Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions 

There are a fixed number of trials. Think of trials as repetitions of an experiment. The letter n denotes the number of trials.

The n trials are independent and are repeated using identical conditions. 

There are only two possible outcomes, called "success" and "failure," for each trial. The letter p denotes the probability of a success on any one trial, and q denotes the probability of a failure on any one trial. p + q = 1

For example, randomly guessing at a truefalse statistics question has only two outcomes. If a success is guessing correctly, then a failure is guessing incorrectly. Suppose Joe always guesses correctly on any statistics truefalse question with a probability p = 0.6. Then, q = 0.4. This means that for every truefalse statistics question Joe answers, his probability of success (p = 0.6) and his probability of failure (q = 0.4) remain the same. So guessing one question is considered a trial. If he guesses n different questions, means the trial is repeated n times and p = 0.6 remains the same for each trial.

Binomial Distribution

$
\mathrm{P}(\mathrm{X}=\mathrm{r})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{p}^{\mathrm{r}} \cdot \mathrm{q}^{\mathrm{n}\mathrm{r}}
$
Where $P(X=r)$ is the probability of $X$ successes in $n$ trials when the probability of success in ANY ONE TRIAL is $p$. And of course $q=(1p)$ and is the probability of a failure in any one trial.

In the experiment, the probability of
At least "r" successes,

$
\mathrm{P}(\mathrm{X} \geq \mathrm{r})=\sum_{\lambda=\mathrm{r}}^{\mathrm{n}} \mathrm{n}_\lambda \mathrm{p}^\lambda \cdot \mathrm{q}^{\mathrm{n}\lambda}
$

At most " $r$ " successes,

$
\mathrm{P}(\mathrm{X} \leq \mathrm{r})=\sum_{\lambda=0}^{\mathrm{r}}{ }^{\mathrm{n}} \mathrm{C}_\lambda \mathrm{P}^\lambda \cdot \mathrm{q}^{\mathrm{n}\lambda}
$
A binomial distribution with $n$Bernoulli trials and probability of success in each trial as $p$, is denoted by $B(n, p)$.
The mean, $\mu$, and variance, $\sigma^2$, for the binomial probability distribution are

$
\mu=n p \quad \text { and } \quad \sigma^2=n p q
$
The standard deviation, $\sigma$, is then

$
\sigma=\sqrt{n p q}
$