binomial distribution

The probability of observing exactly $k$ “positive” events (successes) out of $n$ independent bernoulli experiments, where each event has a probability $p$ of sucesss, is given by the binomial distribution.

$$P(k|p)=(\genfrac{}{}{0ex}{}{n}{k})p^k\cdot (1-p{)}^{n-k}$$

$n\dots$ total number of (bernoulli)events
$k\dots$ number of “positive” events
$p\dots$ probability of positive event
$\left(\genfrac{}{}{0ex}{}{n}{k}\right)\dots$ binomial coefficient

Binomial random variable

$$\begin{array}{r}X\sim \text{binomial}(n,p)⟹P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k\cdot (1-p{)}^{n-k}\end{array}$$

EXAMPLE

$n=2$
$P(X=0)=P(F_1{F}_2)=P(F_1)P(F_2)=(1-p{)}^2$
$P(X=1)=P(S_1{F}_2)+P(F_1{S}_2)=P(S_1)P(F_2)+P(F_1)P(S_2)=2p(1-p)$
$P(X=2)=P(S_1{S}_2)=P(S_1)P(S_2)=p^2$
We can pull them apart because they are independent.
Note: The coefficients are $\left(\genfrac{}{}{0ex}{}{2}{0}\right)=1,\left(\genfrac{}{}{0ex}{}{2}{1}\right)=2,\left(\genfrac{}{}{0ex}{}{2}{2}\right)=1$

Pascals triangle (the binomial distribution) converges to the normal distribution for large $n$.

$$\begin{array}{rlr}X\sim \text{binomial}(n,p)& \to X\sim \text{normal}(np,\sqrt{npq})& \text{if }npq\text{ is not too small}\\ & \to X\sim \text{poisson}(\lambda )& \text{if }np\text{ or }nq\text{ is small}\end{array}$$

where $p$ is the probability of success, $q=1-p$ is the probability of failure, and $n$ is the number of trials.
This is true $npq$ is not too small, i.e. success and failure are not too rare.
If they are, the binomial distribution converges to the poisson distribution, where $\lambda =np$ is the average number of successes.

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References

3b1b, primer