Introduction to Stats · Chapter 3 · Probability Topics

Terminology

Every rule the rest of this chapter uses is stated in a handful of words and symbols — experiment, sample space, event, OR, AND, complement, conditional. This section is where they get their exact meanings.


bookSHelf  ·  Introduction to Statistics  ·  §3.1  ·  a self-paced section

Terminology · bookSHelf Intro Stats§3.1

Outline — by the end of this section you will be able to

Objectives

  1. Describe an experiment, list its sample space, and identify an event as a subset of that sample space definition
  2. Calculate the probability of an event when every outcome is equally likely, and explain what that number means over the long run §3.1.2
  3. Read and write the notation for OR, AND, and complement events, and find each one from a listed sample space §3.1.4
  4. Calculate a conditional probability and explain how conditioning shrinks the sample space §3.1.5
  5. Translate a sentence about a real situation into probability symbols, and translate symbols back into a sentence §3.1.6
Terminology · bookSHelf Intro Stats§3.1

§3.1.1 — Experiments, Outcomes, and Sample Spaces

Three words everything else builds on

Before probability can be a number, it needs a vocabulary. This subsection supplies the three terms that name what you do, what happens, and where you record it.

What you do

An experiment is the procedure — flip a coin, roll a die, draw a card.

Where you record it

The sample space is every possible outcome, listed once each, and an event is any subset of it you care about.

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§3.1.1 — the first term

Experiment

Definition 3.1.1 — Experiment

An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, the experiment is called a chance experiment.


Flipping one fair coin twice is an experiment. So is rolling a die, drawing a card, or selecting one student at random from a roster. The word does not mean a laboratory — it means a procedure you could repeat, whose result you cannot call in advance.

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§3.1.1 — the second term

Outcome

Definition 3.1.2 — Outcome

An outcome is a result of an experiment.

Every flip, roll, or draw produces exactly one outcome — the specific thing that happened, not the list of things that could have.

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§3.1.1 — the third term

Sample Space

Definition 3.1.3 — Sample Space

The sample space of an experiment is the set of all possible outcomes. The uppercase letter SS denotes the sample space.

There are three standard ways to display a sample space: list the outcomes, draw a tree diagram, or draw a Venn diagram. If you flip one fair coin, S={H,T}S = \{H, T\} — two outcomes, and the set of them is the sample space.

Definition 3.1.3: the sample space of two coin flips — every possible result appears in S exactly once.

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Insight Note — the sample space is the whole board

Everything that can happen has to be in S

Everything that can happen has to appear in SS somewhere, exactly once.

If you can describe a result that is not in your list, your list is not the sample space yet.

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§3.1.1 — naming a question about the outcome

Event

Definition 3.1.4 — Event

An event is any combination of outcomes — that is, any subset of the sample space. Uppercase letters such as AA and BB represent events. The probability of an event AA is written P(A)P(A).

The word combination is doing real work: an event does not have to be a single outcome. An event is a question you ask about the result — the answer is yes for some outcomes and no for the others.

Definition 3.1.4: an event is a subset of the sample space — the outcomes that answer yes are the event.

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Your turn — a four-wedge spinner

Try It Now 3.1.1

You spin a spinner with four equal wedges labelled 1, 2, 3, 4. (a) Write the sample space SS. (b) Let AA be the event the spinner lands on an odd number. (c) Let BB be the event it lands on a number greater than 4.


(a) S={1,2,3,4}S = \{1, 2, 3, 4\}. (b) The odd labels are 1 and 3, so A={1,3}A = \{1, 3\}. (c) No label is greater than 4, so B={ }B = \{\ \} — the empty set is still a legitimate event; it just never happens.

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§3.1.2 — Probability and the Law of Large Numbers

Turning outcomes into a number

Once you can name outcomes and events, the next question is how likely each one is. This subsection defines probability itself, the shortcut that works when outcomes are equally likely, and the law that explains why the shortcut works at all.

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§3.1.2 — the long-run definition

Probability of an Outcome

Definition 3.1.5 — Probability of an Outcome

The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive.

P(A)=0P(A) = 0 means AA can never happen; P(A)=1P(A) = 1 means AA always happens. Flip a coin 20 times, then 2,000, then 20,000, tracking the fraction of heads — it wanders early and settles onto 0.5.

Definition 3.1.5 — Every probability is a position on one line from 0 to 1.

Definition 3.1.5: every probability is a position on one line from 0 to 1.

