How to use this menu
Six original tasks: two for criterion A (a short check and an end-of-unit test), one investigation for B, one communication task for C, and two real-life tasks for D that also assess C. Set at least one task for each criterion across the unit; nobody needs all six. The timings spread assessment through the unit rather than leaving it all to the end. Every task fits one lesson of 50 minutes or less.
Descriptors: Year 8 is assessed with the MYP Year 3 descriptors, paraphrased below each task. Check them against your school's current guide.
The menu at a glance
| Task | Criteria | When | Time | Format |
|---|---|---|---|---|
| A1: Quick check: scale, outcomes and trees | A | After lesson 3 | 20 minutes | Individual, written, in class. 15 marks. |
| B1: How far apart are the dice? | B | Mid-unit, after lesson 4 | 45 minutes | In class. Students may build the grid in pairs, but each writes up their own findings. |
| C1: Explain it to someone who missed it | C | After lesson 5 | 35 minutes | Individual, written, in class or as homework. One page of words, a table and a graph. |
| A2: End-of-unit test | A | End of unit, after lesson 6 | 45 minutes | Individual, written, in class, one lesson. 28 marks in three sections. |
| D1: The lucky-draw stall | D and C | End of unit, after lesson 6 | 45 minutes | Individual, in class. A written solution with explanations and a short note to the stall owner. |
| D2: The two-dice game at the fun day | D and C | End of unit, after lesson 6 | 45 minutes | Individual, in class, one lesson. A written solution with explanations and a short reply to the organisers. |
D1 and D2 are alternatives. Both assess criteria D and C, in two different real-life contexts: one is set, not both.
The unit's inquiry
- Key concept. Logic: using chance and counting to decide what is likely, what is fair and what to expect.
- Related concepts. Model, Pattern, Representation
- Global context. Fairness and development: deciding whether a game, a bag of tickets or a plan treats the people involved fairly.
- Statement of inquiry. Counting every outcome and checking the count against what really happens lets us judge whether a chance situation is fair and what to expect from it.
MYP Year 3 descriptors, paraphrased
Each task's clarification, on its student sheet below, says what these levels look like in that task.
| Level | Criterion A: Knowing and understanding |
|---|---|
| 1–2 | Selects suitable mathematics for simple problems in familiar situations and applies it with some success. |
| 3–4 | Selects suitable mathematics for more complex problems in familiar situations and applies it successfully. |
| 5–6 | Selects suitable mathematics for challenging problems in familiar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| 7–8 | Selects suitable mathematics for challenging problems in both familiar and unfamiliar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| Level | Criterion B: Investigating patterns |
|---|---|
| 1–2 | With support, applies problem-solving techniques to find simple patterns and states predictions that fit them. |
| 3–4 | Applies problem-solving techniques to find simple patterns and suggests relationships or general rules that fit the findings. |
| 5–6 | Applies problem-solving techniques to find more complex patterns, describes them as relationships or general rules that fit the findings and verifies them. |
| 7–8 | Selects and applies problem-solving techniques to find complex patterns, describes them as relationships or general rules that fit correct findings, and verifies and justifies them. |
| Level | Criterion C: Communicating |
|---|---|
| 1–2 | Uses limited mathematical language and limited forms of representation; the reasoning is hard to follow. |
| 3–4 | Uses some appropriate mathematical language and forms of representation; the reasoning can be followed and the work is adequately organised. |
| 5–6 | Usually uses appropriate language, moves between forms of representation with some success, and the reasoning is complete and coherent in a mostly logical structure. |
| 7–8 | Consistently uses appropriate language, moves effectively between forms of representation, and the reasoning is complete, coherent and concise in a logical structure. |
| Level | Criterion D: Applying mathematics in real-life contexts |
|---|---|
| 1–2 | Identifies some relevant elements of the real-life situation and applies mathematical strategies to reach a solution with limited success. |
| 3–4 | Identifies the relevant elements, selects adequate strategies and applies them to reach a solution, and describes whether it makes sense in the context. |
| 5–6 | Identifies the relevant elements, selects adequate strategies and applies them to reach a valid solution, explains the degree of accuracy and explains whether the solution makes sense. |
| 7–8 | Identifies the relevant elements, selects appropriate strategies and applies them successfully to reach a correct solution, justifies the degree of accuracy and justifies whether the solution makes sense. |
Task A1: Quick check: scale, outcomes and trees
Why set it. A short check of lessons 1 to 3 while there is still time to reteach: the probability scale, listing outcomes and tree diagrams. Questions 5 and 6 are unfamiliar, so the top band is open to everyone.
When: After lesson 3. Time: 20 minutes. Format: Individual, written, in class. 15 marks.
The student sheet, with its task-specific clarification
When: After lesson 3. Time: 20 minutes. Calculator: no calculator. Criterion: A, knowing and understanding.
Answer every question you can and show your working: a correct method earns credit even if the final answer slips. Questions 5 and 6 are meant to make you think, so try something rather than leaving them blank.
