This is the first of a series of articles I’ll be posting here about some of the most common probability distributions.
Bernoulli random variable
It is a variable that has 2 possible outcomes: “success”, or “failure”. Success occurs with probability \(p\) and failure with probability \(q=1p\)
The distribution of heads and tails in coin tossing is an example of a Bernoulli distribution with \(p=q=1/2\). The Bernoulli distribution is the simplest discrete distribution, and it the building block for other more complicated discrete distributions. The distributions of a number of variate types defined based on sequences of independent Bernoulli trials that are curtailed in some way are summarized in the following list:
 binomial distribution: number of successes in n trials
 geometric distribution: number of failures before the first success
 negative binomial distribution: number of failures before the xth success
Binomial probability distribution
Suppose that n independent Bernoulli trials each one having probability of success p are to be performed. Let X be the number of successes among the n trials.
We say that X follows the binomial probability distribution with parameters
In plain english, the binomial distribution describes the outcome of n independent trials in an experiment. Each trial is assumed to have only two outcomes, either success or failure.
 Probability mass function of X:
where
Binomial Distribution Properties
 Mean: \(\mu = np \)
 Variance: \(\sigma^2=npq\)
 Standard deviation: \(\sigma =\sqrt{npq}\)
 Skewness: \(\alpha_3=\frac{qp}{\sqrt{npq}}\)
 Kurtosis: \(\alpha_4= 3 + \frac{16pq}{npq}\)
 Moment generating function: \(M(t) =(q+pe^t)^n\)
 Characteristic function: \(\phi(\omega)=(q+pe^{i\omega})^n\)
Note: The special case of a binomial distribution with \(n=1\) is also called the Bernoulli distribution.
Binomial Distribution in R
 Generate Binomial random numbers:
In R we use the function rbinom(e,n,p) to generate binomial random numbers.
Parameters:
e: number of experiments you want to simulate
n: number of independent Bernoulli trials in each experiment
p: define the probability of success
Example1: A fair coin is tossed. Let the variable x take values 1 and 0 according to as the toss results in “Head”” or “Tail”. Then, X is a Bernoulli variable with parameter p=1/2. Here, X denotes the number of heads obtained in the toss. Probability of success 1/2 and the probability of failure 1/2
Example2: Run the above experiment 10 times and calculated the expected number of heads. Well, we don’t really need to do the simulation as we already know the E[X] for a binomial distribution is np. So the expected number of heads in 10 experiments is:
let’s run the simulation and calculate the average:
Close enough. To understand why we got 6 instead 5 we need to understand “Law of Large Numbers”. This law explains that in the long run it becomes extremely likely that the proportion of success, X/n, will be as close as you like to the probability of success in a single trial, p.
Let’s run again the same example but 10000 times
let’s run the simulation and calculate the average:
So, 0.4959 is very close to our original p=0.5
Example3: Generate 100 samples of binomial(20,.5). Print the first 3 experiments
The interpretation of the above result is:

In our first experiment we got 13 heads out 20

In our second experiment we got 9 heads out 20

In our third experiment we got 9 heads out 20

Using probability distribution function pbinom() density function dbinom()
Let’s illustrate the above functions with an example:
Example4: What’s the probability of getting exactly 2 heads in 6 tosses of a fair coin?
Let’s do it the hard way: manually
The probability of getting exactly 2 heads in 6 tosses is: 0.23461
Using dbinom() is much simpler: dbinom() calculates the exact probability of success for every x.
Alternatively, we can use the cumulative probability function for binomial distribution pbinom()
Example5: Find the probability that in five tosses of a fair die, a 3 will appear
Let X the number of times a 3 appear in five tosses of a fair die
a) twice: \(P(X=2)\)
b) at most once: \(P(X\leqslant{1})\)
c) at least two times: \(P(X\geqslant{2})=1 P(X\lt{2})\)
Example6: if 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts chosen at random:
a) 1 defective: \(P(X=1)\)
b) 0 defectives \(P(X=0)\)
c) Less than 2 defective: \(P(X \lt 2) = P(X\leqslant{1})\)