# An Introduction to Probability

“Probability Theory Should Be Thrown Under A Bus” — Artifical Intelligence Expert, Carlos E. Perez.

# Why reasoning under undercertainty?

In Knowledge Representation and Reasoning (KR&R) things are very straightforward. It’s easy to represent and formulate propositions without uncertainity. However, issues start to arrise when uncertainity reveals itself.

Take for instance, an expert system designed to replace a doctor. A doctor has no formal rules for diagnosing patients based off of symptons and no formal way to treat the patient. In this instance, the expert system will need to use probability to work out the highest probable chance that patient has a certain disease and the cure to that disease.

There are 3 main reasons as to why we would not want to use propositional logic to deal with a domain like medical diagnosis:

- Uncertainty — Trying to use exact rules to cope with a large domain
- Laziness — It will often take too long to work on a very large list of rules or propositions
- Practical ignorance — Even if strict laws exist, we might be uncertain about a particular subject such as every patient being unique and different.

# High Level Probabillity

Probability provides a way of summarising the uncertainty that comes from our laziness and ignorance. In other words, probability finds out the likeliness that something will happen.

# Discrete probability

Discrete probability is a formalisation of probabilty theroy that describes probability for usage in computers from discrete mathematics.

When solving problems with discrete probability, we start with using probability spaces. A probability space is the pairing (S, P) where:

- S is the sample space of all elementary events X ∈ S. Members of S are called outcomes of the experiment.
- P is the probability distrubtion, that is, assigning a real number P(x) to every elementary event X ∈ S such that it’s probability is between 0 and 1 and ∑P(x) = 1

For point 2, P(x) is read as “the probability of X”. The probability must always be between 0 and 1, or often represented as 0% and 100%.

## Example

```
Imagine flipping a coin. The probability space is (S, P).
The outcome S is
```**S = {H, T}** where S can either be Heads or Tails.
Therefore the probability is
**P(H) = P(T) = 1/2
**The probability for heads is the same as the probability for tails which is the same as a half. In other words, if you flip a coin there is an even chance of it becoming head-side up or tail-side up.

A probability distrubtion is considered uniform if every outcome is equally as likely.

# An introduction to solving probability problems

Many many people including university professors and PhD students cannot solve probability problems. As discussed later in this article the Monty Hall problem is a famous problem and good example of this.

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say №1, and the host, who knows what’s behind the doors, opens another door, say №3, which has a goat. He then says to you, “Do you want to pick door №2?” Is it to your advantage to switch your choice?

This question was sent into Voe Savant who at the time had the highest IQ in the world. Voe Savant replied that there is a 2/3rd chance of winning the car if you switch and 1/3rd if you don’t switch.

Thousands of people argued over the monty hall problem and many university professors of maths said that math illiteracy was rampant in America because the Monty Hall problem solution suggested was wrong.

This problem appeared in every maths class the following week and thousands of readers, many who have PhDs in maths wrote in to explain that Savant was wrong. Even Paul Erdős, one of the worlds most famous mathematicians, said that Savant was wrong.

Unforunately for them, Savant was right. This is a simple probability problem that if defined formally can be explained. Many of the mathematcians used their intuition to solve this problem and not follow the steps in solving a probability problem that will be outlined below.

There are a few steps you need to take before you solve a probability problem to prove you fully understand the problem.

## Sample Space

The sample space is the set which contains all possible outcomes.

So given a coin, the sample space is {heads, tails} as the coin can **only **land on heads or tails.

## Outcome

An outcome consists of all the information of an experiment after the experiment has been performed. When you flip a coin and it lands on heads, the outcome is {heads}.

## Probability space

The probability space is the sample space but every possible outcome has a probability applied to it. With the coin flip the probability space is {(Heads, 0.5), (Tails, 0.5)}.

The total probability of all probabilities in the probabilty space **must **be equal to 1. No single probability can be less than 0 or more than 1.

Many high-achieving students tell me that they try to visualise what they are dealing with as much as possible.

## Example

Let’s suppose we roll a 6-sided dice and we want to work out the probability we get a 4.

- Count the number of possible events. There are
**6**sides to the dice. So there are**6**possible events. - Decide which event you are examining for probability. The problem let’s us know we are trying to roll a
**four**. - Count the number of chances that heads can occur out of the possible events. There is only one side of the die that has 4 dots, so there is only
**1**chance to roll a four out of 6 total chances. - Write the number of chances heads could occur over the number of possible events in a fraction. (
**1/6**)

Although this is a simple problem to solve, it illustrates the important steps to take when solving harder probability problems.

