The binomial (or Bernoulli) distribution

An experiment can be modeled with a binomial distribution whenever:

• there are only two possible events resulting from the experiment: $A, \overline{A}$ (success and defeat).
• the probabilities of every event $A, \overline{A}$ are the same in any happening of the experiment ($p$ and $q = 1-p$, respectively). Namely if a coin is flipped several times, the probability of having 'heads' does not change.
• any realization of the experiment is independent from the rest.

A binomial random variable will give the number of successes when having happened a certain number of experiments.

It turns out to be useful to analyze the number of times that 'heads' is obtained when flipping a coin $n$ times.

The binomial distribution is usually represented by $B (n,p)$, with:

• $n$: number of happenings of the random experiment.
• $p$: probability of success in doing an experiment

So if we want to study the binomial distribution that models $10$ flips of a coin (in which the 'heads' and 'tails' are equally probable) we have:

$$\displaystyle B\Big(10, \frac{1}{2}\Big)$$

The probability function of the binomial distribution is:

$$p(X=k)=\binom{n}{k}p^k\cdot q^{n-k}$$

• $n$: number of experiments
• $k$: number of successes
• $p$: success probability
• $q$: defeat probability

The combinatorial number is defined:

$$\displaystyle \binom{n}{k}= \frac{n!}{k!(n-k)!}$$

Calculate the probability of obtaining $8$ 'heads' when flipping a coin ten times.

Distribution $\displaystyle B\Big(10, \frac{1}{2}\Big)$

number of experiments: $n=10$

number of successful results: $k=8$

probability of each success and each defeat: $\displaystyle p=q=1/2$

$$p(X=8)=\binom{10}{8} \Big(\frac{1}{2}\Big)^8 \Big(\frac{1}{2}\Big)^2 = 0.044$$

what can be interpreted as the product of the possible combinations of $8$ 'heads' and $2$ 'tails' times the probability of extracting $8$ 'heads' times the probability of extracting $2$ 'tails'.

The average of a binomial distribution is:

$$\mu= n \cdot p$$

The variance is:

$$\sigma^2= n \cdot p \cdot q= n \cdot p \cdot (1-p)$$

The standard deviation is:

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