Practice Basic Probability Rules with quiz questions. Log in to track your best streak.

If the probability of event \(X\) is \(\tfrac{2}{3}\), what is the probability that \(X\) does not occur?

Streak 5+

Streak 10+

Streak 15+

Streak 20+

Streak 25+

💡 You can revive any streak of 3 or more using tokens!

Explanation: The complement is \(1 - \tfrac{2}{3} = \tfrac{1}{3}\).

Explore other themes

Basic Probability Rules

Basic Probability Rules Practice Quiz with a Step-by-Step Interactive Lesson

Use the quiz at the top of the page to practice the basic probability rules: probability from 0 to 1, equally likely outcomes, the complement rule, the addition rule of probability, the multiplication rule, and conditional probability. If you want a refresher, click Start lesson to open a step-by-step guide with examples.

How this probability practice works

1. Take the quiz: answer the probability questions at the top of the page.
2. Open the lesson (optional): review the probability rules with worked examples and quick checks.
3. Retry: return to the quiz and apply the formulas immediately.

What you’ll learn in the basic probability rules lesson

Foundations & vocabulary

Experiment, outcome, sample space (what can happen)
Event (a set of outcomes) and probability as a number from 0 to 1
Equally likely outcomes: favorable ÷ total

Complement & certainty

Impossible event: probability \(0\)
Certain (sure) event: probability \(1\)
Complement rule: \(P(A^c)=1-P(A)\)

Addition rule

“A or B” (union): \(P(A\cup B)\)
General rule: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
Mutually exclusive: \(P(A\cap B)=0\), so \(P(A\cup B)=P(A)+P(B)\)

Multiplication & conditional probability

“A and B” (intersection): \(P(A\cap B)\)
Multiplication rule: \(P(A\cap B)=P(A)\,P(B\mid A)\)
Independent events: \(P(A\cap B)=P(A)P(B)\)

Back to the quiz

When you’re ready, return to the quiz at the top of the page and keep practicing the probability rules.

⭐⭐

🎲

Probability
Rules

Step-by-step guide

Tap to open →

Loading…

Basic Probability Rules Lesson

1 / 8

Lesson Overview

Lesson overview

Purpose: Build a clear understanding of basic probability rules and learn reliable formulas you can use in any probability problem.

Success criteria

Identify a sample space and describe an event.
Use the probability scale: \(0\le P(A)\le 1\), with \(P(\emptyset)=0\) and \(P(S)=1\).
Compute probabilities for equally likely outcomes using "favorable ÷ total".
Use the complement rule: \(P(A^c)=1-P(A)\).
Use the addition rule: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) (and the special case for mutually exclusive events).
Use the multiplication rule: \(P(A\cap B)=P(A)P(B\mid A)\), and the independent-events case \(P(A\cap B)=P(A)P(B)\).
Use the conditional probability formula: \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\) when \(P(B)>0\).

Key vocabulary

Experiment: a process with uncertain outcome (flip a coin, roll a die).
Outcome: one possible result of the experiment.
Sample space \(S\): the set of all possible outcomes.
Event \(A\): a set of outcomes (e.g., "roll an even number").
Complement \(A^c\): "not \(A\)".
Union \(A\cup B\): "\(A\) or \(B\)".
Intersection \(A\cap B\): "\(A\) and \(B\)".

Quick pre-check

Pre-check 1: What is the largest value a probability can take?

\(0\) \(0.5\) \(1\) \(2\)

Hint: Probabilities are always between 0 and 1 (inclusive).

Pre-check 2: A fair coin is flipped once. Which set is the sample space?

\(\{H, T\}\) \(\{H\}\) \(\{T\}\) \(\{1,2,3,4,5,6\}\)

Hint: The sample space lists all possible outcomes.

Probability Basics

Outcomes, events, and equally likely probability

Learning goal: Identify outcomes and events, then compute simple probabilities from a sample space.

Key idea

Probability measures how likely an event is. For a finite sample space with equally likely outcomes: \[ P(\text{event})=\frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}. \] Also, the probabilities of all outcomes in the sample space add up to \(1\).

Worked example

Example: Roll a fair six-sided die. Find \(P(\text{roll a number greater than }4)\).

Sample space: \(\{1,2,3,4,5,6\}\).
Favorable outcomes (greater than 4): \(\{5,6\}\) (2 outcomes).
\[ P(\text{greater than }4)=\frac{2}{6}=\frac{1}{3}. \]

Try it

Try it 1: A single fair die is rolled. What is the probability of rolling a 5?

\(\frac{1}{6}\) \(\frac{1}{3}\) \(\frac{1}{2}\) \(\frac{5}{6}\)

Hint: There is 1 favorable outcome (rolling a 5) out of 6 total outcomes.

