A focused Grade 7 learning session for Probability Models using visual models, worked examples, strategy guidance, and a short mastery check.
Student goal: I can solve problems involving probability models.
Probability Models 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 probability models 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 Probability Models, 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 probability models.
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 spinner has \(4\) equal sections labeled \(1, 2, 3, 4\). Which statement is FALSE for this uniform probability model?
Answer D: \(P(5)=\frac{1}{4}\)
\(5\) is not on the spinner, so \(P(5)=0\). The others are correct.
A spinner has \(8\) equal sections: \(4\) red, \(2\) blue, \(1\) green, \(1\) yellow. What is \(P(\text{red})\) for this uniform model?
Answer C: \(\frac{1}{2}\)
There are \(4\) red sections out of \(8\) total, so \(P(\text{red})=\frac{4}{8}=\frac{1}{2}\).
Which spinner represents a non-uniform probability model?
Answer B: Right spinner only
The left spinner has equal sections (uniform). The right spinner has unequal sectors: one 60°, one 120°, one 180° (non-uniform).
A bag contains \(3\) red marbles, \(5\) blue marbles, and \(2\) green marbles. Which probability is correct?
Answer B: \(P(\text{red})=\frac{3}{10}\)
Total marbles: \(3+5+2=10\). Red: \(3\) out of \(10\) gives \(\frac{3}{10}\). (Blue would be \(\frac{5}{10}\), green \(\frac{2}{10}\), not red \(\frac{7}{10}\)).
A spinner is divided into \(6\) equal sections labeled \(A, A, B, C, C, C\). What is \(P(C)\)?
Answer A: \(\frac{1}{2}\)
There are \(3\) sections labeled \(C\) out of \(6\) equal sections, so \(P(C)=\frac{3}{6}=\frac{1}{2}\).
A weather model predicts: \(P(\text{rain})=0.3\), \(P(\text{cloudy})=0.5\), \(P(\text{sunny})=0.2\). Which statement is true?
Answer A: This is a valid non-uniform probability model.
\(0.3 + 0.5 + 0.2 = 1\), so probabilities sum to \(1\) (valid). Outcomes have different probabilities (non-uniform). Also, \(P(\text{not sunny})=0.3+0.5=0.8\).
Does this table represent a valid probability model?
Answer C: Yes, all probabilities are between \(0\) and \(1\).
\(0.4 + 0.35 + 0.25 = 1.0\) and each probability is in \([0, 1]\), so this is valid.
A spinner has \(10\) equal sections. Three are labeled \(W\), four are labeled \(X\), two are labeled \(Y\), and one is labeled \(Z\). Find \(P(W \text{ or } X)\).
Answer C: \(\frac{7}{10}\)
\(W\) has \(3\) sections and \(X\) has \(4\) sections, so \(P(W \text{ or } X)=\frac{3+4}{10}=\frac{7}{10}\).
A card is drawn from a standard deck of \(52\) cards. What is \(P(\text{King})\)?
Answer B: \(\frac{1}{13}\)
There are \(4\) Kings in \(52\) cards, so \(P(\text{King})=\frac{4}{52}=\frac{1}{13}\).
This spinner has three equal 120° sectors. What is \(P(\text{Blue})\)?
Answer A: \(\frac{1}{3}\)
Each sector is \(\frac{120}{360}=\frac{1}{3}\) of the whole circle, so \(P(\text{Blue})=\frac{1}{3}\). Distractors: B is \(\frac{1}{4}\) (too small), C is \(\frac{2}{3}\) (counts two sectors), D is \(\frac{2}{5}\) (incorrect ratio).
A probability model has outcomes \(\{1, 2, 3, 4, 5\}\) with \(P(1)=0.1\), \(P(2)=0.2\), \(P(3)=0.3\), \(P(4)=0.25\), \(P(5)=?\). Find \(P(5)\).
Answer A: \(0.15\)
Probabilities must sum to \(1\): \(0.1 + 0.2 + 0.3 + 0.25 + P(5) = 1 \Rightarrow P(5) = 0.15\).