Part 1. The probability scale
- Write the probability 0.48 as a fraction in its simplest form and as a percentage. (2 marks)
- A bag holds 8 red, 5 blue and 7 yellow counters. One counter is taken without looking. Find the probability that it is blue, and the probability that it is not blue, each as a fraction in its simplest form. (2 marks)
Part 2. Listing outcomes
- A cafe offers 4 drinks and 3 snacks. (a) How many different choices of one drink and one snack are there? (b) The cafe adds a choice of a glass or a cup. How many choices are there now? (2 marks)
- A coin is tossed, then a spinner with red (R), green (G) and blue (B) is spun, then the coin is tossed again. How many outcomes are there in all? List the outcomes that have exactly two heads. (3 marks)
Part 3. Situations you have not met in this form
- Spinner A has the numbers 1, 2 and 3. Spinner B has the numbers 2 and 4. Tia says: “There are 5 outcomes, because 3 + 2 = 5.” Explain her mistake, then list every outcome. (3 marks)
- A locker code has 3 digits, and each digit is 0, 1 or 2. How many codes are there? How many of them use three different digits? Explain how you found each answer. (3 marks)
How your work is judged
| Level | Criterion A in task A1: what it looks like |
|---|---|
| 1–2 | You answer some of questions 1 to 3 correctly, for example the fraction in question 1 or the number of choices in question 3. |
| 3–4 | You answer most of questions 1 to 4 correctly, with a sensible list or a tree for question 4. |
| 5–6 | You answer questions 1 to 4 correctly, with working that shows your method, and make a good start on question 5 or 6. |
| 7–8 | You answer questions 1 to 4 correctly and solve question 5 or 6 with a clear reason that shows why your count is complete. |
Markscheme
| Task A1 question | Answer and marking |
|---|---|
| 1 | 12/25 and 48%. 1 mark each. 48/100 left unsimplified scores 0 for the fraction. |
| 2 | P(blue) = 5/20 = 1/4; P(not blue) = 15/20 = 3/4. 1 mark each. 5/15 for P(not blue) compares blue with the others instead of with all 20 counters. |
| 3 | (a) 12 (b) 24. 1 mark each (4 × 3, then 12 × 2). Adding (7, then 9) is the usual slip. |
| 4 | 12 outcomes; HRH, HGH, HBH. 1 mark for 12 (2 × 3 × 2), 2 marks for all three (1 mark for two of them). HRT and TGH have only one head; 2 + 3 + 2 = 7 adds the branches. |
| 5 | 6 outcomes: (1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4). 1 mark for the explanation (every number on A pairs with every number on B, so multiply 3 × 2; adding gives the total of the numbers, not the pairs), 2 marks for the full list (1 mark with one pair missing). |
| 6 | 27 codes; 6 with all digits different. 1 mark for 27 (3 × 3 × 3). 2 marks for 6: 3 choices for the first digit, 2 left for the second, 1 for the third (1 mark for 6 with no reason). 3 × 2 × 1 is the key move; a full list of 27 is also fine. |
| Marks to a level | Total 15. 1 to 4 marks: 1 to 2. 5 to 8: 3 to 4. 9 to 12: 5 to 6. 13 to 15, with question 5 or 6 fully correct: 7 to 8. Use the marks as a guide, then check the student's profile against the clarification before settling on a level. |
Exemplar answers
| Task A1 response | What it looks like |
|---|---|
| Level 3 to 4 profile | Question 1 right (2); question 2 gives P(blue) = 1/4 but P(not blue) = 5/15 (1); question 3 gives 12 and 7 (1); question 4 gives 12 and lists HRH and HGH (2); question 5 says “5 outcomes” (0); question 6 gives 27 (1). 7 marks, which sits in the 3 to 4 band: counting and listing work when familiar, a product is not yet used in new settings. |
| Level 7 to 8 answer to question 6 | “Each of the 3 places has 3 choices, so 3 × 3 × 3 = 27 codes. For three different digits, I have 3 choices for the first digit, then only 2 are left for the second, then 1 for the third: 3 × 2 × 1 = 6. Check by listing: 012, 021, 102, 120, 201, 210, which is 6.” |
| Level 7 to 8 answer to question 5 | “Tia added the numbers of faces, but every number on spinner A can go with every number on spinner B. A has 3 and B has 2, so there are 3 × 2 = 6 outcomes: (1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4).” |
Differentiation
A way in for anyone stuck
- Give a blank two-way table for question 5 and a blank tree stem for question 4.
- Allow a pictured bag for question 2 (drawn counters) so the 20 can be counted.
- Read question 6 aloud on request and suggest listing the codes that start with 0. Reading is not what is being assessed.
Stretch
- Question 6 again, with the digits 0, 1, 2 and 3. How many codes have three different digits? (24.)
- Four coins are tossed. How many outcomes have at least one head, and how do you know without listing all sixteen? (15: all but TTTT.)
Students learning in English as an additional language
| Word in task A1 | Plain meaning |
|---|---|
| outcome | one possible result of an event |
| probability | a number from 0 to 1 that says how likely an event is |
| fraction in its simplest form | a fraction that cannot be made smaller, such as 1/4 instead of 5/20 |
| at random | chosen without looking, so every one has the same chance |
| list | write down every one, one after another, without missing any |
Sentence starters for task A1:
- There are … outcomes because … × … = …
- The probability is … out of …, which is … in its simplest form.
- Tia’s mistake is that …
Task B1: How far apart are the dice?
Why set it. A real pattern hunt built on lesson 2's grids and lesson 4's counting. The sheet never states the pattern: students organise their results, find it, describe it as a general rule, test it on dice with other numbers of faces and explain why it must hold.
When: Mid-unit, after lesson 4. Time: 45 minutes. Format: In class. Students may build the grid in pairs, but each writes up their own findings.
The student sheet, with its task-specific clarification
In MYP Year 3, support you give (a prompt, a table outline) is allowed at levels 1 to 2 only: note it on the student's work.
When: Mid-unit, after lesson 4. Time: 45 minutes. Calculator: no calculator. Criterion: B, investigating patterns.
Two fair dice are rolled and the difference is found: the bigger score minus the smaller one. So the difference for 2 and 5 is 3, and for 4 and 4 it is 0. Some differences seem more likely than others. This task is about how likely each one is, and about what happens when the dice have a different number of faces.
Part 1. Collect results
- Show all 36 outcomes of two ordinary dice in a grid or table you design yourself, and write the difference for each one.
- For each difference from 0 to 5, count the outcomes and write the probability as a fraction in its simplest form. Check that your counts make 36.
- Which difference is most likely, and which is least likely?
Part 2. Describe what you find
- Look at your counts. Describe what you notice, in words, as a rule another student could use to write the counts without drawing the grid.
- A four-faced die has the faces 1 to 4. Use your rule to predict the number of outcomes for each difference when two of them are rolled. Then list the outcomes to check your predictions.