# Events

Events are often overlooked in probability theory and are not talked about much, so I took it upon myself to expand on what an event is and why they’re important in this section.

An event is a set of outcomes of an experiment in probability. In Bayesian Probabiltiy an event is defined as describing the next possible state space using knowledge from the current state.

An event is often denoted with the charecter ‘e’. Such as the probability being P(e) of an event. Events are alot more important in probability than most people make them out to be.

An event can be the result of rolling a dice such as a “5” or getting a Tail when flipping a coin.

Events can be:

- Independent — Each event is not effected by previous or future events.
- Dependent — An event is affected by other events
- Mutually Exclusive— Events can’t happen at the same time

## Why are events important?

Well, events allow us to do some pretty amazing things with probability. Take for instance, the Monty Hall Problem. Attempt the question below:

One of the doors above contains a fancy sports car, the other 2 doors contain goats. Pick any door you like, go on!

Okay, let’s say you picked #1, the gameshow host will open a door that contains a goat, so let’s say we open door number 3 and it contains a goat. So you know that door 1 is your pick, door 3 is a goat and door 2 is untouched. Note: It doesn’t matter what door you originally picked, what matters is that you pick **a** door and the gameshow host opens a door with a goat in it.

The gameshow then asks: “Are you sure door number 1 is right? Do you want to switch?”

What do you do?

Well, probability states that we should pick door number 2, as in you would switch. Why? well, door number 2 has a 2/3 chance or 77% chance of containing the car and door number 1 (your original pick) has a 33% chance of containing a car.

Whaaaaattt??

This is a famous probability problem called the Monty Hall Problem and it displays how events can affect probabillities. For an explanation of this, watch this Numberphile video below:

# The probability of the complement of an event

The complement of an event is all the other outcomes of an event.

For example, if the event is Tails, the complement is Heads. If the event is {Monday, Tuesday}, the complement is {Wednesday, Thursday, Friday, Saturday, Sunday}.

If you know the probability of p(x) you can find the compliment by doing 1 — P(x). Since all probabillities equal 100%, we can express this as 1.

## Why is the complement useful?

Sometimes it is easier to work out the complement first before the actual probability. For example:

`Work out the probability that when 2 die are thrown that the two scores are different`

A different score is like getting a 2 and 3, or 1 and 6. The set of all possible different scores is quite large, but the complement of all possible different scores (scores are the same) is quite low. In fact, it is:

{ (1, 1), (2, 2), (3,3), (4,4), (5,5), (6,6) }

The total number of different combinations is 6*6 which is 36, so the probability of getting a score that is the same is 6/36 or 1/6. Now we can take 1/6 away from 1 (think of 1 as the universal set) which equates to 5/6.

# The Union of Two Events (Inclusion-Exclusion principle)

If two events are mutually exlusive (they cannot occur at the same time) then the probability of them happening at the same time is 0.

If two events are not mutually exclusive, then the probability of the union of the two events is the probability of both events added together.

*P*(*A*∪*B*) = P(A) + P(B) — *P*(*A*∩*B*)

The reason we take away the intersection of A and B is because P(A) + P(B) contains all that it is in A or B but because of the way union works, there will be an intersection which will make 2 A’s and 2 B’s so we need to take away the intersection to get the probability of each events.

In other words, A contains elements that are in B and B contains elements that are in A. By adding:

A = {1, 2, 3}

B = {3, 4, 5}

|*P*(*A*∪*B*)| = |{1, 2, 3}| + |{3, 4, 5}| — |{1, 2, 3} ∩{3, 4, 5}|

= 3 + 3 -1 = 5

# Disjoint or Mutually Exclusive Events

An event is called disjoint if it cannot happen at the same time as another event. It’s best if we were to show this with an example.

Suppose that I were to roll a fair dice 3 times. S is the set of sequences of events over length three such that { 1..6)³} P(x) = 1/6*6*6 = 1/216 for all x ∈ SWhat is the probability that we roll at least one 6?So because we throw the dice 3 times, let E1 be the probability that the dice roll is a 6, E2 = P(6), E3 = P(6)We would like to work out P(E1∪E2∪E3)

Remember, the union of probabilities is P(A) + P(B) — Intersection of A and B. We want the union of A, b and C which also includes the intersection in the middle. We take away the intersections of A B, A C, B C and add the intersection of all 3 to get the middle part.

You might have noticed that the intersection is 6/216. This may seem confusing because we didn’t hand-define a set for this. Worry not: The formula for intersection is P(A intersection B) = P(A) * P(B).

