Probability

# 10 - Probability/Stats #

Why, where are these used, etc.

bring up music things, part failure rates, tollerences, etc.

## Basics #

For the following I’ll be using a dice roll example, where the events are the total of two die. The Sample Space of this is

$$S = \{2,3,4,5,6,7,8,9,10,11,12\}$$

Note, that 1 isn’t possible as the lowest is both die being ‘1’.

Let’s look at the probability of these outcomes,

die2↓, die1→ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12

While there are 11 (2-12) unique outcomes, there are 36 possible outcomes from the two die, which are shown in the table above.

If you follow the diagonal you can see that there is only one way to get 2 or 12, two ways to get 3 or 11, and so on, this gives this table:

Total of two die 2 3 4 5 6 7 8 9 10 11 12
Probability 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

Now, let’s say we want the probability that we roll an even total (2,4,6,8,10,12) we can just sum their respective probabilities, so $$\frac{1}{36}+\frac{3}{36}+\frac{5}{36}+\frac{5}{36}+\frac{3}{36}+\frac{1}{36} = \frac{18}{36} = \frac{1}{2}$$

Lets go ahead and call this event A, so P[A] = 1/2

Similarly, we can define a new rule, Event B, such that the roll total is greater than 9, that comes out to be 1/6, so P[B]=1/6

### Statistical Independence #

Event A and event B are statistically independent if and only if (iff) $$P[AB] = P[A]P[B]$$

So, here, P[AB], that is the probability that a number is both greater than 9 and the number is even, that would be 1/9. The probability of each event multiplied together, $$P[A]P[B] = \frac{1}{2} * \frac{1}{6} = \frac{1}{54}$$ and, hopefully obviously, that’s not the same as 1/9. Therefore, these events are not statistically independent. This makes logical sense, if you know that the total count of the two die is greater than 9, then you also know that there’s a higher chance that the result is even - of 10,11, and 12, 2/3 of the totals are even. That is dependance.

If instead, we asked, what is the probability that 1 die is a 6 and the the other a 2, both of those would have a probability of 1/6, that is $$P[A]=P[B]=\frac{1}{6}$$ , so now think, in the comined scenario, P[AB] would be the probability of die 1 being a 6, and die 2 being a 2, well, there’s 36 different ways the die can land, and they’re all unique (assuming the die are labeled somehow) so $$P[AB]=\frac{1}{36} = P[A]P[B]$$ - these events are statistically independent.

It’s worth noting, the phsical relationship is not always this clear. In a lot of situations, you’ll just need to do the math and determine if $$P[AB] = P[A]P[B]$$ to check.

### Conditional Probability #

Sometimes, knowing something about one event tells us something about the probability of another event. This can be expressed mathematically. When written down, it looks a bit gross, but it’s actually really easy to understand:

$$P[A|B]P[B]=P[AB]=P[B|A]P[A]$$

This also provides another equation, just by moving things around:
$$P[A|B]=\frac{P[AB]}{P[B]}$$

Here, P[A|B] means, the probability of event A happening given event B has already happened.

NOTE: This means that P[A|B] ≠ P[B|A] (well, not usually). I like to think of it this way: Losing your keys may make you late for work, and being late for work may be from loosing your keys; however, the probability that your late for work GIVEN you’ve lost your keys may be higher or lower than than the probability that you lost your keys GIVEN that you’re late to work.

So, looking at just one side of the above equation, we have $$P[A|B]P[B]=P[AB]$$ , this reads as “The Probability of A given that B has happened times the Probability of B is equal to the probability of of an event that is in both A and B”

The other side of this equation, is just swapping the roles of A and B.

So, let’s go off the above example and assume that you’re really bad at showing up for work:

Late For Work P[A] Lost keys P[B] Both P[AB]
.25 .15 .05

So shoving these values in to the above, we can determine that

\begin{aligned} P[A|B]\times P[B]&=P[AB]\\ P[A|B]\times .15&=.05\\ p[A|B] &= .33 \end{aligned}

So, if you lose your keys, you’ll be late to work 1/3rd of the time.

Alright, but what about a more complicated situation, one wher you have to make multiple decesions! Let’s

Box Box 1 Box 2
Red Ball 90 30
Blue Ball 10 70

Keeping the numbers simple here, lets say you want to know the probability that the ball you picked was from box1, given that you’ve already drawn a blue ball. This is where a Tree diagram comes in handy:

This diagram first helps us expose an assumption in the problem: that the initial choice is a 50/50, split chance between which box we start with. So our first choice is really which box are we picking from. Both have a probability of 1/2. Only then do we pick our ball. This stacking of situtations gives us the term ‘sub experiment’ where you perform each sub experiment in order: so our first sub experiment is picking the box, the second is picking the ball.

