The Bernoulli Distribution
A simplistic overview of the concept of probability in terms of independent trials, providing context to the Binomial Distribution.
Edu Level: Unit1
Date: Apr 1 2026 - 12:45 PM
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Bernoulli Distribution
In simple terms, a Bernoulli trial is a single experiment with two fixed outcomes, either success or failure. It is the fundamental basis for repetitive distributions such as the binomial, geometric and many other distributions you may encounter in your studies of statistics and probability. It is essentially the building block of a multitude of discrete distributions.
A Bernoulli trial with a random variable $X$ where the probability can be expressed as $P(X=x)$, can be defined in terms of its probability of success, $p$ and probability of failure for a single experiment as defined below:
$$X \sim \mathrm{Bernoulli}(p), \forall x \in [0,1]$$
$$
P(X=x) = \begin{cases}
p & \text{if } x = 1
(1-p) & \text{if } x = 0
\end{cases}
$$
Inherently, it is crucial to understand that many critical processes in applied math are repeated failures/successes in experiments, whether this be the quality of items in a production line or answering quiz questions correctly or incorrectly. These general trends which can be measured make up the basis of more complex probability distributions.
There are other noteworthy characteristics of the Bernoulli distribution, as aforementioned, for a random variable X which follows a Bernoulli distribution:
- $$E(X)=p$$
- $$Var(X) = p(1-p)$$
Questions
- A farmer is carrying out a single experiment where he notes that the probability that a carrot rots is 70%. Define the piecewise function, $P(X=x)$ for the single experiment.
- A company, JT Sauces Ltd., notes an identical, independent and recurring trend for the first 50 boxes of pepper sauce created at their factory. The first experiment revealed that a box had a $20.5%$ chance of being defective.Define the piecewise function, $P(X=x)$ for the single experiment.
i. Find the probability that the fourth and fifth boxes are not defective given that the first three boxes are defective.
ii. Prove that $\sum*{i=3}^{n} B_i = \begin{pmatrix} n \ 3 \end{pmatrix}p^3(1-p)^{n-3}$. Consequently, find the probability of $\sum*{i=3}^{n} B_i$ given that $n=5$. Note: $B_i$ is the $i^{th}$ bernoulli trial.