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§3.1.2 — when counting is enough

Equally Likely

Definition 3.1.6 — Equally Likely

Equally likely means that each outcome of an experiment occurs with equal probability.

When every outcome is equally likely, computing a probability becomes counting:

P(A)=number of outcomes in Anumber of outcomes in SP(A) = \frac{\text{number of outcomes in } A}{\text{number of outcomes in } S}

Toss a fair dime and nickel together: S={HH,HT,TH,TT}S = \{HH, HT, TH, TT\}. Let AA be exactly one head — HTHT and THTH qualify, so P(A)=24=0.5P(A) = \frac{2}{4} = 0.5.

Definition 3.1.6 — A fair die: six faces, six identical probabilities.

Definition 3.1.6: a fair die — six faces, six identical probabilities.

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§3.1.2 — why the counting shortcut is trustworthy

Law of Large Numbers

Definition 3.1.7 — Law of Large Numbers

As the number of repetitions of an experiment increases, the relative frequency obtained tends to become closer and closer to the theoretical probability.

The outcomes do not arrange themselves into a pattern to make this happen. Each trial is as unpredictable as the last — what changes is that the accumulated fraction has more trials underneath it, so any one strange run matters less.

Definition 3.1.7: the running fraction of heads settles onto 0.5 as the flips pile up.

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Context Pause — why the long run is the whole point

A single trial tells you nothing

A single flip tells you nothing about whether a coin is fair.

An insurance company cannot predict which house will burn, but it can predict what fraction will, and it prices policies off that fraction. Probability is a statement about many trials, not about the next one.

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Your turn — is 16 rolls enough to judge?

Try It Now 3.1.2

A fair, eight-sided die has faces 1–8. Let AA be rolling a number less than 3. (a) Find P(A)P(A). (b) Marisol rolls 16 times and gets a qualifying number only once, and concludes the die is unfair. Is her conclusion supported?


(a) A={1,2}A = \{1, 2\}, so P(A)=28=0.25P(A) = \frac{2}{8} = 0.25. (b) We'd expect about (0.25)(16)=4(0.25)(16) = 4 qualifying rolls; she got 1. That's a gap, but 16 trials is far too few — the law of large numbers makes no promise about 16 rolls, only about the long run.

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Try it in rāSHio

Watch the fraction settle

Open rāSHio and choose File → Random Numbers… to generate a column of a few hundred whole numbers from 1 to 2, treating 1 as heads.

Read the running share of heads down the column: it lurches over the first handful of draws, then tightens onto 0.5. Generate a short column and a long one and compare — the long one is not more random, it is just further along.

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§3.1.2 — reading it off the panel

rāSHio: File → Random Numbers…

Figure 3.1.1: generating a column of random whole numbers in rāSHio — File → Random Numbers….

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§3.1.3 — When the Outcomes Are Not Equally Likely

The counting shortcut has a condition

Dividing by the size of SS only works when every outcome is equally likely. This subsection names what it means for that condition to fail.

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§3.1.3 — when outcomes are not equally likely

Biased

Definition 3.1.8 — Biased

A coin, die, or other chance device is biased when its outcomes are not equally likely — that is, when at least one outcome occurs with a different probability than another.

Two math professors had their students test the Belgian one-euro coin. In 250 trials, heads came up 56% of the time and tails 44% — though 250 trials is not many, and more repetitions would be needed to draw a firm conclusion.

Definition 3.1.8 — The Belgian euro coin came out 56/44 over 250 trials, not the 50/50 a fair coin predicts.

Definition 3.1.8: the Belgian euro coin came out 56/44 over 250 trials, not the 50/50 a fair coin predicts.

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Insight Note — a biased die still has a sample space

Bias changes how often, not what

Bias does not change what can happen, only how often. S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\} either way.

What breaks is the shortcut of dividing by 6 — you now need the individual probabilities, which later sections give you tools for.

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Your turn — a three-color carnival spinner

Try It Now 3.1.3

Devon's spinner has red taking half the circle, blue and green splitting the rest evenly. (a) Write SS. (b) Devon claims P(red)=13P(\text{red}) = \frac{1}{3} — why is that wrong? (c) Find each probability.


(a) S={red,blue,green}S = \{\text{red}, \text{blue}, \text{green}\}. (b) Dividing by 3 assumes the three colors are equally likely, and they are not. (c) Probability follows area: P(red)=12P(\text{red}) = \frac{1}{2}, P(blue)=P(green)=14P(\text{blue}) = P(\text{green}) = \frac{1}{4}.

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§3.1.4 — OR, AND, and the Complement

Combining events three ways

Once you can name single events, you need a way to combine them. Three combinations cover almost everything: one event or the other, both at once, and everything an event leaves out.

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§3.1.4 — either one, or both

“OR” Event

Definition 3.1.9 — “OR” Event

An outcome is in AA OR BB if it is in AA, in BB, or in both. Also written ABA \cup B (union). Unlike everyday English, mathematics includes the overlap.

Let A={1,2,3,4,5}A = \{1,2,3,4,5\}, B={4,5,6,7,8}B = \{4,5,6,7,8\}. Then AB={1,2,3,4,5,6,7,8}A \cup B = \{1,2,3,4,5,6,7,8\} — 4 and 5 are not listed twice; a set records what is in it, not how many ways an element got there.

Definition 3.1.9: A OR B covers both circles, and the overlap is shaded once, not twice.

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§3.1.4 — only the overlap

“AND” Event

Definition 3.1.10 — “AND” Event

An outcome is in AA AND BB if it is in both AA and BB at the same time. Also written ABA \cap B (intersection).

Using the same two sets from Definition 3.1.9, AB={4,5}A \cap B = \{4, 5\} — only the elements that appear in both lists.

Definition 3.1.10 — A AND B is only the lens: the outcomes in both sets at the same time.

Definition 3.1.10: A AND B is only the lens — the outcomes in both sets at the same time.

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§3.1.4 — everything an event leaves out

Complement

Definition 3.1.11 — Complement

The complement of event AA, denoted AA', consists of all outcomes NOT in AA. Because every outcome is in AA or AA' and never both,

P(A)+P(A)=1P(A) + P(A') = 1

Let S={1,,6}S = \{1,\ldots,6\}, A={1,2,3,4}A = \{1,2,3,4\}. Then A={5,6}A' = \{5,6\}, and 46+26=1\frac{4}{6} + \frac{2}{6} = 1.

Definition 3.1.11 — A and its complement tile the whole sample space, so their probabilities add to 1.

Definition 3.1.11: A and its complement tile the whole sample space, so their probabilities add to 1.

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Insight Note — the complement is the cheap route

Count backwards for “at least one”

“At least one” problems are usually miserable to count directly and easy to count backwards.

The complement of “at least one head” is “no heads at all”, which is a single outcome. Subtract from 1 and you are done.

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Your turn — union, intersection, complement

Try It Now 3.1.4

Let S={1,,10}S = \{1,\ldots,10\}, A={2,4,6,8,10}A = \{2,4,6,8,10\}, B={1,2,3,4,5}B = \{1,2,3,4,5\}. Find (a) AA OR BB and its probability, (b) AA AND BB and its probability, (c) BB' and its probability.


(a) AB={1,2,3,4,5,6,8,10}A \cup B = \{1,2,3,4,5,6,8,10\}, P=810=0.8P = \frac{8}{10} = 0.8. (b) AB={2,4}A \cap B = \{2,4\}, P=210=0.2P = \frac{2}{10} = 0.2. (c) B={6,7,8,9,10}B' = \{6,7,8,9,10\}, P=510=0.5P = \frac{5}{10} = 0.5 — check: 0.5+0.5=10.5 + 0.5 = 1.

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§3.1.5 — Conditional Probability

When you already know something

Sometimes you learn something before you have to make a judgement. Conditional probability is how that extra information gets built into the number.

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§3.1.5 — conditioning shrinks the sample space

Conditional Probability

Definition 3.1.12 — Conditional Probability

P(AB)P(A \mid B) is the probability AA occurs given that BB already has. Conditioning reduces the sample space to BB:

P(AB)=P(A AND B)P(B),P(B)>0P(A \mid B) = \frac{P(A \text{ AND } B)}{P(B)}, \qquad P(B) > 0

Roll a die: let AA be 2-or-3, BB be even, so B={2,4,6}B = \{2,4,6\}. Inside BB, only 2 qualifies, so P(AB)=13P(A \mid B) = \frac{1}{3} — by the formula, 1/63/6=13\frac{1/6}{3/6} = \frac{1}{3} too.

Definition 3.1.12: conditioning on B discards the outcomes outside B, so the denominator drops from 6 to 3.

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Context Pause — a test result is a condition

P(AB)P(A \mid B) is not P(A)P(A)

A screening test that is positive does not tell you the probability of having the disease — it tells you the probability given a positive test.

