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