Basic Probability Rules Practice Quiz with a Step-by-Step Interactive Lesson
Use the question set below 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.
Answer the question set and review your mistakes at the end.
How this probability practice works
1. Take the practice set: answer the probability questions below.
2. Open the lesson (optional): review the probability rules with worked examples and quick checks.
3. Retry: return to the question set and apply the formulas immediately.
What you will 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 \(\div\) 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 \(P(A\cup B)=P(A)+P(B)\)
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?
Hint: Probabilities are always between 0 and 1 (inclusive).
Pre-check 2: A fair coin is flipped once. Which set is the sample space?
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?
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?
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 \(B\)?
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?
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?
Hint: Mutually exclusive means no overlap, so you can add the probabilities.
Try it 2: Two events are mutually exclusive. What is \(P(A\cap B)\)?
Hint: "Mutually exclusive" means they cannot occur together, so the intersection has probability \(0\).
Summary
General addition rule: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).
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?
Hint: Multiply \(\tfrac{1}{2}\) by \(\tfrac{1}{2}\).
Try it 2: If events \(A\) and \(B\) are independent, which statement is true?
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 \(P(T\mid C)\).
Try it: If \(P(B)=0.2\) and \(P(A\cap B)=0.05\), what is \(P(A\mid B)\)?
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?
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?
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?
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 \(1\)?
Hint: Probabilities are proportions, so they cannot exceed \(1\).
Final recap
Probability values satisfy \(0\le P(A)\le 1\). Impossible: \(0\). Certain: \(1\).
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.
Practice set
Grundläggande sannolikhetsregler practice questions with instant score
Answer all 10 questions below, then get your final score and a mistake review at the end so you know exactly what to improve.
0/10answered
Question 1Not answered
Vad är sannolikheten för en händelse som säkert inträffar?
Correct answer: C. \(1\)
Explanation: En säker händelse har sannolikheten \(1\).
Question 2Not answered
Om sannolikheten för en händelse \(A\) är \(0.3\), vad är sannolikheten att \(A\) inte inträffar?
Correct answer: B. \(0.7\)
Explanation: Kompletteringsregeln: \(P(\neg A) = 1 - 0.3 = 0.7\).
Question 3Not answered
Vad är sannolikheten för en omöjlig händelse?
Correct answer: A. \(0\)
Explanation: En omöjlig händelse har sannolikheten \(0\).
Question 4Not answered
Vilket av dessa tal kan vara sannolikheten för en händelse?
Correct answer: C. \(0.8\)
Explanation: Sannolikheter måste ligga mellan \(0\) och \(1\), inklusive ändpunkterna.
Question 5Not answered
Om ett experiment har tre lika sannolika utfall, vad är sannolikheten att något utfall inträffar?
Correct answer: C. \(1\)
Explanation: Den totala sannolikheten över alla utfall summerar till \(1\).
Question 6Not answered
Vilken har högre sannolikhet: en säker händelse eller en omöjlig händelse?
Correct answer: D. Den säkra händelsen har sannolikheten \(1\)
Explanation: En säker händelse har sannolikheten \(1\), medan en omöjlig händelse har sannolikheten \(0\).
Question 7Not answered
Om sannolikheten för händelse \(B\) är \(0.6\), vad är sannolikheten för att inte \(B\)?
Explanation: Enligt definitionen kan sannolikheter inte överstiga \(1\).
Question 9Not answered
Om två ömsesidigt uteslutande händelser vardera har sannolikheten \(0.2\), vad är sannolikheten att någon av dem inträffar?
Correct answer: A. \(0.4\)
Explanation: För ömsesidigt uteslutande händelser adderar man sannolikheterna: \(0.2 + 0.2 = 0.4\).
Question 10Not answered
Händelserna \(C\) och \(D\) har vardera sannolikheten \(0.5\), och båda kan inträffa samtidigt. Vad är det största möjliga värdet för sannolikheten att någon av dem inträffar?
Correct answer: D. \(1\)
Explanation: Det maximala värdet uppstår när de är ömsesidigt uteslutande: \(0.5 + 0.5 = 1\).