Try it 2: An experiment has four possible outcomes, each equally likely. What is the sum of their probabilities?

\(1\) \(4\) \(0.25\) \(2\)

Hint: The total probability across the entire sample space is always \(1\).

Summary

For equally likely outcomes, use \(P=\frac{\text{favorable}}{\text{total}}\).
The probabilities of all outcomes in the sample space sum to \(1\).

Complement Rule

Complements: "not \(A\)"

Learning goal: Use the complement rule to find probabilities quickly and avoid double counting.

Key idea

The complement of event \(A\), written \(A^c\), means "not \(A\)". Because either \(A\) happens or it doesn't (no overlap and no missing outcomes): \[ P(A)+P(A^c)=1 \quad \Rightarrow \quad P(A^c)=1-P(A). \]

Worked example

Example: If \(P(A)=0.3\), find \(P(A^c)\).

Use the complement rule:
\[ P(A^c)=1-P(A)=1-0.3=0.7. \]

Try it

Try it 1: If the probability of event \(B\) is \(0.6\), what is the probability of not \(B\)?

\(0.6\) \(0.4\) \(1.6\) \(0\)

Hint: \(P(B^c)=1-P(B)\).

Try it 2: If the probability of an event is \(p\), what is the sum of the probability of the event and its complement?

\(p\) \(1\) \(2p\) \(0\)

Hint: \(A\) and \(A^c\) cover the whole sample space with no overlap.

Summary

The complement rule is \(P(A^c)=1-P(A)\).
\(P(A)+P(A^c)=1\) always.

Addition Rule

The addition rule: "\(A\) or \(B\)"

Learning goal: Find probabilities of unions ("or") using the general addition rule and the mutually exclusive shortcut.

Key idea

"\(A\) or \(B\)" means the union \(A\cup B\). If \(A\) and \(B\) can both happen, we must subtract the overlap to avoid double counting: \[ P(A\cup B)=P(A)+P(B)-P(A\cap B). \] If \(A\) and \(B\) are mutually exclusive (cannot happen together), then \(P(A\cap B)=0\), so: \[ P(A\cup B)=P(A)+P(B). \]

Worked example

Example: Suppose \(P(A)=0.4\), \(P(B)=0.3\), and \(P(A\cap B)=0.1\). Find \(P(A\cup B)\).

\[ P(A\cup B)=0.4+0.3-0.1=0.6. \]

Try it

Try it 1: Two events are mutually exclusive. If their probabilities are \(0.25\) and \(0.5\), what is the probability that either happens?

\(0.25\) \(0.50\) \(0.75\) \(1.00\)

Hint: Mutually exclusive means no overlap, so you can add the probabilities.

Try it 2: Two events are mutually exclusive. What is \(P(A\cap B)\)?

\(0\) \(P(A)+P(B)\) \(1\) \(P(A)\cdot P(B)\)

Hint: "Mutually exclusive" means they cannot occur together, so the intersection has probability \(0\).

Summary

General addition rule: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).
Mutually exclusive shortcut: \(P(A\cup B)=P(A)+P(B)\).

Multiplication Rule

The multiplication rule: "\(A\) and \(B\)"

Learning goal: Compute intersection probabilities ("and") and recognize the independent-events shortcut.

Key idea

"\(A\) and \(B\)" means the intersection \(A\cap B\). The multiplication rule connects intersection and conditional probability: \[ P(A\cap B)=P(A)\,P(B\mid A). \] If \(A\) and \(B\) are independent, then \(P(B\mid A)=P(B)\), so: \[ P(A\cap B)=P(A)P(B). \]

Worked example

Example: Two fair coins are flipped. Find \(P(\text{two heads})\).

Each flip has \(P(H)=\tfrac{1}{2}\), and the flips are independent.
\[ P(\text{two heads})=\tfrac{1}{2}\cdot\tfrac{1}{2}=\tfrac{1}{4}. \]

Try it

Try it 1: If two fair coins are flipped independently, what is the probability of two heads?

\(\frac{1}{2}\) \(\frac{1}{3}\) \(\frac{1}{4}\) \(\frac{1}{8}\)

Hint: Multiply \(\tfrac{1}{2}\) by \(\tfrac{1}{2}\).

Try it 2: If events \(A\) and \(B\) are independent, which statement is true?

\(P(A\cap B)=P(A)+P(B)\) \(P(A\cap B)=P(A)\cdot P(B)\) \(P(A\cap B)=P(A)-P(B)\) \(P(A\cap B)=\dfrac{P(A)}{P(B)}\)

Hint: Independence means knowing \(A\) doesn't change the probability of \(B\).

Summary

General multiplication rule: \(P(A\cap B)=P(A)P(B\mid A)\).
If independent: \(P(A\cap B)=P(A)P(B)\).

