Intro to Probability

Definitions, axioms, and examples

This page covers…

Basic definitions and vocabulary for probability (outcome space, events, probability, impossible events etc)
Probabilities of equally likely outcomes
Axioms of probability
Set notation and Venn diagrams
The box model
Addition rule (probability of the union of mutually exclusive events)

Concept Acquisition

Definition of equally likely outcomes, outcome space, events, impossible event
Definition of probability of an event and first two rules of probability
Unions, intersections, Venn diagrams
Mutually exclusive events and the addition rule

Tool Acquisition

Rules of probability
Using a box of tickets to model equally likely outcomes
New functions: sample() and replicate
Using simulations to estimate probabilities with functions listed above

Concept Application

Computing probabilities in various scenarios
Using R to simulate probabilities

Much of the field of statistics concerns itself with taking a sample of data to try to learn about a larger population. For example, an ice cream company can ask 100 volunteers if they prefer flavor A or flavor B, and they hope to learn which flavor would sell better if offered to everyone in grocery stores.

This is generalization, and it is perhaps the most central task of statistics. Probability is a crucial lodestone in the statistician’s work. This lesson begins our explorations of probability.

In order to be taken seriously as we extend our statements about data sets to larger populations, we need to be careful about how we collect data, and then about how we generalize our findings. For example, we are awash in polls these days, as we head towards the midterms. You may have observed that some polling companies are more accurate than others in their estimates and predictions, and consequently people pay more attention to them.

Below is a snapshot of rankings of polling organizations from the well-known statistician Nate Silver’s Substack newsletter Silver Bulletin.¹, and one can imagine that not many take heed of the polling done by the firms with C or worse grades. According to the website, the rankings are based on the polling organization’s “historical accuracy and transparency.”

When we want to understand voter opinions based on a small poll, or understand the efficacy of a new drug based on a clinical trial of a few hundred participants, or any time we examine a sample of data and make conclusions about a larger group, we have to understand the variations in our estimates introduced by the random nature of taking a sample.

For example, if we conducted the same poll twice, we would get a different group of voters each time, and therefore a different estimate of what the overall population’s opinion is. How different could these estimates be? How concerned should we be? A core part of a statistician’s work is to quantify uncertainty and communicate it clearly to others.

To understand this variation and uncertainty, though, we first have to understand how probability was involved in collecting the data. In these notes, we are going to describe methods by which we can obtain representative samples from the target population. Since these methods are based on probability, we need to first understand the basics of probability or randomness.

We will begin our study of probability right where the subject began: in the salons of 17th century France where high society men and women gathered to gamble, usually with dice and coins.

But wait, you say, dice have been around for thousands of years… well, that is quite true. One of the oldest surviving dice artifacts is the Harappan dice, (the left most image below) - cubical terracotta dice, from the Indus Valley Civilization, about 2500 - 1900 BCE. In the middle we have the famous “Kempten Astragaloi”, ancient knucklebones discovered in Bavaria, dating back to the 1st century BCE – 1st century CE. If we venture into mythologies and legends, we have the famous “Shakuni’s dice” (on the right) which were part of a pivotal game in the ancient Indian epic Mahabharata.

We should say, more correctly, that the study of modern probability began with a French nobleman (and gambler), the Chevalier de Méré, and his gambling woes.

De Méré’s Paradox

Seventeenth century French gamblers would bet on anything. In particular, they would bet on a fair six-sided die landing on 6 at least once in four rolls. Antoine Gombaud, aka the Chevalier de Méré, was a gambler who also considered himself something of a mathematician. He computed the chance of getting at least one six in four rolls as 2/3 by computing $(4 \times (1/6) = 4/6)$. He won quite often by betting on this event, and was convinced his computation was correct. Was it? (Spoiler alert: No! It was wrong!)

The next popular dice game was betting on at least one double-six in twenty-four rolls of a pair of dice. De Méré knew that there were 36 possible outcomes when rolling a pair of dice, and therefore, assuming all the outcomes were equally likely, he computed that the chance of a double six was 1/36. Using this he concluded that the chance of at least one double six in 24 rolls was the same as that of at least one six in four rolls, that is, 2/3 ($=24 \times 1/36$). He happily bet on this event (at least one double six in 24 rolls) but to his shock, lost more often than he won!

