A focused Grade 7 learning session for Sample Spaces for Compound Events using visual models, worked examples, strategy guidance, and a short mastery check.
Student goal: I can solve problems involving sample spaces for compound events.
Sample Spaces for Compound Events is part of Grade 7 Probability. This activity focuses on the central idea students need before they move into independent practice.
Students will build Grade 7 mastery of sample spaces for compound events using models, equations, and test-style reasoning.
| What to notice | How to use it |
|---|---|
| Numbers, units, labels, diagrams, or graphs | Identify what each part represents before calculating. |
| The relationship in the problem | Connect the model to an equation, comparison, or explanation. |
Use the figures and worked questions in the practice set below as the visual model for this skill.
Read the first practice question slowly. Decide what the question is asking, choose a strategy, solve, and then compare your reasoning with the answer explanation.
For Sample Spaces for Compound Events, a strong worked example should show the setup, the calculation or model, and a final answer that matches the context.
Many Grade 7 test questions use real situations. Look for quantities, units, rates, measures, expressions, or data values that connect the math to the story.
After solving, explain what the answer means in the situation, not just the number you calculated.
I can solve problems involving sample spaces for compound events.
Use the warmup, guided question, practice set, and exit questions below to check whether this skill is ready or needs more practice.
Use scratch paper. The explanation matters more than simply seeing whether the choice is correct.
A coin is flipped and a six-sided die is rolled. How many outcomes are in the sample space?
Answer C: \(12\)
The coin has \(2\) outcomes and the die has \(6\). By the counting principle: \(2\times6=12\) total outcomes.
Two coins are flipped. Which tree diagram shows all possible outcomes?
Answer C: \(4\) outcomes
The tree shows all branches: HH, HT, TH, TT. That is \(4\) equally likely outcomes.
A spinner is divided into 3 equal sections (Red, Blue, Yellow), and a fair coin is flipped. How many total outcomes are possible?
Answer C: \(6\)
The spinner has \(3\) outcomes and the coin has \(2\) outcomes. Using the counting principle: \(3 \times 2 = 6\) total outcomes.
A student chooses a shirt from 5 options and pants from 3 options. How many different outfits are possible?
Answer B: \(15\)
By the counting principle: \(5\) shirts \(\times\) \(3\) pants \(= 15\) different outfits.
A fair die is rolled twice. What is the total number of outcomes in the sample space?
Answer D: \(36\)
First roll has \(6\) outcomes; second roll has \(6\) outcomes. Total: \(6 \times 6 = 36\) outcomes.
A menu offers 4 drinks, 6 main courses, and 3 desserts. How many complete meals can be ordered if you choose one of each?
Answer C: \(72\)
By the counting principle: \(4 \times 6 \times 3 = 72\) complete meal combinations.
Two spinners are spun. The first spinner has sections labeled 1, 2, and 3. The second spinner has sections labeled A and B. How many outcomes are in the sample space?
Answer B: \(6\)
First spinner: \(3\) outcomes; second spinner: \(2\) outcomes. Total: \(3 \times 2 = 6\) outcomes.
A letter is drawn from the word MATH, and then a number is chosen from 1 to 5. How many outcomes are in the sample space? Letters: M, A, T, H \quad Numbers: 1, 2, 3, 4, 5 \par
Answer C: \(20\)
Letters: \(4\) (M, A, T, H); numbers: \(5\) (1 to 5). By the counting principle: \(4 \times 5 = 20\) outcomes.
A coin is flipped and a number is chosen from a spinner with sections 1, 2, 3, 4. How many favorable outcomes result in a heads and an odd number?
Answer B: \(2\)
Total outcomes: \(2 \times 4 = 8\). Favorable (Heads and odd): H1, H3. That is \(2\) favorable outcomes.
A die is rolled and a spinner with sections Red, Green, Blue is spun. What is the probability of rolling an even number and landing on Red?
Answer A: \(\frac{1}{6}\)
Total outcomes: \(6 \times 3 = 18\). Favorable (even and Red): 2R, 4R, 6R = \(3\) outcomes. Probability: \(\frac{3}{18} = \frac{1}{6}\).
A letter is randomly selected from the word CAT, and a coin is flipped. Which table correctly represents the sample space?
Answer D: \(6\) outcomes
The table shows \(3\) rows (letters: C, A, T) and \(2\) columns (coin: H, T), giving \(3 \times 2 = 6\) outcomes.