Part 3. Test your rule
- Two eight-faced dice are rolled. Without listing all 64 outcomes, find the number of outcomes with a difference of 3, and the probability that the difference is 5 or more.
- Choose a number of faces of your own from 3 to 12, not 4, 6 or 8. Collect enough results to test whether your rule, or a rule like it, works for your dice. Say what stays the same and what changes.
Part 4. Explain
- Explain why your rule must work for dice with any number of faces, not only the ones you tried. Use your grid in your explanation.
- Write the probability of a difference of 0 for two dice with n faces each. What happens to this probability as n gets larger? Explain why.
How your work is judged
| Level | Criterion B in task B1: what it looks like |
|---|---|
| 1–2 | With help, you list the outcomes, count some of the differences and say something you notice about your counts. |
| 3–4 | You complete the table of counts, suggest a rule that fits the six-faced dice and use it to predict the four-faced die. |
| 5–6 | You organise your results clearly, describe a general rule that fits the counts, and check it by listing the four-faced and one other die. |
| 7–8 | You organise your results, describe a general rule that fits all your findings, make correct predictions for the eight-faced die, test it on dice of your own, and explain why it must hold for any number of faces. |
Markscheme
| Task B1 question | Answer and marking |
|---|---|
| The pattern (do not share it before the task) | For n-faced dice, a difference of 0 has n outcomes and a difference of k (from 1 to n − 1) has 2(n − k) outcomes. For n = 6: 6, 10, 8, 6, 4, 2 (total 36). The count for 0 is the exception: it is n, not 2n, because (a, a) is one outcome, not two. Accept any correct description that handles the 0 case, in words or symbols. |
| 1 | A grid with the first die across and the second down, each cell showing the difference (0 on the main line, 1 on the lines either side, and so on up to 5 in two corners). 1 mark for a complete correct grid. |
| 2 | 0: 6, 1/6; 1: 10, 5/18; 2: 8, 2/9; 3: 6, 1/6; 4: 4, 1/9; 5: 2, 1/18. The counts total 36. 1 mark for the counts, 1 for the simplified fractions. |
| 3 | Most likely 1 (10 of 36); least likely 5 (2 of 36). A difference of 0 is not the most likely, which is the surprise for many students. |
| 4 | Any correct description, for example “a difference of 0 has 6, then 1 has 10, and each difference after that has 2 fewer outcomes than the one before, down to 2”. A description of the counts only for 1 to 5 earns the lower band; the 0 case should be handled separately for the top band. |
| 5 | Four-faced: 0: 4, 1: 6, 2: 4, 3: 2 (total 16). Prediction from the rule, then a list of the 16 outcomes to check. |
| 6 | Eight-faced: 0: 8, 1: 14, 2: 12, 3: 10, 4: 8, 5: 6, 6: 4, 7: 2 (total 64). A difference of 3 has 10 outcomes. Difference 5 or more: 6 + 4 + 2 = 12, so 12/64 = 3/16. |
| 7 | Credit a test that gets at least four differences right for a die with a number of faces of the student's choice and says clearly whether the rule held. For 5 faces: 5, 8, 6, 4, 2 (total 25). The count for 0 is the number of faces and the others run down by 2; what changes is the starting count and the number of differences. |
| 8 | The grid has n cells on the main line (difference 0). For a difference k from 1, there are n − k cells on the line on one side of it and n − k on the other, so 2(n − k). Expect this at 7 to 8 only. A check that works: n + 2(n − 1) + … + 2 × 1 = n + n(n − 1) = n2. |
| 9 | n/n2 = 1/n. There are always n outcomes with a difference of 0 out of n2; with more faces there are many more outcomes, but only n of them are doubles, so the share shrinks. |
Exemplar answers
| Task B1 response | What it looks like |
|---|---|
| Level 3 to 4 | “Counts for differences 0 to 5: 6, 10, 8, 6, 4, 2. After 10 the numbers go down. For the four-faced die I predict 4, 6, 4, 2 and the list agrees.” Both sets of counts right and a rule that fits, but it is not written so that it could be used for the eight-faced die, and no other die is tested. |
| Level 7 to 8 | “For a die with n faces, a difference of 0 has n outcomes. Any other difference k has 2 × (n − k): one lot above the doubles line and one below. For 8 faces, difference 3 is 2 × 5 = 10 and 5 or more is 6 + 4 + 2 = 12, so 12/64 = 3/16. I tested 5 faces: 5, 8, 6, 4, 2 gives 25 = 5 × 5. The rule works because each line of the grid that is k away from the doubles line has n − k cells, and there is one on each side. Zero is different because there is only one doubles line.” |
Differentiation
A way in for anyone stuck
- Give a blank six by six grid for Part 1 and let the student choose how to show the differences.
- Suggest making the grid for the four-faced die first, because it is smaller, then the six-faced one.
- Prompt with a question, never the rule: “Which difference has the most cells, and what is next to it?”
Stretch
- Compare the difference of two six-faced dice with the total. Which is more likely, a difference of 1 or a total of 7, and by how much? (5/18 against 1/6: 10 outcomes against 6.)
- For six-faced dice, which differences have more outcomes than a difference of 0, and what happens to the count at a difference of 3? (1 and 2 have more; 3 has the same, 6.)
Students learning in English as an additional language
| Word in task B1 | Plain meaning |
|---|---|
| difference | the answer when you take the smaller number from the bigger one |
| face | one side of a die, with a number on it |
| predict | say what will happen before you check |
| rule | a sentence that tells you what happens every time |
Sentence starters for task B1:
- I noticed that the counts …
- My rule is: …
- I tested my rule on a die with … faces and it worked / did not work because …
Task C1: Explain it to someone who missed it
Why set it. Criterion C on its own: language, a table, a graph and a line of reasoning a classmate can follow, built around three real errors in reading experimental results from lesson 5.
When: After lesson 5. Time: 35 minutes. Format: Individual, written, in class or as homework. One page of words, a table and a graph.
The student sheet, with its task-specific clarification
When: After lesson 5. Time: 35 minutes. Calculator: no calculator. Criterion: C, communicating.
Mei was away for lesson 5. She copied this working from a classmate, Matteo, and it has mistakes in it.
Matteo’s working
A spinner has five equal sections numbered 1 to 5. I spun it 20 times and got a 5 three times.
(a) The experimental probability of a 5 is 3/20, which is 0.15. This is less than 1/5 = 0.2, so the spinner is unfair.
(b) I spun it 80 more times and got a 5 seventeen times. 17/80 = 0.2125. To find the probability for all of my spins I average the two: (0.15 + 0.2125) ÷ 2 = 0.18125.
(c) A 5 came up too rarely so far, so it is due: it will come up more often in the next 80 spins to catch up.
Write a one-page explanation for Mei. She should be able to follow it on her own, without asking anyone.
Part 1. Put it right
- Find every mistake in Matteo’s working. For each one, say what he did, why it is wrong, and write what he should have said or done.
Part 2. Teach the idea
- Matteo kept a running count of his spins, and after the first 100 he spun 100 more. He had a 5 on 3 of the first 20 spins, 9 of the first 50, 20 of the first 100 and 38 of the first 200. Show these results in a table with the experimental probability of a 5 for each number of spins.
- Draw a graph with the number of spins across and the experimental probability up. Add a line at 0.2. Describe, in a sentence for Mei, what the graph shows about few spins and many spins.
Part 3. Check your explanation
- Write one sentence that Mei could use to decide whether a result from a small number of trials is enough to say a spinner or a die is unfair.
- Read your page as if you were Mei. Check that you used the words trial, outcome, experimental probability and theoretical probability, that fractions and decimals are labelled, and that the table and graph have headings and units.
How your work is judged
| Level | Criterion C in task C1: what it looks like |
|---|---|
| 1–2 | Some of your explanation can be followed, but key steps or words are missing, and the table or graph is incomplete or unlabelled. |
| 3–4 | You use some correct words and a table or graph with labels; your steps are complete and in a sensible order, so Mei could follow most of it. |
| 5–6 | Your table and graph agree with each other and with what you say, you usually use the right words, and the page is mostly well ordered, with each mistake explained before the next. |
| 7–8 | Your words, table and graph tell the same story and each adds something; you always use the key words correctly, label every quantity and axis, and the page is ordered so that Mei could learn from it alone. |
Markscheme
| Task C1 question | Answer and marking |
|---|---|
| The mistake in (a) | 20 spins is too few to call the spinner unfair: 3/20 = 0.15 is a little below 0.2, and 3 fives against an expected 4 is a small gap that chance can easily make. Look for the reason that the number of trials is small, not just that 0.15 is not 0.2. |
| The mistake in (b) | Averaging two fractions treats 20 spins and 80 spins as equal. All the spins together are 100 and fives total 3 + 17 = 20, so the experimental probability is 20/100 = 0.2. His 0.18125 is too low because the 20-spin run counts too much. |
| The mistake in (c) | Spins have no memory: the chance of a 5 on each new spin is still 1/5 whatever came before. More spins settle the relative frequency near 0.2 because the new spins dilute the early ones, not because they make up for them. |
| Table (Part 2) | 20 spins: 3, 0.15; 50 spins: 9, 0.18; 100 spins: 20, 0.20; 200 spins: 38, 0.19. Headings with units; probability as decimals or fractions, labelled. |
| Graph (Part 2) | Four points at (20, 0.15), (50, 0.18), (100, 0.20), (200, 0.19), scales labelled and the line at 0.2. Look for a sentence such as “with few spins the result is further from 0.2; with more spins it gets closer, but it does not have to land exactly on it”. |
| The sentence (Part 3) | Any sentence such as “a few trials can be far from the theoretical probability, so use many trials before deciding”. |
| How to level | Level the page as a whole against the clarification, not by counting corrections. Notation, labelling and organisation decide between 5 to 6 and 7 to 8 more often than the corrections do. Mathematical slips count only where they break the line of reasoning Mei has to follow; accuracy itself is criterion A. |
Exemplar answers
| Task C1 response | What it looks like |
|---|---|
| Level 3 to 4 extract | “(a) wrong because 20 spins is not many. (b) You should add them: 3 + 17 = 20 and 20 + 80 = 100 so it is 20/100. (c) Wrong. Table: spins 20, 50, 100, 200, probability 0.15, 0.18, 0.2, 0.19.” The corrections are right but the reasons are short, the graph has no axis labels and the table has no heading for the probabilities. |
| Level 7 to 8 extract | “Mistake 2: Matteo averaged 0.15 and 0.2125, but the first run is only 20 spins and the second is 80, so they should not count the same. Add the fives and the spins: (3 + 17) out of (20 + 80) = 20/100 = 0.2. Mistake 3: each spin is independent, so a 5 is never due; the experimental probability moves towards 0.2 because the later spins are added to the early ones. The graph shows 0.15 at 20 spins and 0.19 at 200 spins, closer to 0.2 but not exactly on it.” Each correction has a stated reason and the table, graph and sentences agree. |
Differentiation
A way in for anyone stuck
- Give the table outline (headings for spins, fives and experimental probability) and the axes for the graph, and let students fill them.
- Let students explain aloud to a partner first, then write down what they said.
- Offer the frame: “Matteo wrote …. This is wrong because …. He should have ….”
Stretch
- Explain why more trials make the experimental probability tend to settle, but do not make it exactly equal to the theoretical probability.
- Write a second version of the page for a student two years younger. What did you change, and why?
Students learning in English as an additional language
| Word in task C1 | Plain meaning |
|---|---|
| trial | one go at the experiment, such as one spin |
| experimental probability | how often something happened out of the number of trials |
| theoretical probability | the probability worked out by counting equally likely outcomes |
| unfair | not equally likely for every outcome |
| average | adding values and dividing by how many there are |
Sentence starters for task C1:
- Matteo wrote …, but …
- All together there were … spins and … fives, so …
- The graph shows that when there are more spins, …
Task A2: End-of-unit test
Why set it. The whole unit at three levels of demand. Section 3 is unfamiliar on purpose: it is where the evidence for 7 to 8 comes from. Use the parallel version on this page if a student has met the sheet before.
When: End of unit, after lesson 6. Time: 45 minutes. Format: Individual, written, in class, one lesson. 28 marks in three sections.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 6. Time: 45 minutes. Calculator: no calculator. Criterion: A, knowing and understanding.
Three sections, from familiar questions to ones you have not met in this form. Show your working everywhere: it earns credit even when an answer slips.
Section 1. Familiar questions
- A bag holds 6 red, 9 green and 5 white counters. One is taken at random. Write the probability that it is green as a fraction in its simplest form and as a percentage. (2 marks)
- Find the probability that the counter is not white. (1 mark)
- A fair coin is tossed and a fair five-sided spinner with the numbers 1 to 5 is spun. (a) Show every outcome in a list or table and say how many there are. (b) Find the probability of a tail and an even number. (3 marks)
- An event has a probability of 0.06. Say in words how likely it is, and write the probability that it does not happen as a percentage. (2 marks)
- Two fair dice are rolled and the scores are added. Find the probability that the total is 9 or more. (2 marks)
Section 2. More demanding questions
- A family chooses a meal: rice, noodles or bread, then chicken, tofu or egg, then mango or pineapple for dessert. Draw a tree diagram or a list. How many different meals are there? How many have tofu and mango? (4 marks)
- A drawing pin is dropped 80 times and lands point up 52 times. Write the experimental probability of point up as a fraction in its simplest form and as a decimal. How many times would you expect point up in 200 more drops? (3 marks)
- Omar rolls a fair die 240 times and gets a five 47 times. How many fives would you expect? Does this show that the die is unfair? Explain. (3 marks)
Section 3. Situations you have not met in this form
- Lucia and Omar roll two fair four-faced dice, with the faces 1 to 4. Lucia scores a point when the product of the two numbers is even; Omar scores when it is odd. Is the game fair? Show how you know. (4 marks)
- A stall owner says that a lucky-draw bag gives a prize with probability 0.2. Rosa simulates one draw with a random digit from 0 to 9, where 0 or 1 means a prize. In 50 random digits she gets a prize 14 times. Explain why her rule suits the bag, say how many prizes she expects, and say what her result suggests and what she could do next. (4 marks)
How your work is judged
| Level | Criterion A in task A2: what it looks like |
|---|---|
| 1–2 | You answer some of the Section 1 questions correctly, for example a probability as a fraction or the number of outcomes. |
| 3–4 | You answer most of Section 1 and make a start on Section 2: a tree or list, an experimental probability or an expected number. |
| 5–6 | You answer Section 1 and most of Section 2 correctly with working shown, and make a good start on Section 3. |
| 7–8 | You answer Sections 1 and 2 correctly and solve both questions in Section 3, with reasons that show your answer works, including a clear comment on what chance can and cannot show. |
Markscheme
| Task A2 question | Answer and marking |
|---|---|
| 1 | 9/20 = 45%. 1 mark each. Counters 6 + 9 + 5 = 20. 9/11 compares green with the other colours instead of all 20 counters. |
| 2 | 15/20 = 3/4. 1 mark (1 − 5/20, or 6 + 9 = 15). |
| 3 | (a) 10 outcomes: H1, H2, H3, H4, H5, T1, T2, T3, T4, T5. (b) 2/10 = 1/5 (T2 and T4). 2 marks for a complete list or table with 10, 1 mark for (b). |
| 4 | Very unlikely, close to impossible; P(not) = 0.94 = 94%. 1 mark each. |
| 5 | 10/36 = 5/18. Totals of 9, 10, 11 and 12 come from 4 + 3 + 2 + 1 = 10 of the 36 outcomes. 1 mark for 10 of 36, 1 mark for the simplified fraction. Counting only the total of 9 (4/36) is the usual slip. |
| 6 | 18 meals; 3 with tofu and mango (rice, noodles or bread). 2 marks for a correct tree or list (1 mark if a stage is missing), 1 for 18 (3 × 3 × 2), 1 for 3. |
| 7 | 52/80 = 13/20 = 0.65; expect 130 (0.65 × 200). 1 mark for 13/20, 1 for 0.65, 1 for 130. 52/200 is a common slip. |
| 8 | 40 fives expected (240 ÷ 6). 47 is 7 more, but a run of 240 rolls can differ from the expected number by that much by chance: it is not enough to say the die is unfair. 1 mark for 40, 1 for comparing 47 with 40, 1 for a cautious, reasoned conclusion (more rolls would help). |
| 9 | No, it is not fair. 16 outcomes. The product is odd only when both numbers are odd: (1, 1), (1, 3), (3, 1), (3, 3), so 4 outcomes, P(odd) = 4/16 = 1/4 for Omar, and 12/16 = 3/4 for Lucia. 1 mark for the 16 outcomes, 1 for 4 odd outcomes, 1 for both probabilities, 1 for the conclusion. |
| 10 | Two digits out of ten (0 and 1) is 2/10 = 0.2; 10 prizes expected in 50; 14 is 4 more, which is plausible by chance and does not show the owner is wrong; repeat with many more digits. 1 mark each. Accept a different cautious comment with a reason. |
| Marks to a level | Total 28. 1 to 6 marks: 1 to 2. 7 to 13: 3 to 4. 14 to 21: 5 to 6. 22 to 28, with at least 5 of the 8 marks in Section 3: 7 to 8. Use the marks as a guide, then check the student's profile against the clarification before settling on a level. |
Exemplar answers
| Task A2 response | What it looks like |
|---|---|
| Level 3 to 4 profile | Section 1: question 1 right (2), question 2 right (1), question 3 lists 10 outcomes but gives 2/5 (2), question 4 gives “unlikely” and 6% (1), question 5 gives 10/36 unsimplified (1). Section 2: question 6 gives 18 with an incomplete tree (1), question 7 gives 52/80 = 0.65 but does not simplify the fraction (1), question 8 gives 40 and nothing more (1). Section 3: question 9 lists 16 outcomes (1), question 10 gives 10 expected (1). 12 marks. |
| Level 7 to 8 answer to question 9 | “There are 16 outcomes, 4 × 4. A product is odd only if both numbers are odd, and there are two odd numbers, 1 and 3, so 2 × 2 = 4 outcomes: (1, 1), (1, 3), (3, 1), (3, 3). So Omar has 4/16 = 1/4 and Lucia has 12/16 = 3/4. Lucia is three times as likely to score, so the game is not fair.” |
| Level 7 to 8 answer to question 10 | “Digits 0 and 1 are 2 of the 10 digits, 2/10 = 0.2, matching the owner’s probability. In 50 digits I expect 50 × 0.2 = 10 prizes. She got 14, 4 more. A short run can differ from the expected number by this much by chance, so it does not show the owner is wrong. I would run many more digits, such as 500, and see whether the share of prizes settles near 0.2.” |
Differentiation
A way in for anyone stuck
- Give a blank two-way table for question 3 and a stem for the tree in question 6.
- Allow a labelled 6 by 6 grid for question 5 and a blank 4 by 4 grid for question 9.
- Read question 10 aloud on request and suggest listing the ten digits first. Reading is not what is being assessed.
Stretch
- Question 9 again with the faces 1 to 6. Is the game fair now, and who is favoured? (No: 9 odd of 36 outcomes, P(odd) = 1/4, so Lucia is still favoured.)
- In question 10, Rosa uses 500 random digits instead of 50. How many prizes does she expect? (100.) Would 120 prizes then be as easy to get by chance as 14 prizes in 50 digits? Explain in terms of the share of prizes. (Both are a share of 0.24 or 0.28 against 0.2; accept a reasoned answer that longer runs stay nearer the expected share, so 120 of 500 is a bigger surprise.)
Students learning in English as an additional language
| Word in task A2 | Plain meaning |
|---|---|
| product | the answer when you multiply two numbers |
| fair | the same chance of winning for everyone |
| expect | say how many times you think something will happen |
| simulate | use random numbers to act out a real event |
Sentence starters for task A2:
- There are … outcomes in all and … of them …, so …
- I expect … because … × … = …
- This does / does not show that … because …
Task D1: The lucky-draw stall
Why set it. Theoretical probability, expected numbers, simulation and judging a result inside one real decision about a temple fair stall. An alternative to D2: set one of them.
When: End of unit, after lesson 6. Time: 45 minutes. Format: Individual, in class. A written solution with explanations and a short note to the stall owner.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 6. Time: 45 minutes. Calculator: no calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
At a temple fair, a stall owner runs a lucky draw. A bag holds 50 tickets: 5 gold, 10 silver and 35 blank. Each player pays 20 coins, takes one ticket at random and puts it back. A gold ticket wins a prize that costs the owner 60 coins, and a silver ticket a prize that costs 15 coins. The owner expects 400 players in the evening and wants a profit of at least 4000 coins.
Part 1. The chances
- Find the probability of a gold ticket, of a silver ticket, of any prize and of no prize. Give each as a fraction in its simplest form.
- Describe how to simulate one draw with a random digit from 0 to 9, so that gold, silver and blank come up with the right probabilities.
Part 2. The evening
- Out of 400 players, how many gold and silver tickets should the owner expect? Work out the money taken in, the cost of prizes and the profit.
- Before the fair, the owner drew a ticket 50 times, replacing it each time, and got 12 prizes. Compare this with what you expect. Does it show something is wrong with the bag?
Part 3. A change
- The owner thinks of swapping 5 blank tickets for silver ones. Work out the new probability of a silver ticket, and the new expected profit for 400 players. Does it meet the target?
- How reliable is your expected profit for the evening? Say what could make the real profit different, and what you would tell the owner to do.
- Write a short note to the owner with your recommendation, so that someone who has not seen your working can understand it.
How your work is judged
| Level | Criterion D in task D1: what it looks like |
|---|---|
| 1–2 | You find some of the right numbers, for example the probability of a gold ticket or the number of prizes, but do not reach a decision. |
| 3–4 | You find the probabilities, the expected prizes and the profit, and say whether the swap meets the target and whether your answers make sense. |
| 5–6 | You choose suitable methods for every part, reach correct answers, compare the 12 prizes with what you expect, explain how reliable the expected profit is and explain whether the swap makes sense. |
| 7–8 | You choose and use suitable methods successfully throughout, reach correct answers, justify how reliable your expected profit is and why, and justify whether the swap and your recommendation make sense for a real evening. |
| Level | Criterion C in task D1: what it looks like |
|---|---|
| 1–2 | Some working is shown, but it is hard to see how you reached the probabilities or the profit. |
| 3–4 | Your working for the probabilities and the money can be followed, with some correct notation and units (coins). |
| 5–6 | Your working is clear and complete, usually with correct units and fraction notation, and your note to the owner is mostly well organised. |
| 7–8 | Your working is clear, complete and well ordered, units and notation are always correct, and your note to the owner makes sense on its own. |
Markscheme
| Task D1 question | Answer and marking |
|---|---|
| 1 | Gold 5/50 = 1/10; silver 10/50 = 1/5; any prize 15/50 = 3/10; no prize 35/50 = 7/10. The four add to 1 as gold + silver + none (1/10 + 2/10 + 7/10). 1 mark for gold and silver, 1 for the other two. |