What was going on? Why did he win more often than lose in the first situation, and lose more often than win in the second? De Méré got so desperate after losing a lot of money that he consulted one of the great thinkers of that time, Blaise Pascal. Pascal got intrigued by the problem, and began a famous correspondence with another mathematician of the day, Pierre Fermat - and modern probability was born!

We will see later how to compute the probability of de Méré winning his bet, but for now we can estimate (guess at) the value by simulating rolling a die many times (1000 times) and looking at the proportion of times we see at least one six in 4 rolls of a fair die (game 1), and do the same with at least one double-six in 24 rolls (game 2).

Number of simulations = 1000

  prop_wins_game_1
1            0.514

  prop_wins_game_2
1            0.487

You can see here that the poor Chevalier wasn’t as good a mathematician as he imagined himself to be, and didn’t compute the chances correctly. The simulated probabilities are nowhere close to his own calculation of 2/3 for both games!

By the end of this unit, you’ll be able to conduct simulations like these yourself in R. For today, we are going to begin by introducing the conceptual building blocks behind probability.

Basics

First, let’s establish some vocabulary—it’s much easier if all of us are thinking of the same meanings when we use probabilistic terms:

Experiment: An action, involving chance, that can result in a number of possible outcomes (results of the experiment). For example, a coin toss is an experiment, and the possible outcomes are the coin landing heads or tails. Note that we can have finitely many outcomes or infinitely many outcomes.

Outcome space: This is just a set of everything that could happen. The collection of all possible outcomes of an experiment is called an “outcome space” aka “sample space,” and we denote it by the upper case Greek letter $\Omega$ (“Omega”). We will denote a set by enclosing the elements of the set in braces: $\{ \}$.

For example, if we toss a coin, then the corresponding outcome space is $\Omega = \{\text{Heads, Tails}\}$.
If we roll a die, then the corresponding outcome space $\Omega = \{1, 2, 3, 4, 5, 6\}$.

Event: Technically an “event” is just a subset of all possible outcomes. When we run an experiment, depending on the specific outcome, one or more events might “happen.” For example, in rolling a die, we could define event “A” as “rolling a 6” and define event “B” as “rolling an even number.” If we roll a 1, neither event happens. If we roll a 2, event B happens, but not event A. If we roll a 6, both A and B happen. We usually denote events by upper case letters from the beginning of the alphabet: $A, B, C, \ldots$. Because an event is a subset of the outcome space, we denote this by writing $A \subset \Omega$.

P(A): For any event $A$, we write the probability of $A$ as $P(A)$. When all outcomes are equally likely*, we can define the probability of an event $A$ as $P(A) = \dfrac{\text{ size of }A}{\text{ size of }\Omega}$. For example, if our experiment or action is rolling a fair six-sided die, then $\Omega$ has 6 outcomes, and its size is 6. If $A$ is the event that we roll a number less than 3, then $A = \{1, 2\}$, and $P(A) = \dfrac{2}{6} = \dfrac{1}{3}$. Of course, in this particular example, we are assuming here that all the outcomes are equally likely (defined below).

* Technically there must also be a finite number of outcomes. This is nuanced, and we will return to it when we get to continuous probability distributions.

Equally likely outcomes: This means exactly what you think it might: When all the possible outcomes in a finite outcome space of size $n$ happen with the same probability, which is $\displaystyle \frac{1}{n}$. Rolling a die is a classic example: each of the six possible outcomes has a 1/6 probability of occurring.

Let’s say that there are $n$ possible outcomes in the outcome space $\Omega$, and an event $A$ has $k$ possible outcomes out of those $n$. If all the outcomes are equally likely to happen (as in a die roll or coin toss), then we say that the probability of $A$ occurring is $\displaystyle \frac{k}{n}$.

\[ P(A) = \frac{k}{n} \]

Example: Tossing a fair coin

Suppose we toss a fair coin. What is the chance of the coin landing heads? Most people would guess 50%, reasoning that there are two possible things that can happen, and if the coin is fair, then they are both equally likely, making the probability of heads 1/2 or 50%.

Here, we have thought about an event: the coin landing heads. We observed that there is one outcome in that event, and two (equally likely) outcomes in the outcome space, so we say the probability of the event, $P(\text{Heads})$, is 1/2.

Example: Tossing a fair six-sided die²

Consider rolling a fair six-sided die: six outcomes are possible so $\Omega = \{1, 2, 3, 4, 5, 6\}$. Since the die is fair, each outcome is equally likely, with probability $= \displaystyle \frac{1}{6}$. We can list the outcomes and their probabilities in a table.

Outcome	$1$	$2$	$3$	$4$	$5$	$6$
Probability	$\displaystyle \frac{1}{6}$	$\displaystyle \frac{1}{6}$	$\displaystyle \frac{1}{6}$	$\displaystyle \frac{1}{6}$	$\displaystyle \frac{1}{6}$	$\displaystyle \frac{1}{6}$

Let $A$ be the event that an even number is rolled. Then the set $A$ can be written $\{2,4,6\}$. Since all of these outcomes are equally likely:

\[P(A) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6}\]

Axioms of probability

In order to compute the probabilities of events, we need to set some basic mathematical rules called axioms (which are intuitively clear if you think of the probability of an event as the proportion of the outcomes that are in it). There are three basic rules that we use to compute probabilities, and here are the first two:

Axiom 1: The chance of any event is at least $0$: $P(A) \ge 0$ for any event $A$. In other words, there are no negative probabilities.
Axiom 2: The chance of an outcome being in $\Omega$ is $1$: $P(\Omega) = 1$. This follows by definition - $\Omega$ is defined as the set of every single thing that could happen. Hence, whatever happens must be in $\Omega$. In the case of equally-likely outcomes, we also see that the probability of $\Omega$ is the number of outcomes in $\Omega$ divided by $n$, which is $n/n = 1$.

Before we write the third rule, we need some more definitions and notation:

Impossible event: An event with no outcomes in it. Denoted by either empty braces $\{\}$ or the symbol for the empty set $\emptyset$. Note that this is a symbol representing a set, and is not the same thing as the number 0. The probability of the impossible event is $0$.

For example, if you roll a standard 6-sided die, the event of “rolling an 8” is impossible. There is no 8 anywhere on a standard die, so 8 is not in the outcome space.

Union of events: Given events $A$, $B$, we can define a new event called $A$ or $B$, which consists of all the outcomes that are either in $A$ or in $B$ or both. This is also written as $A \cup B$, read as “$A$ union $B$.”

Example: let’s say you roll a standard 6-sided die, and define these two events:

$A = \{1,2,3\}$ (the roll is 3 or less)
$B = \{1,3,5\}$ (the roll is odd)

Then the event $A \cup B = \{1,2,3,5\}$. You’ll notice that the number 2 is only in $A$, the number 5 is only in $B$, and the numbers 1 and 3 are in both $A$ and $B$. The union of $A$ and $B$ contains all of these numbers.

Put differently, in order for a die roll to be 3-or-less, odd, or both, we need to roll a 1, 2, 3, or 5.

Intersection of events: Given events $A$, $B$, we can define a new event called $A$ and $B$, which consists of all the outcomes that are in both $A$ and $B$. This is also written as $A \cap B$, read as “$A$ intersect $B$.”

Same example: let’s say you roll a standard 6-sided die, and define these two events:

$A = \{1,2,3\}$ (the roll is 3 or less)
$B = \{1,3,5\}$ (the roll is odd)

Then the event $A \cap B = \{1,3\}$. These are the only outcomes in both events.

Put differently, in order for a die roll to be both odd and 3-or-less, we need to roll a 1 or a 3.

Now we consider events that don’t intersect or overlap at all, that is, they are disjoint from each other, or mutually exclusive:

Mutually exclusive events: If two events $A$ and $B$ cannot both happen at once, we say that the events are mutually exclusive. Mutually exclusive events are sets that do not overlap at all, and therefore have no outcomes in common.

A die roll cannot be both even and odd at the same time. A single coin flip cannot be both heads and tails. Such events are mutually exclusive.

If $A$ and $B$ are mutually exclusive, then we know that if one of them happens, the other one cannot. If a die roll is even, then it is definitely not odd. We denote this by writing $A \cap B = \emptyset$ and read this as “$A$ intersect $B$ is empty”. Therefore, for mutually exclusive events:

\[P(A \cap B) = P(\emptyset) = 0\]

The event that a die roll is even $\{2,4,6\}$ and the event that a die roll is odd $\{1,3,5\}$ are mutually exclusive. They have no outcomes in common, and therefore cannot both happen.

However, the event that a die roll is even $\{2,4,6\}$ and the event that a die roll is a prime number $\{2,3,5\}$ are not mutually exclusive. The number 2 is both even and prime, so both events could happen.

Here’s another example that might interest soccer fans: The event that Arsenal wins the English Premier League (EPL) this season, and the event that Manchester City wins the EPL this season, are mutually exclusive, since you cannot have two teams winning the premier league in the same season. However, the event that Arsenal wins the EPL this season and the event that Manchester City wins the EPL next season are not mutually exclusive. Both could happen.

Now for the third axiom:

Axiom 3: If $A$ and $B$ are mutually exclusive ($A \cap B = \emptyset$), then

\[P(A \cup B) = P(A) + P(B)\]

That is, for two mutually exclusive events, the probability that either of the two events might occur is the sum of their probabilities. This is called the addition rule.

For example, consider rolling a fair six-sided die, and the two events $A$ and $B$, where $A$ is the event of rolling a multiple of $5$, and $B$ is the event that we roll a multiple of $2$.

The only outcome in $A$ is $5$, while $B$ has $\{2, 4, 6\}$. $P(A) = 1/6$, and $P(B) = 3/6$. Since $A \cap B =\emptyset$ ($A$ and $B$ have no outcomes in common), we have that

\[P(A \cup B) = P(A) + P(B) = \frac{1}{6} + \frac{3}{6} = \frac{4}{6}\]

You can only add probabilities like this IF the events are mutually exclusive. Why? Consider:

Event A: a die roll is 2 or higher.
- That is, $A = \{2,3,4,5,6\}$
- So $P(A) = 5/6$
Event B: a die roll is 4 or lower.
- That is, $B = \{1,2,3,4\}$
- So $P(B) = 4/6$

These events are NOT mutually exclusive. They can both be true if we roll a 2, 3, or 4.

If you try to compute $P(A \cup B)$ (“A or B”) by adding $P(A) + P(B)$, you get $5/6 + 4/6 = 9/6$ which is 150%. You can’t have a > 100% chance of anything happening! (See Axiom 2).

Why didn’t this work? Well, you’ve double-counted the numbers that satisfy both events (2, 3, and 4). You can only add probabilities directly if you have nothing to double-count, which only happens when the events are mutually exclusive.

The complement rule

Here is an important consequence of axiom 3. Let $A$ be an event in $\Omega$. The complement of $A$, written as $A^C$, consists of all those outcomes in $\Omega$ that are not in $A$. For example, if we roll a die and define event $A$ as rolling a 3 or higher $\{3, 4, 5, 6\}$, then the complement $A^C$ is rolling a 2 or lower $\{1, 2\}$. This leads to the following complement rule:

\[P(A) + P(A^C) = 1\]

Any event will either happen ($A$) or it won’t ($A^C$). One of those has to be true. A coin flip has to be either Heads or “not Heads.” A die roll must be either “greater than 3” or “not greater than 3.” This is because $A \cup A^C = \Omega$, and $A \cap A^C = \emptyset$.

It may feel silly, but this is a surprisingly useful rule. Sometimes a probability is really hard to calculate, but its complement is easy to calculate. So you calculate the complement instead, and subtract from 1.

An example using the axioms and our favorite birds (penguins, not shoebills)

Consider the penguins dataset, which has 333 observations, of which 146 are Adelie penguins and 68 are Chinstrap penguins. Suppose we pick a penguin at random. What is the probability that we would pick an Adelie penguin? What about the probability of picking a Gentoo penguin?

Check your answer

Let $A$ be the event of picking an Adelie penguin, $C$ be the event of picking a Chinstrap penguin, $G$ be the event of picking a Gentoo penguin.

Assuming that all the penguins are equally likely to be picked, we see that then $P(A) = 146/333$, and $P(C) = 68/333$. While we could subtract a bit to find $P(G)$, we’ll do it with the complement rule.

Since only one penguin is picked, we see that $A, C$, and $G$ are mutually exclusive. This means that $P(A)+P(C)+P(G) = 1$, since $A, C$, and $G$ together make up all of $\Omega$.

Rearranging the algebra a little, we unveil what is, essentially, the complement rule:

\[P(G) = 1 - (P(A) + P(C)) = 1 - P(G^C) = 1 - 214/333 = 119/333\]

WARNING!!

We use $A$ to denote an event or a set, while $P(A)$ is a number - you can think of $P(A)$ as representing the relative size of $A$. This means that the following types of statements don’t make sense as we haven’t defined what it means to add sets or union numbers etc.:

$P(A) \cup P(B)$ or $P(A) \cap P(B)$
$A + B$, or $A - B$, or $A \times B$ etc

Some Examples

1. Tossing a fair coin

Suppose we plan to toss a coin twice, but first we want to write down all of the equally likely outcomes. What is $\Omega$? And looking at this $\Omega$, what is the chance of at least one head?

Solution: $\Omega = \{HH, HT, TH, TT\}$, where $H$ represents the coin landing heads, and $T$ represents the coin landing tails. Note that since we can get exactly one head and one tail in two ways, we have to write out both ways so that all the outcomes are equally likely.

Now, let $A$ be the event of getting at least one head in two tosses. We can do this by listing the outcomes in $A$: $\{HH, HT, TH\}$ and so $P(A) = 3/4$.

Alternatively, we can consider $A^C$ which is the event of no heads, so $A^C = \{TT\}$ and $P(A^C) = 1/4$.

By the complement rule, $P(A) = 1- P(A^C) = 1-1/4 = 3/4$.

Now you try: Let $\Omega$ be the outcome space of tossing a coin three times. What is the probability of at least one head? What about exactly one head?

Check your answer

$\Omega = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \}$.

Let $A$ be the event of at least one head. Then $A^C$ is the event of no heads, so $A^C = \{TTT\}$, and $P(A^C) = 1/8$. Therefore $P(A) = 1-1/8 = 7/8$. Note that this is much quicker than listing and counting the outcomes in $A$.

If $B$ is the event of exactly one head, then $B = \{HTT, THT, TTH\}$ and $P(B) = 3/8$.

2. A box of tickets

Consider the box above which has five almost identical tickets. The only difference is the value written on them. Imagine that we shake the box to mix the tickets up, and then draw one ticket without looking so that all the tickets are equally likely to be drawn³.

What is the chance of drawing an even number?

Check your answer

Solution:

Let $A$ be the event of drawing an even number. Then $A = \{2, 2, 4\}$: we list 2 twice because there are two tickets marked 2, making it twice as likely as any other number. $P(A) = 3/5$.

Note: a “set” in mathematics does not allow for repeated elements. But here, we actually don’t have repeated elements - there are three distinct tickets! We simply wrote their values in set notation for simplicity. There are two different tickets with a 2 on them, so it is a set (of tickets).

Venn Diagrams

We often represent events using Venn diagrams. The outcome space $\Omega$ is usually represented as a rectangle, and events are represented as circles inside $\Omega$. Remember, $\Omega$ is literally everything that could possibly happen. Here is a Venn diagram showing two events $A$ and $B$, their intersection, and their union:

Here is a Venn diagram showing two mutually exclusive events (no overlap):

The Box Model

One of the simplest ways of understanding probabilities is through the box model⁴ like the one earlier. A box model consists of a box with numbered tickets, from which tickets are drawn.

To specify a box model, we have to say:

What tickets go in the box (that is, what are the numbers on the tickets, and how many of each)
How many tickets will we draw from this box
Whether the tickets will be drawn with or without replacement.
- The difference is this: after drawing a ticket from the box, do we put it back in the box before we draw again (“with replacement”) or leave it out of the box (“without replacement”)?
- If with replacement, note that we could draw the same ticket multiple times.
- If without replacement, note that we will never draw the same ticket twice. And the box changes between draws as tickets are removed.

A box of tickets

We saw this box earlier. As long as we shake the box up, and don’t look when we draw, each ticket has an equally likely chance to be picked. For example, the chance of drawing a ticket labeled “4” is one in five, as there are five tickets to choose from and only one is labeled “4” etc. We will come back to this box in the next chapter. For now, what is the chance of drawing a ticket labeled “2”?

With or without replacement?

When we make multiple draws at random with replacement, we draw one ticket, and put it back before drawing another ticket. For the box in this section, if we draw twice, both times we will draw from the following box:

If we draw twice at random without replacement, then our second draw is from a different box, illustrated below:

The purpose of creating a box model is to understand the variability in a “random process” such as tossing a coin over and over and counting the number of heads, or sampling voters in Texas to ask them if they will vote for James Talarico or Ken Paxton in November’s Senate elections. We do this by having the tickets represent the possible outcomes associated with the random process (of say, sampling voters), and using the variability in the tickets that are drawn, to understand the variability of the random process we are actually interested in.

Let’s review what we need in order to create a box model that will model a random process. We need to specify:

The box and tickets (what tickets go in the box, and the values written on them)
The number of draws
Whether or not we replace tickets between draws
How we will summarize the values on the drawn tickets (usually take the sum of the tickets drawn or the average of tickets drawn)

Note that these tickets should be identical in every way, except the value written on them, and assume that the tickets in the box are equally likely to be drawn (we draw without looking at the box).

Setting up a box model: examples

Let’s practice creating boxes to model common scenarios. For each of the cases below, we should specify the four items above.

Example 1: Tossing a fair coin and counting the number of heads

Tossing a fair coin once

What box and which tickets would you use to model a single coin toss?

Check your answer

We could draw one ticket from the box but these are not numbered!

Or we could draw one ticket from the box

with the ticket $\fbox{1}$ representing the coin landing heads, and $\fbox{0}$ representing the coin landing tails. This change of numbering the ticket representing the outcome we are counting with a 1 and all other tickets with a 0 is very important, since we are classifying the outcomes into two categories, and counting the instances of one of them (in this case, the coin landing heads).

Tossing a fair coin twice

How would you use a box model to simulate this?

Check your answer

We could use the same box as above:

Remember that the ticket $\fbox{1}$ represents the outcome of a toss landing heads and $\fbox{0}$ represents a toss landing tails (because we are counting the number of heads). We let the box be as above, and draw two tickets at random with replacement from this box. To count the number of heads in two tosses, we add the draws.

Tossing an unfair coin

How would you use the box model for tossing a coin which has chance of 5/6 landing heads, and counting the number of heads?

Check your answer

We do the same trick, of representing the outcome of heads with $\fbox{1}$, and tails with $\fbox{0}$. Since the coin lands heads 5 times out of 6, the box is given by:

We draw as many times as the number of tosses, replacing the ticket each time,and sum up the values on the tickets drawn.

Example 2: Rolling a fair die and summing the spots

⁵

Rolling a fair die once

How would you use a box to represent this experiment?

Check your answer

The box will have six tickets as shown below, and we would draw one ticket from this box. The chance of any one of the tickets is $1/6$.

Rolling a fair die twice

How would you use a box to represent this experiment?

Check your answer

We would use the same box as in the previous example, draw twice at random with replacement from this box, then add the draws.

Example 3: Betting on red in roulette

⁶

An American roulette wheel has 38 pockets, of which 18 are red, 18 black, and 2 are green. The wheel is spun, and a small ball is thrown on the wheel so that it is equally likely to land in any of the 38 pockets. Players bet on which colored or numbered pocket the ball will come to rest in. If you bet one dollar that the ball will land on red, and it does, you get your dollar back, and you win one dollar, so your gain is $1. If it doesn’t, and lands on a black or green number, you lose your dollar, and your “gain” is -$1. What would be the box model for your gain from a single spin?

Gain from a single spin

How would you set up a box to represent this experiment?

Check your answer

We would use the box shown below marked “Gain”. In the top box, there are 38 tickets, each representing a numbered pocket on the wheel. Draw one ticket to represent the ball landing in a particular pocket. In the second row each ticket represents a colored pocket, and we see that there are 18 red pockets, 18 black, and 2 green. We only care about the outcome ball lands in red pocket, and our gain from this. We would draw once from it, with the ticket marked $\fbox{+1}$ representing our gain if the ball lands on a red pocket, and the ticket $\fbox{-1}$ representing our loss if the ball doesn’t land on red. We can see that the chance of drawing a $\fbox{+1}$ is 18 out of 38.

Gain from 10 spins

How would we use a box model to represent our net gain or winnings from 10 spins (we bet $1 on each spin, and either lose it or get it back plus $1)?

Check your answer

Use the same box as above, with 18 tickets marked $\fbox{+1}$ and 20 marked $\fbox{-1}$, draw 10 times with replacement and sum the draws.

Example 4: Antarctic penguins

You have seen the penguins data, which measured 8 variables on a sample of 333. Here the box will contain 333 tickets, and we will draw 50 tickets at random without replacement. Note that each ticket in the box represents one penguin in the sample and has information for multiple variables measured on that penguin, and we would consider only the information we are interested in. Suppose we wanted sex, species, and flipper length, to compare the flipper lengths of males and females, grouped by species.

Check your answer

The box would have 333 tickets, one for each penguin. Each ticket would have three things written on it: the species, sex, and flipper length in mm of the penguin. We would draw 50 tickets at random without replacement. Below is an example ticket. The left hand side has values for all 8 variables for one of the sampled penguins, but we only care about 3 of the variables. So when we draw our tickets from the box, we can think about them looking like the right hand side ticket. Now we can summarize the sample data to look at the average flipper length for males and females of each species.

Example 5: Weighing a donkey in the Kenyan countryside

⁷

Donkeys are economically vital in rural Kenya, being cheaper than horses or cattle, and playing all kinds of roles - used for transport, farming etc. When they fall sick, it is crucial for veterinarians to know their weight to dose their medications correctly. The problem is that they don’t usually have scales. A study showed that the vet can obtain the weight indirectly using a statistical method called a nomogram, that is, the weight of the donkey is estimated using measurements that were easier to obtain. The study obtained the data by measuring donkeys at 17 sites in 2010⁸.

Four body measurements were made for each donkey: liveweight (kg), heart girth (cm), height (cm), and length (cm), and in addition, sex and body condition score (BCS) were recorded (the BCS categorized the donkeys according to their condition, ranging from “emaciated” to “obese”).

Suppose I want to sample 50 donkeys from this dataset, to compute their average weight and average girth (girth or heart girth is the circumference of the body, measured just behind the front legs.)

How would you use a box model to represent this process (sampling donkeys)?

Check your answer

The box has 544 tickets, one for each donkey. The ticket has information from all 8 variables recorded (3 categorical and 5 numerical). We only need 2 variables, so the tickets will have those two variables recorded. We draw 50 tickets from this box at random without replacement, and compute 2 averages from the draws.

Summary

In this lecture, we introduced equally likely outcomes, and defined the outcome space of an experiment.
Then, using equally likely outcomes, we defined the probability of an event as the ratio of the number of outcomes in the event to the number of total outcomes in the outcome space.
We wrote down the axioms (fundamental rules) of probability, after defining unions, intersections, and mutually exclusive events and Venn diagrams.
We also encountered an extremely useful construct: The box model. You will see how useful this is already when you go through the tutorial, but we will see the real power of the box model when we study variability.

Continue on to the tutorial ⮕

Footnotes

You might have heard of a website called FiveThirtyEight, which was begun by Nate Silver. That was where he originally published his pollster rankings, but ABC News (and Disney+) bought FiveThirtyEight, and then eventually shut it down. Nate Silver had moved away by then, and begun his newsletter.↩︎
The singular is die and the plural is dice. If we use the word “die” without any qualifiers, we will mean a fair, six-sided die.↩︎
We call the tickets equally likely when each ticket has the same chance of being drawn. That is, if there are $n$ tickets in the box, each has a chance of $1/n$ to be drawn. We also refer to this as drawing a ticket uniformly at random, because the chance of drawing each ticket is the same, or uniform.↩︎
The box model was introduced by Freedman, Pisani, and Purves in their textbook Statistics↩︎
Photo via unsplash.com↩︎
Photo via unsplash.com↩︎
https://people.maths.bris.ac.uk/~mazjcr/#donkeys ↩︎
K. Milner and J.C. Rougier, 2014. How to weigh a donkey in the Kenyan countryside. Significance, 11(4), 40–43. 74, 115 https://rss.onlinelibrary.wiley.com/doi/full/10.1111/j.1740-9713.2014.00768.x ↩︎

Concept Acquisition

Tool Acquisition

Concept Application

De Méré’s Paradox

Basics

Example: Tossing a fair coin

Example: Tossing a fair six-sided die2

Axioms of probability

An example using the axioms and our favorite birds (penguins, not shoebills)

Some Examples

1. Tossing a fair coin

2. A box of tickets

Venn Diagrams

The Box Model

A box of tickets

With or without replacement?

Setting up a box model: examples

Example 1: Tossing a fair coin and counting the number of heads

Example 2: Rolling a fair die and summing the spots

Example 3: Betting on red in roulette

Example 4: Antarctic penguins

Example 5: Weighing a donkey in the Kenyan countryside

Summary

Footnotes

Example: Tossing a fair six-sided die²