Confusing P(AB)P(A \mid B) with P(A)P(A) is one of the most expensive mistakes in medicine and law, and it is the whole reason this notation exists.

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Your turn — why the two answers differ

Try It Now 3.1.5

A fair die is rolled. AA = roll greater than 3, BB = roll is odd. (a) Find P(A)P(A). (b) Find P(AB)P(A \mid B). (c) Why do the two answers differ?


(a) A={4,5,6}A=\{4,5,6\}, P(A)=12P(A)=\frac{1}{2}. (b) B={1,3,5}B=\{1,3,5\}; only 5 also satisfies AA, so P(AB)=13P(A\mid B)=\frac{1}{3}. (c) Knowing the roll is odd removes 4 and 6, two of the three faces that made AA true.

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§3.1.6 — Understanding Terminology and Symbols

Two phrasings that look alike and are not

“Of all the …”

“Of all the women, the probability a student has short hair” names the reduced sample space: P(short hairwoman)P(\text{short hair} \mid \text{woman}).

“… and …”

“The probability a student is a woman and has short hair” is computed over everyone: P(woman AND short hair)P(\text{woman AND short hair}).

Those two sentences describe different numbers, and the only difference in the English is a handful of small words. When it is already clear the process is “rolling a die,” P(rolling a 1)P(\text{rolling a } 1) shortens to P(1)P(1).

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Your turn — every symbol, off one sample space

Try It Now 3.1.6

SS is all ordered pairs (1–3, 1–4). AA = sum is even, BB = first number is prime. Find SS, AA, BB, P(A)P(A), P(B)P(B), AA AND BB, AA OR BB, their probabilities, AA', P(A)P(A'), and both conditionals.


12 pairs total. P(A)=612=12P(A) = \frac{6}{12} = \frac{1}{2}, P(B)=812=23P(B) = \frac{8}{12} = \frac{2}{3}. P(A AND B)=13P(A \text{ AND } B) = \frac{1}{3}, P(A OR B)=56P(A \text{ OR } B) = \frac{5}{6}. P(A)=12P(A') = \frac{1}{2}, and P(A)+P(A)=1P(A)+P(A')=1. P(AB)=48=12P(A \mid B) = \frac{4}{8} = \frac{1}{2}, P(BA)=46=23P(B \mid A) = \frac{4}{6} = \frac{2}{3} — not equal.

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§3.1.6 — a full worked pass

Example 3.1.1 — Reading Every Symbol Off One Sample Space

SS = whole numbers 1 to 19. AA = even numbers, BB = numbers greater than 13. Find every quantity from Try It Now 3.1.6, for this SS.


19 outcomes. P(A)=919P(A) = \frac{9}{19}, P(B)=619P(B) = \frac{6}{19}. A AND B={14,16,18}A \text{ AND } B = \{14,16,18\}, P=319P = \frac{3}{19}; P(A OR B)=1219P(A \text{ OR } B) = \frac{12}{19}. P(A)=1019P(A') = \frac{10}{19}, summing to 1. P(AB)=12P(A \mid B) = \frac{1}{2}, P(BA)=13P(B \mid A) = \frac{1}{3} — not equal: the reduced sample space differs each way.

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Try it in rāSHio

Check the overlap before you subtract it

Open rāSHio and choose Calculators → Probability to check the answers above without redoing the arithmetic. Enter the sample-space size and each event count, and the panel returns the AND together with both conditional values.

The point is not to skip the counting — that is the skill this section teaches — but to check the overlap before you use it in P(A OR B)=P(A)+P(B)P(A AND B)P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B), where subtracting it is the step most often missed.

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§3.1.6 — reading it off the panel

rāSHio: Calculators → Probability

Figure 3.1.2: checking the intersection and the conditional probabilities in rāSHio — Calculators → Probability.

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Your turn — ten numbers, ten questions

Try It Now 3.1.7

A number is chosen from 1–10. TT=5, AA=more than 6, BB=odd, CC=composite, DD=multiple of 3. Find SS, each event, AA', ABA \mid B, BAB \mid A, AA OR BB, AA OR BB', and every probability.


P(T)=110P(T)=\frac{1}{10}, P(A)=25P(A)=\frac{2}{5}, P(B)=12P(B)=\frac{1}{2}, P(A)=35P(A')=\frac{3}{5}, P(AB)=25P(A\mid B)=\frac{2}{5}, P(BA)=12P(B\mid A)=\frac{1}{2}, P(A OR B)=710P(A \text{ OR } B)=\frac{7}{10}, P(A OR B)=710P(A \text{ OR } B')=\frac{7}{10} — two different events, same probability. P(C)=12P(C)=\frac{1}{2}, P(D)=310P(D)=\frac{3}{10}.