| 2 | One digit for gold, two for silver, seven for blank: for example 0 is gold, 1 and 2 are silver, 3 to 9 are blank (digits are equally likely, so these share out 1/10, 2/10 and 7/10). Accept any split with 1, 2 and 7 digits. |
| 3 | Gold 40, silver 80 (120 prizes). Money in 400 × 20 = 8000 coins. Prizes cost 40 × 60 = 2400 and 80 × 15 = 1200, so 3600 coins. Profit 4400 coins. |
| 4 | Expect 50 × 3/10 = 15 prizes; 12 is 3 fewer. A run of 50 draws can be this far from 15 by chance, so it does not show the bag is wrong; more draws, or counting the tickets, would be better evidence. Look for the cautious conclusion. |
| 5 | Silver tickets become 15: P = 15/50 = 3/10. Silver prizes 120 × 15 = 1800; gold unchanged 2400; prizes cost 4200; profit 8000 − 4200 = 3800 coins, which misses the 4000 target by 200. |
| 6 | Expected values are an average over many evenings, not a promise. More silver or gold winners than expected, or fewer players, changes the profit. Each extra gold winner costs the owner 60 coins beyond expectation. At 4400 expected there is a margin of 400 over the target; the owner should keep the bag as it is. |
| 7 (criterion C) | The note gives the probability of a prize, the expected profit and the recommendation, with units, in sentences that make sense without the working. |
Exemplar answers
| Task D1 response | What it looks like |
|---|---|
| Level 3 to 4 (D) | “Gold 5/50, silver 10/50, any prize 15/50. 400 players, 120 prizes, 8000 in, profit 4400. After the swap the profit is 3800, which is less than 4000, so no.” Numbers right, but the simplest-form fractions, the 12 out of 50 comparison and any comment on reliability are missing. |
| Level 7 to 8 (D and C) | “Digits 0 for gold, 1 and 2 for silver, 3 to 9 for blank. I expect 15 prizes in 50 draws; 12 is 3 fewer, which chance alone can produce, so I would not change the bag on that. Expected profit is 4400 coins. After swapping 5 blanks, silver is 3/10, the profit is 3800, which misses 4000. The 4400 is an expected value: if more people win gold the profit falls, so I recommend keeping the bag, with a margin of about 400 coins.” |
Differentiation
A way in for anyone stuck
- Give a frame for Part 2: a three-row table headed tickets, number expected and cost.
- Allow a drawn row of 50 tickets so that the counts can be seen.
- For Part 3, ask: “If more people than expected win gold, what happens to the profit?”
Stretch
- How many blank tickets could be swapped for silver and still give a profit of at least 4000 coins? (3: each swap adds 8 silver winners, 120 coins; 4400 − 360 = 4040, but 4 swaps give 3920.)
- With the 15 silver tickets, what ticket price would give the owner exactly 4000 coins profit for 400 players? (20.5 coins: (4200 + 4000) ÷ 400.)
Students learning in English as an additional language
| Word in task D1 | Plain meaning |
|---|---|
| profit | money taken in minus money paid out |
| ticket | a small piece of paper drawn from the bag |
| replace | put back, so the bag is the same again |
| recommend | say what someone should do, and why |
Sentence starters for task D1:
- The probability of … is … out of …, which is …
- I recommend … because …
Task D2: The two-dice game at the fun day
Why set it. Sample spaces, theoretical probability, expected numbers and experimental evidence inside one real decision about a charity game. An alternative to D1: set one of them.
When: End of unit, after lesson 6. Time: 45 minutes. Format: Individual, in class, one lesson. A written solution with explanations and a short reply to the organisers.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 6. Time: 45 minutes. Calculator: no calculator. Criteria: D, applying mathematics in real-life contexts; C, communicating.
A community centre holds a charity fun day. One game costs 10 coins to play. The player rolls two fair dice once. The player wins if the total is 2, 3, 11 or 12, and the organisers then pay the winner 50 coins. The organisers expect 360 players and want to raise at least 800 coins for the charity from this game.
Part 1. The game as it stands
- Show the outcomes that win. Find the probability that a player wins, as a fraction in its simplest form.
- Out of 360 players, how many winners should the organisers expect? Work out the money taken in, the prizes paid and the profit.
- Does the game meet the 800 coins target? By how much does it miss or beat it?
Part 2. Evidence and risk
- Rosa watched 60 plays and saw 13 winners. Write the experimental probability and compare it with the probability you found. Does this show that the game is not what the organisers think?
- Suppose that on the day 75 of the 360 players win. Work out the profit. What does this tell the organisers about the expected profit you found?
Part 3. A better game
- Keep the price and the prize the same, and choose new winning totals for the game. Show that your choice should meet the 800 coins target, using the probability and the expected profit.
- Write a short reply to the organisers with your recommendation, how reliable it is, and anything the numbers leave out.
