At a glance. One lesson of about 60 minutes. It assumes fraction multiplication and the sample-space work in Lesson 3. Students draw and label two- and three-stage trees, multiply along routes, add routes, and distinguish with and without replacement.
Calculator: GDC optional. A GDC can multiply fractions and check a tree, but every product here is small and is better done by hand, so the optional section only checks.
What students will learn
Know
- The probabilities on branches from one point add to 1.
- Multiply along a route, add across routes.
- Without replacement the second-stage fractions change.
Understand
- A route is an “and then” event, so its probability is a product.
- An event that happens along several routes needs the routes added.
- At least one is the complement of none.
Do
- Draw a tree for two or three stages and label every branch.
- Find the probability of one route, of several routes, and of at least one.
- Say whether a situation is with or without replacement.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: tree diagrams with and without replacement, products and sums of routes (questions 1 to 8).
- Criterion B. Investigating patterns: the first figure. Ask students to predict how the second-stage branches change without replacement, and the second figure's growth of at least one with more trials.
- Criterion C. Communicating: labelling every branch and showing which routes were added (the worked example and question 4).
- Criterion D. Applying mathematics in real-life contexts: draws from a drum, selecting names, seeds that may or may not grow.
The one thing to do with the figure
Make students commit before they slide. In the first figure, ask “Without replacement, will P(same colour) go up, down or stay the same?” and take a show of hands before pressing the button.
In the second figure, ask how many trials of a 20% chance are needed for at least one success to be more likely than not. Students can test two, three and four trials.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 7 min | Hook: the drum at the fair. Collect predictions about with and without replacement. |
| 10 min | First figure and key idea: the tree and the two rules. Settle the hook. |
| 10 min | Second figure, the warning card and the worked example; pairs finish the faded example. |
| 5 min | The check. If many choose C or D, return to the bag and count what is left. |
| 18 min | Your turn, Levels 1 to 3, at each student's own pace. |
| 5 min | Something to think about, then the exit ticket, question 9. |
The answers
| Warm-up | 1/2; 7/12; 0.2. |
|---|---|
| Hook | Once the first ticket is known to be a winner, not the same: with w winners, the student who keeps it has (w − 1)/99 and the friend still has w/100. The friend is more likely to win two prizes: w/100 × w/100 against w/100 × (w − 1)/99. (If nothing is known about the first ticket, the second wins with probability w/100 for both; question 8 shows this, 63/90 = 7/10.) |
| Worked example | P(one of each) = 24/56 = 3/7. |
| Faded example | P(exactly one grey) = 42/100 = 21/50. |
| 1. A second-stage branch | B. A ignores the removal; C changes only the top; D changes only the bottom. |
| 2. Two independent events | 0.1225. 0.35 × 0.35. |
| 3. Without replacement, one route | 30. 6 × 5 over 10 × 9. |
| 4. Adding two matching routes | 54. 42 + 12. |
| 5. At least one, by complement | 0.9775. 1 − 0.15 × 0.15. |
| 6. Different colours, without replacement | 48. 24 + 24. |
| 7. A three-stage route | 336. 8 × 7 × 6. |
| 8. A later stage from all routes | 63. 42 + 21. |
| 9. Both the same, without replacement | A. B puts the name back; C is the first stage only; D is the second stage only. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| A, C or D in question 1 | Ignores the removal, or changes only the top or only the bottom. | Use physical counters: take one out and count again. |
| 0.7, 0.4225 or 0.5775 in question 2 | Adds along a route, or gives neither or at least one. | Ask: do both have to happen? So is it an “and” or an “or”? |
| 36, 12 or 48 in question 3 | Put the biscuit back, found both plain, or found one of each. | Cross out the biscuits on a drawing as they are taken. |
| 42, 65 or 56 in question 4 | Found only one route, used replacement, or found one of each. | Write the two routes as a list before any arithmetic. |
| 1.7, 0.7225 or 0.0225 in question 5 | Added, found both succeed, or gave the complement. | Ask what a probability above 1 would mean. |
| 24, 56 or 6 in question 6 | Found one route, or a same-colour route. | Colour the two different-colour routes on the tree. |
| 512, 56 or 1320 in question 7 | Used replacement, stopped early, or gave the bottom. | Draw a three-stage tree for just the all-red route. |
| 42, 21 or 60 in question 8 | Found one of the two routes, or only the second-stage branch. | Highlight every route that ends in red on the second draw. |
| B, C or D in question 9 | Put the name back, or gave only the first or second stage. | Say it aloud: first a girl, and then a girl. |
Questions worth asking
- Why do the branches from one point add to 1? Listen for “one of them must happen: the branches cover every possible result”.
- Without replacement, why is the second-stage fraction different after red than after blue? The bag is different: one fewer counter and, for the colour drawn, one fewer of that colour.
- When do two events need multiplying, and when adding? “And then” multiplies; “or” adds.
- Why is at least one easier as 1 minus none? At least one covers many routes, while none is a single route.
Differentiation
A way in for anyone stuck
- Give a partly completed tree with the first stage filled in.
- Use a checklist: do the branches from each point add to 1? Did I multiply along and add between?
- Use real counters in a bag and let students take them out without looking.
Stretch
- Draw a three-stage tree for three fair coins and find P(at least two heads). Then find it again as 1 minus P(at most one head).
- Show that P(second red) equals P(first red) for any bag drawn without replacement.
Students learning in English as an additional language
“Replacement” means putting something back. “With replacement” = put it back; “without replacement” = keep it out. “Route” is a path from the start to the end of the tree, “branch” is one line of it.
| Word | Plain meaning | Example |
|---|---|---|
| tree diagram | a diagram that shows every route through several stages, with a probability on each branch | the picture for two draws from a drum |
| branch | one line of a tree diagram, labelled with a probability | the line marked 3/7 |
| route | a path from the start to the end of a tree; one possible outcome | red then blue |
| with replacement | each item is put back before the next draw, so the stages are independent | draws from a bag where the sweet is returned |
| without replacement | an item is kept out, so the next stage has changed | names pulled from a hat and set aside |
| independent | the result of one stage does not change the probabilities of the next | a coin toss and a die roll |
Sentence starters for explaining aloud or in writing:
- The probability on this branch is … because ….
- I multiplied along the route because both events have to ….
- I added these routes because the event can happen ….
Two things not to say
“Just add the probabilities.” Along a branch, both have to happen, so they are multiplied.
“The bottom number stays the same.” Without replacement the number of items goes down by 1 on the second stage.
Questions to set
Extra practice in three tiers, including one question from earlier work for spacing. Every answer has been checked.
1Fluency
The idea on its own, with friendly numbers. Set these first.
- IndependentA fair coin is tossed twice. Draw a tree and find P(two tails).
1/2 × 1/2 = 1/4. - Without replacementA bag has 3 red and 2 blue counters. Two are taken. P(both blue)?
2/5 × 1/4 = 2/20 = 1/10.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- ContextOn a bus in Lagos, each of 3 independent passengers has a 0.6 chance of paying by card. What is P(at least one pays by card)?
1 − 0.4 × 0.4 × 0.4 = 1 − 0.064 = 0.936. - Earlier work (Lesson 3)A fair spinner has 8 equal sectors numbered 1 to 8. P(a number greater than 5)?
3/8.
3Challenge
Reasoning, generalizing and working backwards.
- ReasoningWithout replacement from 5 red and 5 blue counters, show that P(second is red) = 1/2.
5/10 × 4/9 + 5/10 × 5/9 = 20/90 + 25/90 = 45/90 = 1/2. - Working backwardsA bag has 5 counters, some red and the rest blue. P(both red) without replacement is 3/10. How many are red?
r(r − 1)/20 = 3/10 gives r(r − 1) = 6, so r = 3.
Practicalities
Works on a phone or a shared screen. Nothing a student types is saved or sent anywhere.
Both figures work with the keyboard (tab to a slider or button, then the arrow keys or Enter) and by touch. Under each one, “Show me the steps” gives the same states in words.
No device? Use a real bag of counters or paper slips and have pairs run 20 draws each way.