Conditional Probability

Conditional probability: \(P(A\mid B)\)

Learning goal: Compute conditional probability and connect it to the multiplication rule.

Key idea

Conditional probability means "the probability of \(A\) given \(B\) happened." When \(P(B)>0\): \[ P(A\mid B)=\frac{P(A\cap B)}{P(B)}. \] This also rearranges to the multiplication rule: \[ P(A\cap B)=P(B)\,P(A\mid B). \]

Worked example

Example: In a survey, \(P(C)=0.50\) like coffee, and \(P(T\cap C)=0.30\) like tea and coffee. Find \(P(T\mid C)\).

\[ P(T\mid C)=\frac{P(T\cap C)}{P(C)}=\frac{0.30}{0.50}=0.60. \]

Try it

Try it: If \(P(B)=0.2\) and \(P(A\cap B)=0.05\), what is \(P(A\mid B)\)?

\(0.10\) \(0.25\) \(0.20\) \(0.05\)

Hint: Use \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\).

Worked solution

\[ P(A\mid B)=\frac{0.05}{0.2}=0.25. \]

Summary

Conditional probability: \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\) when \(P(B)>0\).
It connects directly to the multiplication rule for intersections.

Putting It Together

Combine rules and check your answers

Learning goal: Use complements and independence to solve "at least one" probability questions and keep results within \([0,1]\).

Key idea

A powerful strategy is to use a complement: \[ P(\text{at least one}) = 1 - P(\text{none}). \] This is often simpler than counting many cases directly.

Worked example

Example: Two fair coins are flipped. Find \(P(\text{at least one head})\).

"At least one head" is the complement of "no heads" (which means two tails).
\[ P(\text{no heads})=P(TT)=\tfrac{1}{2}\cdot\tfrac{1}{2}=\tfrac{1}{4}. \] \[ P(\text{at least one head})=1-\tfrac{1}{4}=\tfrac{3}{4}. \]

Try it

Try it 1: Two fair coins are flipped. What is the probability of at least one head?

\(\frac{1}{4}\) \(\frac{1}{2}\) \(\frac{3}{4}\) \(1\)

Hint: Use \(1-P(\text{no heads})\). The only "no heads" outcome is \(TT\).

Try it 2: Which of these is the probability of an impossible event?

\(0\) \(0.5\) \(1\) \(-0.2\)

Hint: Impossible means it cannot happen at all.

Summary

Use complements to simplify: \(P(\text{at least one})=1-P(\text{none})\).
Always check that your final probability is between \(0\) and \(1\).

Applications & History

Why probability rules matter

Learning goal: Connect probability rules to everyday decisions, games, and data — and learn a little history behind probability.

Where you use probability

Games and puzzles: dice, cards, and fair decision-making.
Risk and planning: weather chances, budgeting uncertainty, safety decisions.
Science and data: experiments, sampling, and statistics.
Technology: reliability, quality control, and randomized algorithms.

Worked example: drawing a card

Example: A standard deck has 52 cards with 4 aces. Find \(P(\text{ace})\).

\[ P(\text{ace})=\frac{4}{52}=\frac{1}{13}. \]

Try it

Try it 1: A standard deck has 52 cards. What is the probability of drawing an ace?

\(\frac{1}{13}\) \(\frac{1}{52}\) \(\frac{1}{4}\) \(\frac{4}{13}\)

Hint: There are 4 aces out of 52 cards. Simplify the fraction.

Fun facts (a little history)

Origins: Modern probability grew from questions about games of chance studied by mathematicians like Pascal and Fermat.
Notation: Many probability rules look like set math: "or" is \(A\cup B\), "and" is \(A\cap B\), and "not" is \(A^c\).
Big idea: The same basic rules power advanced topics like statistics, machine learning, and decision-making under uncertainty.

Try it 2: Can a probability ever be greater than \(1\)?

Yes, probabilities can be any number. No, a probability is always between \(0\) and \(1\). Only for complements. Only for independent events.

Hint: Probabilities are proportions, so they cannot exceed \(1\).

Final recap

Probability values satisfy \(0\le P(A)\le 1\). Impossible: \(0\). Certain: \(1\).
Equally likely outcomes: \(P=\dfrac{\text{favorable}}{\text{total}}\).
Complement rule: \(P(A^c)=1-P(A)\).
Addition rule: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) (mutually exclusive: subtract 0).
Multiplication rule: \(P(A\cap B)=P(A)P(B\mid A)\) (independent: \(P(A)P(B)\)).
Conditional probability: \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\), for \(P(B)>0\).

Next step: Close this lesson and try your quiz again. If you miss a question, reopen the book and review the page that matches the probability rule you need.