🍈 Zettelkasten

A Discrete Distribution used to model Independent Bernoulli Trials . Uses Binomial Theorem in tandem with probabilities to get our probability distributions.

PMF

$X \sim Binomial (n, p) ⟹ p_{X} (x) = (x n) p^{x} (1 - p)^{n - x}$

$n$ is the number of trials
$p$ is the probability of success for a trial

Expectation

$E (X) = n p$

Variance

$V (X) = n p (1 - p)$

Examples

Example 1

heads probability 50%, tails probability 50%. probability distribution of flipping 3 times Note that h + t = 1. 100% divided up into probability sets $h = 1/2, t = 1/2$ $(h + t)^{3}$

Solution

$(h + t)^{3} = h^{3} + 3 (h^{2}) (t) + 3 (h) (t^{2}) + t^{3}$ = (1/8) + (3/8) + (3/8) + (1/8) first term is P(no tails), P(no tails) = 1/8 P(one tail) = 2nd term. P(one tails) = 3/8 P(second tail) = 3rd term. P(second tails) = 3/8 P(third tail) = 1/8

Example 2

Flip a coin 4 times. GIve the probability distribution where ‘T’ is the successful outcome T = the number of Tails. What is the expected number of ‘T’s?

T = t	P(T = 1)
0	1/16
1	= $_{4} C_{1} * (\frac{1}{2})^{1} * (\frac{1}{2})^{3} = \frac{4}{16}$
2	= $_{4} C_{2} * (\frac{1}{2})^{2} * (\frac{1}{2})^{2} = \frac{6}{16}$
3	4/16
4	1/16
This follows pascal triangle you see! If you add up all the probabilities, they should equal 1.
1/16 + 4/16 + 6/16 + 4/16 + 1/16 = 16/16 = 1!
The expected number is 2.

Example 3

Game consists of rolling a pair of dice 10 times. for each sum that equals 6,7,8 or 2, you win 1 dollar. it costs 5 dollars to play the game. Is this a fair game?

Solution

n = 10 p = … cases for 7 : {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} = 6 cases cases for 6 : {(1,5),(2,4),(3,3),(4,2),(5,1)} = 5 cases cases for 8 : {(2,6),(3,5),(4,4),(5,3),{6,2}} = 5 cases 6 + 5 + 5 total possibilities = 6^2 = 36 p = 16/36

E(X) = np = 10 (4/9) = 40/9 = 4.44

4.44 is smaller than 5 dollar. not fair!