1 Probability Foundations and Sample Spaces

A structured introduction to probability that develops sample spaces, events, probability rules, counting methods, dependence, and reliable problem-solving techniques.

Experiments and Sample Spaces

Probability provides a mathematical way to describe uncertainty. A probability experiment produces one result from a set of possible results, and the result of one trial is called an outcome.

A well-constructed must satisfy three requirements:

  • Complete: It includes every outcome that could occur.

  • Mutually exclusive: One trial cannot produce two different listed outcomes at the same time.

  • Appropriately detailed: The outcomes contain enough information to answer the question.

Examples include:

  • One coin toss: S={H,T}S=\{H,T\}.

  • One standard die roll: S={1,2,3,4,5,6}S=\{1,2,3,4,5,6\}.

  • Two coin tosses when order is recorded: S={HH,HT,TH,TT}S=\{HH,HT,TH,TT\}.

For a multistage experiment, ordered outcomes help prevent omissions. For example, (2,5)(2,5) and (5,2)(5,2) are different outcomes when the first and second die rolls are recorded separately.

Takeaway: Start every probability problem by identifying the experiment, the recorded outcome, and the complete .

Events and Set Operations

Events describe outcomes of interest and are represented as subsets of the . Suppose one die is rolled and S={1,2,3,4,5,6}S=\{1,2,3,4,5,6\}. Let A={2,4,6}A=\{2,4,6\} represent an even roll and B={4,5,6}B=\{4,5,6\} represent a roll greater than 3.

The main set operations are:

  • Complement: AcA^c contains outcomes in SS that are not in AA. Here, Ac={1,3,5}A^c=\{1,3,5\}.

  • Union: A∪BA\cup B contains outcomes in at least one of the two events.

  • Intersection: A∩BA\cap B contains outcomes in both events. Here, A∩B={4,6}A\cap B=\{4,6\}.

  • Empty : ∅\varnothing contains no outcomes and represents an impossible .

  • Certain : SS contains every possible outcome and represents an that must occur.

Two events are mutually exclusive, or disjoint, when they cannot occur together. This is expressed as A∩B=∅A\cap B=\varnothing. For example, rolling an even number and rolling an odd number on one die roll are mutually exclusive.

In ordinary language, “or” usually corresponds to a union, while “and” corresponds to an intersection. Always translate the wording into set notation before calculating.

Takeaway: notation makes it possible to represent complements, alternatives, simultaneous conditions, impossible outcomes, and certain outcomes precisely.

Probability Models and Equally Likely Outcomes

A probability model combines a with probabilities assigned to events. These assignments follow three axioms:

  1. Nonnegativity: P(A)≥0P(A)\ge 0.

  2. Normalization: P(S)=1P(S)=1.

  3. Additivity for disjoint events: If A∩B=∅A\cap B=\varnothing, then P(A∪B)=P(A)+P(B)P(A\cup B)=P(A)+P(B).

These properties imply that every has probability between 0 and 1:

0≤P(A)≤1.0\le P(A)\le 1.

A probability of 00 represents an impossible , while a probability of 11 represents a certain .

When all outcomes in a finite are equally likely, the probability of an is the favorable count divided by the total count:

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

For a fair die, three of the six outcomes are even, so

P(even)=36=12.P(\text{even})=\frac{3}{6}=\frac{1}{2}.

This counting ratio cannot be used automatically when outcomes are not equally likely, such as with a biased die.

Takeaway: Before using favorable outcomes divided by total outcomes, verify that the individual outcomes are equally likely.

Core Probability Rules

Several rules organize probability calculations.

The complement rule is

P(Ac)=1−P(A).P(A^c)=1-P(A).

It is especially useful for phrases such as “not,” “none,” and “at least one.” If the probability that a package arrives on time is 0.920.92, then

P(late)=1−0.92=0.08.P(\text{late})=1-0.92=0.08.

For any two events, the is

P(A∪B)=P(A)+P(B)−P(A∩B).P(A\cup B)=P(A)+P(B)-P(A\cap B).

The intersection is subtracted because adding P(A)P(A) and P(B)P(B) counts outcomes in both events twice. For one die roll, if AA is an even result and BB is a result greater than 3, then

P(A∪B)=36+36−26=46=23.P(A\cup B)=\frac{3}{6}+\frac{3}{6}-\frac{2}{6}=\frac{4}{6}=\frac{2}{3}.

If the events are mutually exclusive, their intersection has probability zero, so their probabilities can be added directly.

The multiplication rule connects intersections with :

P(A∩B)=P(A)P(B∣A)=P(B)P(A∣B).P(A\cap B)=P(A)P(B\mid A)=P(B)P(A\mid B).

When P(A)>0P(A)>0, is

P(B∣A)=P(A∩B)P(A).P(B\mid A)=\frac{P(A\cap B)}{P(A)}.

Knowing that AA occurred restricts attention to the outcomes inside AA. For example, if a card is known to be a heart in a standard deck, then

P(ace∣heart)=113,P(\text{ace}\mid\text{heart})=\frac{1}{13},

because one of the 13 hearts is an ace.

Takeaway: Use complements for “not” and “at least one,” the for “or,” and the multiplication rule with for “and.”

and Dependence

Two events are independent when knowing that one occurred does not change the probability of the other. Algebraically,

P(A∩B)=P(A)P(B).P(A\cap B)=P(A)P(B).

When P(A)>0P(A)>0, an equivalent statement is

P(B∣A)=P(B).P(B\mid A)=P(B).

Successive tosses of a fair coin are commonly modeled as independent. Drawing two cards without replacement generally creates dependence because the first card changes the composition of the deck. Sampling with replacement often produces because the available objects are restored before the next selection.

Mutually exclusive and independent describe different relationships. If AA and BB are mutually exclusive, then

P(A∩B)=0.P(A\cap B)=0.

If both events have positive probability, then P(A)P(B)>0P(A)P(B)>0, so they cannot also be independent.

Takeaway: Ask whether earlier information changes later probabilities. If it does, the events are dependent; if it does not, may be appropriate.

Counting and Observed Probability

Counting correctly is often the central challenge in a probability problem. First determine whether order matters.

The applies when a process has several stages. If there are mm choices for the first step and nn choices for the second, there are

m×nm\times n

possible ordered outcomes. A meal with 4 entrees and 3 drinks has

4×3=124\times 3=12

possible combinations.

A tree diagram displays the stages of a multistep experiment. Each branch represents a choice, and each complete path represents one outcome. Tree diagrams are particularly useful when the number of choices or the probabilities change from one stage to the next.

Order matters when assigning a president and a vice president, because the assignments differ. Order does not matter when choosing a two-person committee, because the same two people form the same committee regardless of selection order.

uses a model, while uses observations. If a coin lands heads 47 times in 100 tosses, its observed relative frequency for heads is

P^(heads)=47100=0.47.\widehat P(\text{heads})=\frac{47}{100}=0.47.

Repeated trials do not guarantee that an empirical proportion equals the , but relative frequencies often stabilize near the model probability as the number of trials becomes large.

Takeaway: Decide whether order matters, choose a systematic counting method, and distinguish model-based probabilities from observed relative frequencies.

A Practical Problem-Solving Strategy

A reliable solution can be organized as a sequence of decisions:

  1. Identify the experiment. Determine what process is being performed.

  2. Define one outcome. Specify exactly what is recorded on one trial.

  3. Construct the . List outcomes, make a table, draw a tree, or apply a counting principle.

  4. Define the . Translate the question into a set such as AA, A∪BA\cup B, or A∩BA\cap B.

  5. Check assumptions. Ask whether outcomes are equally likely, whether sampling is with or without replacement, and whether order matters.

  6. Select a rule. Consider direct counting, the complement rule, the , , or the multiplication rule.

  7. Calculate and simplify. Exact fractions are often useful; decimals and percentages can aid interpretation.

  8. Check the result. Every probability must lie between 00 and 11, and the result must fit the context.

For two fair coin tosses,

S={HH,HT,TH,TT}.S=\{HH,HT,TH,TT\}.

The of getting at least one head is {HH,HT,TH}\{HH,HT,TH\}, so direct counting gives

P(at least one head)=34.P(\text{at least one head})=\frac{3}{4}.

The complement method is shorter: the complement is the single outcome TTTT, so

P(at least one head)=1−P(TT)=1−14=34.P(\text{at least one head})=1-P(TT)=1-\frac{1}{4}=\frac{3}{4}.

Common errors include using an incomplete , confusing an outcome with an , assuming equal likelihood without justification, failing to subtract overlap, treating dependent events as independent, and ignoring order.

Takeaway: Clear definitions and assumption checks usually prevent more errors than complicated algebra does.