**Example question**

`Given 4 coins, what is the probability that at least 3 of them come up tails?`

The event that at least 3 coins come up tails is the union of five disjoint events, that all coins come up tails (1 disjoint event) and that 4 specified coins (4 disjoint events) come up heads. This may sound confusing, so I’ll explain it visually. Feel free to skip the next paragraph if you’re not confused.

A disjoint event means that events cannot happen at the same time. The first disjoint event is “what if all coins come up tails?” That is that the 5 coins {T, T, T, T, T}. The other 4 events are what if one specified coin comes up heads? So the first disjoint event is {H, T, T, T}, the second is {T, H, T, T} etc. Since we need **at least **3 coins to be tails, {H, H, T, T} isn’t valid.

The union of 5 disjoint events is the probability of each event happening added together.

First, lets find out the probability that any probability within this space is possible. The problem space is {H, T} over 4 different coins. Each coin has a 1/2 chance of being heads or tails and there are 4 coins so 4 * 1/2 or 1/2 + 1/2 + 1/2 + 1/2 is 1/16 chance of any possible outcome in the state space.

Therefore the probability of an event is P(1/16)

Know that we know how likely it is to get any combination of {H, T} over the 4 coins we can use this to work out how likely it is to get the 5 disjoint events. Since each event is disjoint, one event does not effect another so it’s just a case of 1/16 * 5 (for the 5 disjoint events) which results in 5/16.

Thus the probability of at least 3 coins coming up tails is 5/16.

## Question

Without looking back up, what is the formula for working out the probability of the union of 2 events? This question is designed to help you retain that information in your head.

# Conditional Probability

Conditional probability is where an event can only happen if another event has happened. Let’s start with an easy problem:

John's favourite programming languages are Haskell and x86 Assembley. Let A represent the event that he forces a class to learn Haskell and B represent the event that he forces a class to learn x86 Assembley.On a randomly selected day, John is taken over by Satan himself, so the probability of P(A) is 0.6 and the probability of P(B) is 0.4 and the conditional probability that he teaches Haskell, given that he has taught x86 Assembley that day is P(A|B) = 0.7.Based on the information, what is P(B|A), the conditional that John teaches x86 Assembley given that he taught Haskell, rounded to the nearest hundredth?

The probability of P(A and B) = P(A|B) * P(B) read “|” as given, as in, “A|B” is read as “A given B”. It can also be written as P(B|A) * P(A).

The reason it is P(A|B) * P(B) is because given the probability of “Given the probability that B happens, A happens” and the probability of B is P(B). (A|B) is a different probability to P(B) and P(A and B) can only happen if P(B) happens which then allows P(B|A) to happen.

So we can transform this into a mathematical formula:

P(A and B) = P(A|B) * P(B) = 0.7 * 0.5 = 0.35Solving it P(B|A) * P(A) P(A) = 0.5 So 0.6 * P(B|A) Now we don't know what P(B|A) is, but we want to find out. We know that P(B|A) must be a part of P(A and B) because P(A and B) is the probability that both of these events happen so...P(A and B) = 0.35 0.35 = P(B|A) * 0.5 With simple algebraic manipulation 0.35/0.5 = P(B|A) P(B|A) = 0.7

For a visual explanation of conditional probability, watch this video by Khan Academy

Or this video

## Bayes Therom

Bayes Therom allows us to work out the probability of events given prior knowledge about the events. It is more of an observation than a therom, as it correctly works all the time. Bayes therom is created by Thomas Bayes, who noted this observation in a notebook. He never published it, so he wasn’t recgonised for his famous therom during his life time. Thomas Bayes was born, lived and died in Royal Tunbridge-wells, Kent — the greatest county in the UK.

The probability of A given B is the probability of B given A (note: it’s reversed here) times by the probability of A divided by the probability of B.

Of course this sounds confusing, so it may help to see an example.

Suppose a new strand of mexican black tar heroin is found on the streets and the police want to identify whether someone is a user or not. The drug is 99% sensitive, that is that the proportion of people who are correctly identified as taking the drug.The drug is 99% specific, that is that the proportion of people who are correctly identified as not taking the drug.Note: there is a 1% false positive rate for both users and non users.Suppose that 0.5% of people at John Moores takes the drug. What is the probability that a randomly selected John Moores student with a positive test is a user?

Once you have all the information, it’s simply a case of substituting in the values and working it out.

