseased person or 0 with a probability of 0.1.

## 2. When to use Bernoulli distribution?

For a random variable to be described by the Bernoulli distribution:

1. The random variable can take only one of two possible outcomes. We call one of these outcomes a success and the other, a failure.
2. The probability of success, denoted by p, is the same in every Bernoulli trial.
3. The trials are independent, meaning that the outcome in one trial does not affect the outcome in other trials.

We can determine the Bernoulli distribution from the results of different Bernoulli trials.

### – Example 1

You are tossing a coin. The random variable equals to 1 if you get a head and 0 if you get a tail.

You tossed the coin 100 times and get the following results:

0 1 0 1 1 0 1 1 1 0 1 0 1 1 0 1 0 0 0 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 1 1 0 1 0 0 0 1 0 1 1 1 0 1 1 1 0 0 0 0 1 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 0 0 1 0 0 1.

What is the Bernoulli distribution for this coin?

You can use that data to estimate the probability mass function (or the probability distribution) for tossing this coin.

1. We construct a frequency table for each outcome.

 Outcome frequency 0 53 1 47

2. Add another column for the probability of each outcome.

Probability = frequency/total number of data = frequency/100.

 Outcome frequency probability 0 53 0.53 1 47 0.47

The probabilities are >= 0 and sum to 1.

This is a likely fair coin where the probability of heads nearly equals the probability of tails = 0.5.

We do not get exactly 50 heads and 50 tails due to randomness in the process but we get a good approximation to the probability of the fair coin = 0.5.

3. Use the table to plot the Bernoulli distribution for that coin:

We have two outcomes:

• Head or 1 with a probability of 0.47.
• Tail or 0 with a probability of 0.53.

### – Example 2

You screened 50 individuals from a certain population for the presence of hypertension and get the following results:

 ID condition 1 normotensive 2 normotensive 3 normotensive 4 normotensive 5 normotensive 6 normotensive 7 normotensive 8 normotensive 9 normotensive 10 normotensive 11 hypertensive 12 normotensive 13 normotensive 14 normotensive 15 normotensive 16 normotensive 17 normotensive 18 normotensive 19 normotensive 20 hypertensive 21 normotensive 22 normotensive 23 normotensive 24 hypertensive 25 normotensive 26 normotensive 27 normotensive 28 normotensive 29 normotensive 30 normotensive 31 hypertensive 32 normotensive 33 normotensive 34 normotensive 35 normotensive 36 normotensive 37 normotensive 38 normotensive 39 normotensive 40 normotensive 41 normotensive 42 normotensive 43 normotensive 44 normotensive 45 normotensive 46 normotensive 47 normotensive 48 normotensive 49 normotensive 50 normotensive

What is the estimated Bernoulli distribution for hypertension in this population?

1. We construct a frequency table for each outcome.

 Outcome frequency hypertensive 4 normotensive 46

2. Add another column for the probability of each outcome. As we are interested in hypertension, so we denote hypertensive persons as 1 and normotensive persons as 0.

Probability = frequency/total number of data = frequency/50.

 outcome frequency probability 1 4 0.08 0 46 0.92

The probabilities are >= 0 and sum to 1.

3. Use the table to plot the Bernoulli distribution for hypertension: