4.5 Binomial Distribution Lesson 05 | ND111 - Data Science II

4.5 Binomial Distribution `Lesson 05`

In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome: a random variable containing a single bit of information: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q = 1 − p). A single success/failure experiment is also called a Bernoulli trial or Bernoulli experiment and a sequence of outcomes is called a Bernoulli process; for a single trial, i.e., n = 1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance. Wikipedia

Example 1: 5 coin flips. \(P(\text{two HEADS})\).

The order of the HEADS do not matter.

What’s the number of combinations?

Flip 1	Flip 2	Flip 3	Flip 4	Flip 5
H	?	?	?	?
H	H	?	?	?
5	4	1	1	1

You can place H in 5 places, after place the first H you only have 4 places. For this reason, there are 20 posibilities.

\[ P_1 = 5 * 4 = 20 \]

Bear in mind, the H’s are equal and you can swap each one.

\(H_1\)	\(H_2\)
?	?
2	1

So there are two possible H’s to insert in the \(H_1\)

\[ P_2 = 2 * 1 = 2 \]

What this \(P_2\) means?

You have two equals instances so the \(P_1\) has double entries. The good part is the \(P_2\) is used to “fix” it.

\[ P = \frac{P_1}{P_2} = \frac{20}{2} = 10 \]

Example 2: 10 coin flips. \(P(\text{four HEADS})\).

The order of the HEADS do not matter.

What’s the number of combinations?

Flip 1	Flip 2	Flip 3	Flip 4	Flip 5	Flip 6	Flip 7	Flip 8	Flip 9	Flip 10
H	?	?	?	?	?	?	?	?	?
H	H	H	T	H	T	T	T	T	T
10	9	8	1	7	1	1	1	1	1

You can place H in 10 places, after place the first H you only have 9 places. For this reason, there are (\(10*9*8*7\)) posibilities.

\[ P_1 = 10*9*8*7 = 5,040 \]

Bear in mind, the H’s are equal and you can swap each one.

\(H_1\)	\(H_2\)	\(H_3\)	\(H_4\)
?	?	?	?
4	3	2	1

So there are two possible H’s to insert in the \(H_1\)

\[ P_2 = 4 * 3 * 2 * 1 = 24 \]

What this \(P_2\) means?

You have two or more equals instances so the \(P_1\) has “double” entries. The good part is the \(P_2\) is used to “fix” it.

\[ P = \frac{P_1}{P_2} = \frac{5,040}{24} = 210 \]

4.5.1 Equation

Founded on the examples above, it is possible to write a equation, given 10 flips (\(k\)) and an expected 3 heads (\(n\)).

\[ P = \frac{P_1}{P_2} = \frac{10 * 9 * 8}{\underbrace{3 * 2 * 1}_{3!}}\]

Let’s multiply by (\(7 * 6 * 5 * 4 * 3 * 2 * 1\)) or simply by \(7!\).

\[ P = \frac{P_1}{P_2} = \frac{10 * 9 * 8 * 7!}{3! * 7!} = \frac{\overbrace{10!}^{k!}}{\underbrace{3!}_{n!} * \underbrace{7!}_{(k-n)!}} = \frac{k!}{n!(k-n)!} \tag{1}\]

Equation (1) is also noted as:

\[ C_{n,k} = \begin{pmatrix} n \\ k \end{pmatrix} \tag{2}\]

Equation (2) will only calculate the number of combinations. We can aggregate the probability.

\(P(H)\): for heads;
\(P(T)\): for tails;

Given the 10 coins flips, the probability for a single instance, no matter the order:

Obs.: Do not confound with permutation notation.

\[ P_{k,n} = P(H)^n * P(T)^{k-n}\] \[ P_{10,3} = P(H)^3 * P(T)^{7}\tag{3}\]

For a fair coin.

\[ P_{10,3} = 0.5^3 * 0.5^{7} = 0.000976563 \tag{4}\]

The value of \(P(10,3)\) is for a single time, I know there are many instances where could happen 3 heads, for this reason I use the combination.

\[ C_{10,3} = \begin{pmatrix} 10 \\ 3 \end{pmatrix} = \frac{10!}{7!*3!} = 120\]

The probability to happen 3 heads in 10 flips coins is:

\[ P(10|3) = C_{10,3} * P_{10,3} = 120 * 0.000976563 = 0.1171875 \]

Expanding this concept to a all around equation:

\[ P(k|n) = \underbrace{C_{k,n}}_{\text{note 1}} * \underbrace{P_{k,n}}_{\text{note 2}} \tag{5}\]

note 1: Probability to occur the given combination (3 heads and 7 tails) over the all combinations (\(10^2 = 1,024\));
note 2: Probability based on the coins (heads and tails probabilities).

4.5.1.1 Additional Info

Do not confound Permutation with Combination.

Combination: When the order does not matter;
Permutation: When the order is important.