Binomial Distribution

Interpretation

boolean, success or failure
- getting heads or not
- being universal blood donor or not
- winning lotto or not
needs to be independent
- if when one trial is done, the percentages change, its not independent and not a binomial
  - X would have the hypergeometric distribution
a distribution of the number of successes in n independent Bernoulli Trials
- or a sequence of bernoulli trials

A random variable $Y = \sum_{i} X_{i}$ follows a binomial distribution, counting the number of successes of the random variable $X$ that follows the Bernoulli Distribution

Probability Distribution Function

A random variable $X$ has a binomial distribution, denoted by $X \sim B i n (n, p)$ if and only if it has the probability distribution function

f (x) = P (X = x) = (\binom{n}{x}) p^{x} (1 - p)^{n - x}

for $0 \leq x \leq n,$ with:

$f (x) \geq 0$
$\sum_{x} f (x) = 1$

where $n$ represents the number of total trials, $p$ the probability of "success" and $x$ the number of "successes"

$p^{x} (1 - p)^{n - x}$ represents the probability of a specific ordering of $x$ successes and $n - x$ failures
$(\begin{matrix} n \\ x \end{matrix})$ represents number of ways it can happen

Mean

\begin{aligned} E (X) & = \sum_{x} x f (x) \\ = \sum_{x = 0}^{n} x (\binom{n}{x}) p^{x} (1 - p)^{n - x} \\ = \sum_{x = 1}^{n} x \frac{n!}{x! (n - x)!} p^{x} (1 - p)^{n - x} \\ = \sum_{x = 1}^{n} \frac{n (n - 1)!}{(x - 1)! (n - x)!} p \cdot p^{x - 1} (1 - p)^{n - x} \\ = n p \sum_{x = 1}^{n} \frac{(n - 1)!}{(x - 1)! (n - 1 - (x - 1))!} p^{x - 1} (1 - p)^{n - 1 - (x - 1)} \\ = n p \sum_{x = 1}^{n} B i n (n - 1, p) \\ μ & = n p \cdot 1 = n p \end{aligned}

Variance

V a r (X) = n p (1 - p)

If $X$ represents the number of successes in $n$ trials, then $Y = \frac{X}{n}$ represents the Sample Proportion of successes

Using Chebyshev's Theorem we get the law of large numbers:

lim_{n \to \infty} P (| y - p | \leq c) = 1

if we repeat the trial for sufficient large number, the proportion of "success", $y = \frac{X}{n}$ will be very close to $p = P (success)$

Moment-generating function

M_{X} (t) = [1 + p (e^{t} - 1)]^{n}

can be approximated with the poisson distribution (which is a special case of the binomial)
- when $n \to \infty$ and $p \to 0$
  - reasonable if $n > 50$ and $n p < 5$
- useful when we don't wanna deal with big factorials
  - or if you dont know n and p, but know the mean

examples

1
- fair dice rolled 10 times
- probability of three coming up exactly 2 times
- $* P (X = 2) = (\begin{matrix} 10 \\ 2 \end{matrix}) (\frac{1}{6})^{2} (1 - \frac{1}{6})^{10 - 2} *$
2
- n = 3, we want p(X=2)
- pp(1-p) = p(1-p)p = (1-p)pp = p^2(1-p)
- p(X=2) = 3p^2(1-p)
- 3 = combination formula of (3 2)
3
- 25 babies, 20% chance of acne
- P of exactly one having acne
- $P (X = 1) = (\begin{matrix} 25 \\ 1 \end{matrix}) (\frac{1}{5})^{1} (1 - \frac{1}{5})^{25 - 1}$
- = 0.0236....
- P that no more than 2 babies have acne?
  - P(X<=2) = P(X = 0) + P(X = 1) + P(X = 2)
  - $P (X \leq 2) = (\begin{matrix} 25 \\ 0 \end{matrix}) ({0.20}^{0} (1 - 0.20)^{25} + (\begin{matrix} 25 \\ 1 \end{matrix}) ({0.20}^{1} (1 - 0.20)^{24} + (\begin{matrix} 25 \\ 2 \end{matrix}) ({0.20}^{2} (1 - 0.20)^{23}$
  - = 0.0982
- P that at least one of the babies has acne
  - P(X>=1) = P(x=1) + ... + P(X=25)
  - P(X>=1) = 1 - P(X=0) = 1 - 0.8^25 = 0.996...