Below is a video explaining Bayes Therom intuitively with real-world examples along with the history behind it as well as the philosophy of Bayes Therom:

For a very good article on Baye’s Therom click here:

I know it seems like I haven’t bothered with this section, but basically these experts of mathematics can distill this information better than I ever could.

# Random Variables

A random variable is a function, it is not random or a variable.

A random variable does not need to specify the sample space S directly but assign a probability that a variable (X) takes a certain value. Unlike previous probability where we needed to define a sample space, we only care about the probability itself.

Random variables are often written as P(f=r) where f is the event name and r is the probability.

The probably has to be between 0 and 1, like all probability values.

We write NOT (using any notation you may wish) (F=r) for the event that F is every variable apart from R.

An example of this

```
P(Die=1) = 1/6
The probability that this die takes the value 1 is 1/6
NOT P(Die=1) is the event that the die is
(Die=2) OR (Die=3) OR (Die=4) OR (Die=5) Or (Die=6)
```

The compliment of P(f=r) — the notation used to represent random variables, is 1 — P(f=r), where 1 is 100% or just 1.

We sometimes use symbols (words) instead of numbers to represent random variables. This is really useful. Let’s say the weather can be 1 of 4 states, sunny, rain, cloudy, snow. Thus, instead of assinging Weather = 1 we could write Weather = sunny.

Sometimes it lengthy to write up all the probabilities such as P(Weather = sunny) = 0.7 or P(Weather = rain) = 0.3. If the values are fixed in order then we could write P(Weather) = (0.7, 0.3)

We use bold-face **P **to indicate that the result is a vector of numbers representing the indivudal values of Weather. An example of this is: **P**(Weather) = (0.7, 0.3)

# Joint Probability Distributions

A joint probaiblity distrubition allows you to have multiple random variables, typically 50 or 100 but our examples will include fewer.

This is a joint probability distrubiton for tooth cavities and the weather. Cavity is a boolean value, it is either 0 or 1 and there are 4 options for the weather. If we want to create a joint probability distrubtion of **P**(Weather, Cavity) we would make the table above.

The probability for weather = sunny, and cavity = 1 is 0.144. The probability of the joint distrubtion sums to 1.

# Full joint probability distrubtion

We call it a full joint probability distrubtion if everything that is relevevant in the domain is included. Unlike the above example, cavities and weather aren’t in the same domain.

Suppose you have K random variables in a domain and they are all Boolean, either containing 1 or 0, how many entries does your table have?

Answer: 2 to the power of k.

## Marginalization

One can computer the marginal probabilities of random variables by summing the variables. For example in the above example, if one wanted to sum the probability of P(Cavity=1) then you will sum all the probabilities where the cavity is equal to 1.

P(Cavity = 1) = 0.108 + 0.012 + 0.072 + 0.008 = 0.2

## Conditional / Posteior Probability

We can calculate the conditional / posteior probability of a full joint distribution the same way we would do it normally.

`P(F,G) = P(F|G) * P(G)`

Note that (F, G) signifies F and G.

## (Bayesian) Probablistic Inference

Bayesian Inference is particularly useful because if you have alot less data than you wanted you could use bayesian inference to juice every last bit of information out of the data.

## Example of Bayesian Inference

Let’s say you’re at a nightclub and you notice someone drop their phone. You see them walking away, they look like this from behind:

Do you call out “Excuse me sir” or “Excuse me mam”? Depending on your area, you may think this is a woman.

What if you follow them and see them line up outside the men’s toilets? Well, you would assume they were male.

Bayesian Inference allows us to capture this notion of “guesstimating” probabilities in mathematics.

## Independence to the rescue

If you know the domain, you can take out independent values that do not effect the final product to dramatically reduce the problem. For example, if you are given a full probability distrubtion with the probability of the weather and the probability within the dental domain you can take out the probability of the weather as it is not relevant to dental surgeries.

# Expected Value

The expected value is exactly what it sounds like, what do you expect the value to be? You can use this to work out the average score of a dice roll over 6 rolls, or anything really relating to probability where it has a value property.

Given the outcomes=(1, 2) and the probabilities=(1/8, 1/4) the expected value, E[x] is E[x] = 1(1/8) + 2(1/4) = 0.625.

## Linearity of Expected Value

E[F+G] = E[F] + E[G].

## Multiplication Rule of Expected Value

E[F x G] does not equal E[F] * E[G]

For more information about random variables, make sure to check out this short blog post on them:

If you want to learn about how to use probability for machine learning, this post is perfect!

**Further learning**

https://brohrer.github.io/how_bayesian_inference_works.html