How your work is judged
| Level | Criterion D in task D2: what it looks like |
|---|---|
| 1–2 | You find some of the right numbers, for example how many ways to win, but do not reach a decision. |
| 3–4 | You find the probability, the expected winners and the profit, say whether the game meets the target and whether your answers make sense. |
| 5–6 | You use a full list of outcomes and expected numbers correctly, compare Rosa's 60 plays with the theory, explain how reliable the expected profit is and explain whether your new game makes sense. |
| 7–8 | You choose and use suitable methods successfully throughout, reach correct answers, justify how reliable the expected profit is, and justify whether your new game is a sensible choice for a real fun day. |
| Level | Criterion C in task D2: what it looks like |
|---|---|
| 1–2 | Some working is shown, but it is hard to see how you found the probability or the profit. |
| 3–4 | Your working for the winners and the money can be followed, with some correct notation and units (coins). |
| 5–6 | Your working is clear and complete, usually with correct units and notation, and your reply to the organisers is mostly well organised. |
| 7–8 | Your working is clear, complete and well ordered, units and notation are always correct, and your reply to the organisers makes sense on its own. |
Markscheme
| Task D2 question | Answer and marking |
|---|---|
| 1 | 6 winning outcomes of 36: total 2 (1, 1); total 3 (1, 2), (2, 1); total 11 (5, 6), (6, 5); total 12 (6, 6). P(win) = 6/36 = 1/6. 1 mark for the list or grid, 1 for the probability. |
| 2 | Expect 360 × 1/6 = 60 winners. Money in 360 × 10 = 3600 coins; prizes 60 × 50 = 3000 coins; profit 600 coins. |
| 3 | No: 600 is 200 coins short of 800. |
| 4 | 13/60 is about 0.22, against 1/6 which is about 0.17; expected 10 winners in 60 plays, seen 13. A run of 60 plays can differ from the expected number by 3 by chance, so it is not enough to say the game is not as thought; more plays would help. |
| 5 | 75 × 50 = 3750 coins in prizes; 3600 − 3750 = −150, a loss of 150 coins. The expected profit is an average: 15 more winners than expected turn a profit into a loss, so the organisers should not count on it. |
| 6 | Any winning set with at most 5 of the 36 outcomes. The most that meets the target is 56 winners in 360 (3600 − 800 = 2800, and 2800 ÷ 50 = 56), a probability of at most 56/360, which is 5.6 of the 36 cells, so at most 5 cells. For example total 8 (5 outcomes): P = 5/36, expect 360 × 5/36 = 50 winners, prizes 2500 coins, profit 1100 coins. Total 6 also has 5. 6 outcomes or more give at least 60 winners and miss the target. Look for P and the profit worked out, not only the choice. |
| 7 (criteria C and D) | A strong reply gives the new game, its probability and its expected profit (1100 coins for a total of 8), says that this is an average and that the real number of winners varies from fun day to fun day, so there is some risk of falling short or of a loss even with a good margin, and may suggest a margin or a higher price. |
Exemplar answers
| Task D2 response | What it looks like |
|---|---|
| Level 3 to 4 (D) | “Winning outcomes: 1+1, 1+2, 2+1, 5+6, 6+5, 6+6, so 6 out of 36. 360 players, 60 winners, 3600 in, 3000 out, profit 600. Not 800. Winning on 8 is 5/36 so it should be more.” The numbers are right but the new profit is not worked out and Rosa's 60 plays are compared without a comment on chance. |
| Level 7 to 8 (D and C) | “Winning on a total of 8: 5 outcomes, 5/36. I expect 360 × 5/36 = 50 winners, prizes 2500, profit 1100 coins, which meets 800 with 300 to spare. This is an expected value only: with 75 winners in the original game they would lose 150 coins, so the profit is not certain. I recommend total 8, and a little margin by keeping a reserve for a bad run.” |
Differentiation
A way in for anyone stuck
- Give a blank six by six grid with the totals filled in for question 1.
- Let students list the money in, the prizes paid and the profit in a three-line table.
- For Part 3, ask: “How many winners can the organisers afford, and how many of the 36 outcomes is that?”
Stretch
- Another organiser suggests winning only on a total of 7. Is that a better or worse game for the organisers than the original one? (The same: 6/36 = 1/6, so the same expected profit of 600 coins.)
- Keep the original winning totals and the 10 coins price, and cut the prize to 40 coins. What is the expected profit? (60 winners × 40 = 2400; 3600 − 2400 = 1200 coins, which meets the target.)
Students learning in English as an additional language
| Word in task D2 | Plain meaning |
|---|---|
| winner | a player who gets the prize |
| total | the answer when you add the two scores |
| raise | collect money for a purpose |
| target | the amount you hope to reach |
Sentence starters for task D2:
- There are … winning outcomes out of 36, so …
- I expect … winners, so the profit is …
- This is only expected because …
Practicalities
Every task is original and written for this unit; none is copied or adapted from a published book.
The student sheets, with their task-specific clarifications, are on this teacher page only, so every task stays unseen until you set it. Share a sheet's clarification with the class beforehand if you want them to know what each level looks like. Markschemes and exemplars are only on this page.
To print a sheet, open it on the unit page, then print the page. No task needs a calculator: the lessons do not use one and the numbers are chosen to be done by hand.
If a student may already have met A1 and A2 (a sibling, a previous class), use this parallel version (same marks, same markscheme notes). Q1 4 red, 11 green and 5 white counters: P(green) 11/20 = 55%. Q2 P(not white) 15/20 = 3/4. Q3 a coin and a three-sided spinner (1 to 3): 6 outcomes; tail and an odd number 2 of 6 = 1/3. Q4 probability 0.03: very unlikely; P(not) 97%. Q5 two dice, a total less than 5: 6/36 = 1/6. Q6 3 drinks, 2 snacks, 4 fruits: 24 meals; coconut water and mango 2. Q7 90 drops, 63 point up: 63/90 = 7/10 = 0.7; 0.7 × 200 = 140. Q8 150 rolls: 25 fives expected; 31 is 6 more, not enough to call the die unfair. Q9 two three-faced dice (1 to 3), Lucia scores on an even product, Omar on an odd one: 9 outcomes, 4 odd (1 and 3 both ways) and 5 even, so 5/9 against 4/9: not fair. Q10 digits 0, 1 and 2 are a prize (probability 0.3); 40 digits: 12 expected, 8 seen, plausibly chance, so repeat with more digits. For an unseen A1 (same marks, same markscheme notes): Q1 0.45 (9/20; 45%). Q2 6 red, 9 blue and 5 yellow counters (P(blue) 9/20; P(not blue) 11/20). Q3 5 drinks and 2 snacks (10), then a glass or a cup (20). Q4 a coin, then a spinner with 1 to 4, then the coin again (16 outcomes; exactly one head: 8 of them, H1T to H4T and T1H to T4H). Q5 spinner A 1, 2, 3, 4 and spinner B 1 and 5; Tia says 6 because 4 + 2 (8 outcomes: (1, 1), (1, 5), (2, 1), (2, 5), (3, 1), (3, 5), (4, 1), (4, 5)). Q6 a bike lock with 2 dials, each 1 to 5 (25 codes; 20 with two different digits, 5 × 4).
The Year 3 descriptor bands on this page are band summaries written for this site; check them against the current IB MYP Mathematics guide before relying on them for reporting.