This lets us figure out the conditional probabilities super easily, as all that’s needed is to look at the respective branch. For example, in the above we can see P[Red|Box1] is 9/10. Another nice thing is by mulitplying across the branches we can get the probability of the entire ‘system’ easily. Note the ‘,’ instead of a ‘|’ in the diagram below. This is saying that these have both happened not implying conditional probability. That’s why adding all of these up will add up to 1 (100%) as it’s a look at all of the possible events.

So now we know the probability of a blue ball overall: 0.4 (.05 + .35 )(1), the probability of a bule ball in box 1 (.1), what is the probability that we actually picked from box 1? We know P[Blue|Box1], but what is P[Box1|Blue] ? Well, naturally, there’s a handly formula, known as “Bayes Rule” for this situation:

$$P[A|B]=\frac{P[B|A]P[A]}{P[B]}$$

Applying this, we can see that we get

\begin{aligned} 0.1&=\frac{P[B|A]0.4}{0.5}\\ (0.1/0.4)*.5&=P[B|A]\\ P[B|A]&=.125 \end{aligned}

so, there’s a 12.5% chance that the box we picked the blue ball from was Box1.

Just to check outselves, what’s the chance that the box was Box2?

\begin{aligned} 0.7&=\frac{P[B|A]0.4}{0.5}\\ (0.7/0.4)*.5&=P[B|A]\\ P[B|A]&=.875 \end{aligned}

And this works out, adding to 100%.

There are a few more things to note regarding conditional probability:

• if P[A|B] is 0, the two events are mutally exclusive. This happens in dumb situations like “Given you’ve rolled a 2, what is the probability you rolled a 3” but also more complex events may mean that this is less obvious, so it’s nice to be able to math it out.

[TODO]

Tree diagram with more branches, some ‘incomplete branches’

fair/unfair coin:

/C1-H-H-H \C2-H-H-H

Sub experiments & Tree diagrams

Counting methods

combinations and permutations

Looking at binary, arrangements of bits

with or without replacement

## Random Variables #

Probability Mass Function (PMF)

Types of RV’s

### Bernoulli #

$$\begin{cases} q=1-p & \text{if }k=0 \\ p & \text{if }k=1 \end{cases}$$
Tossing a coin- if the coin is fair p and q both equal 1/2. This is literally just a true/false question and the probability of a given answer.

### Geometric #

Original Image by Skbkekas - Own work, CC BY 3.0, link

In probability theory and statistics, the geometric distribution is either one of two discrete probability distributions:

• The probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set { 1, 2, 3, … }

• The probability distribution of the number Y = X − 1 of failures before the first success, supported on the set { 0, 1, 2, 3, … }

Which of these one calls “the” geometric distribution is a matter of convention and convenience.

### Binomial #

Original Image by Tayste - Own work, Public Domain, link

Sum of n, independent Bernoulli trials

### Pascal (Negative Binomial) #

From Introduction to Probability by Hossein Pishro-Nik, CC BY-NC-ND 3.0 … technically I’m abusing the licence a bit, but the ‘derivative’ here is just a CSS invert, you can open the image in a new tab to see the ‘original’

Number of trials until k’th success. While technically the Pascal distribution is just a specific subset of Negitive Binomial distributions, you’ll probably only ever use the Pascal distribution. Be aware the more general case exists though, as someday it may pop up.

### Discrete Uniform #

Original Image by IkamusumeFan - Own work, CC BY-SA 3.0, link

It’s a uniform distribution. Rolling a fair dice? Great, you have 1/6 chance to get a 1,2,3,4,5, or 6. So your ‘a’ and ‘b’ are 1 and 6, ‘n’ is the number of points, so b-a+1

### Poisson #

Original Image by Skbkekas - Own work, CC BY 3.0, link

Probability of a number of events occuring in a fixed amount of time (or space, or whatever) provided the events happen with a constant average rate and the events are independent.

Cumulative Distribution Function (CDF)

functions of Random Variables

Families of continuous RVs

Conditional Probability Mass Fn & Conditional Expected Value

Gaussian Random Variables / Normal RVs (same thing)

## Chaos #

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