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§3.1.6 — eleven events, one die

Example 3.1.2 — Naming Eleven Events on One Die

A fair die is rolled. TT=two, AA=even, BB=less than four, NN=prime, II=seven (impossible). Find every event, its complement, both conditionals, AND, OR, OR-with-complement, and every probability.


P(T)=P(A)=P(B)=P(A)=12P(T)=P(A)=P(B)=P(A')=\frac{1}{2} or 16\frac{1}{6} as fits; P(AB)=P(BA)=13P(A\mid B)=P(B\mid A)=\frac{1}{3}; P(A AND B)=16P(A \text{ AND } B)=\frac{1}{6}; P(A OR B)=56P(A \text{ OR } B)=\frac{5}{6}; P(A OR B)=23P(A \text{ OR } B')=\frac{2}{3}; P(N)=12P(N)=\frac{1}{2}; and P(I)=0P(I)=0 — a six-sided die has no seven, so II is the empty set.

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Your turn — a two-way table

Try It Now 3.1.8 — Table 3.1.1

TeaCoffee
Men2226
Women1636

100 people. MM=man, WW=woman, TT=tea, CC=coffee.

Find P(M)P(M), P(W)P(W), P(T)P(T), P(C)P(C), P(M AND T)P(M \text{ AND } T), P(W AND C)P(W \text{ AND } C), P(M OR W)P(M \text{ OR } W), P(M)P(M'), and the conditionals.

Margins: 0.48, 0.52, 0.38, 0.62. Cells: P(M AND T)=0.22P(M \text{ AND } T)=0.22, P(W AND C)=0.36P(W \text{ AND } C)=0.36. P(M OR W)=1P(M \text{ OR } W)=1, P(M)=0.52P(M')=0.52. P(TM)=22480.4583P(T\mid M)=\frac{22}{48}\approx 0.4583, P(CW)=36520.6923P(C\mid W)=\frac{36}{52}\approx 0.6923.

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§3.1.6 — reading probabilities off a table

Example 3.1.3 — Table 3.1.2

Right-handedLeft-handed
Males439
Females444

100 people, by sex assigned at birth and handedness.

MM=male, FF=female, RR=right-handed, LL=left-handed. Find the margins, the AND cells, the OR unions, P(M)P(M'), and the conditionals.

P(M)=0.52P(M)=0.52, P(F)=0.48P(F)=0.48, P(R)=0.87P(R)=0.87, P(L)=0.13P(L)=0.13. P(M AND R)=0.43P(M \text{ AND } R)=0.43. P(RM)=43520.8269P(R\mid M)=\frac{43}{52}\approx0.8269, P(LF)=4480.0833P(L\mid F)=\frac{4}{48}\approx0.0833 — the same cell, 4, divided by two different reduced sample spaces.

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Key Terminology — the vocabulary this section put to work

experiment — a planned operation carried out under controlled conditions; a chance experiment's result is not predetermined.

outcome — a single result of an experiment.

sample space — the set of all possible outcomes, denoted SS.

event — any combination of outcomes; a subset of the sample space.

probability — the long-term relative frequency of an outcome, between 0 and 1 inclusive.

equally likely — every outcome occurs with the same probability.

law of large numbers — as repetitions increase, observed relative frequency approaches theoretical probability.

biased — a device whose outcomes are not equally likely.

union (OR event)ABA \cup B, the outcome is in AA, in BB, or in both.

intersection (AND event)ABA \cap B, the outcome is in both at once.

complementAA', the outcome is not in AA; P(A)+P(A)=1P(A)+P(A')=1.

conditional probabilityP(AB)P(A \mid B), the probability of AA on the reduced sample space BB.

3.1
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§3.1 — conclusions

What §3.1 leaves you with

The core idea

Every probability question is built from the same small kit: an experiment produces outcomes in a sample space SS; an event is a subset of SS; OR, AND, and complement combine events; and conditioning shrinks SS to whatever you already know.

Watch for

The counting shortcut P(A)=ASP(A) = \frac{|A|}{|S|} needs equally likely outcomes — check that before you use it. And read conditioning language carefully: “of all the …” is not the same claim as “… and …”.

Next: §3.2 — Independent and Mutually Exclusive Events, where these same OR and AND values decide whether two events affect each other at all.