123i - Tree Diagrams for Combined Events

Lesson 123 — Tree Diagrams for Combined Events: Guided Practice

Strand: Probability | Descriptor: AC9M8P02 | Duration: 45 minutes

Learning Intentions

• To construct tree diagrams to show all outcomes of two successive events.
• To calculate probabilities from tree diagrams for events with and without replacement.

Success Criteria

I can:

1. Construct a tree diagram for two successive events, labelling every branch with its probability.
2. Multiply along branches to find the probability of a combined outcome.
3. Add relevant branch probabilities to find the probability of an “or” / “exactly one” outcome.
4. Explain how and why probabilities change on later branches when selecting without replacement.

Warmup

(5 minutes — mini whiteboards, rapid-fire)

1. A bag has red and blue counters ( total). A counter is drawn and put back. What is on the second draw?
2. Now suppose the counter is not put back, and the first counter drawn was red. How many counters remain, and how many are red?
3. Evaluate .
4. True or false: “Removing a counter without replacing it can change the probabilities for the next draw.”

Answers: 1. (unchanged); 2. remain, are red; 3. ; 4. True — this is the central idea of today’s lesson.

Activities

Activity 1 — Explicit Instruction: Tree Diagrams with Replacement (10 min)

I do / We do / You do.

I do: A bag has red and blue counters. A counter is drawn, replaced, then a second is drawn. Build the tree:

• Stage 1: ,
• Stage 2 (unchanged, since the counter was replaced): , from either branch

Multiply along each path:

Check: . ✓ “Every set of branches leaving a point must sum to — and so must every complete path through the tree.”

We do: A spinner has and , spun twice (a spinner has no memory, so this is automatically “with replacement” reasoning). Build the tree and find .

You do: A weighted coin has . It is tossed twice. Build the tree and find (a) (b) .

(Answers: (a) ; (b) .)

Activity 2 — Explicit Instruction: Tree Diagrams without Replacement (10 min)

I do: The same bag ( red, blue, total) — but now the first counter is not replaced before the second draw. Stage 1 is unchanged, but stage 2 now depends on stage 1:

• If stage 1 was red: remain ( red, blue) ,
• If stage 1 was blue: remain ( red, blue) ,

Check the sum: . ✓

We do: Using the same without-replacement tree, together find .

You do: A box has chocolates: dark, milk. Two are eaten one after another, without replacement. Build the tree, then find (a) (b) .

(Answers: (a) ; (b) .)

Activity 3 — Inquiry: Does Replacement Matter? (16 min)

Pairs, then whole-class share.

A jar has green and yellow jelly beans ( total). Two beans are drawn one after another.

(a) If each bean is replaced before the next draw, find . (b) If beans are not replaced, find . (c) Which situation gives the higher probability of both green? Explain, using the idea of a changing denominator. (d) Harder: without replacement, find , and compare it with the with-replacement version from Activity 1’s style of calculation. Which is larger? Is this what you expected?

Socratic scaffolding for part (d):

Prompt	Purpose
Understand the problem	You need under both replacement rules, then to compare them.
Devise a plan	Use the complement: find first, for each rule.
Carry out — with replacement	, so .
Carry out — without replacement	, so .
Compare	Without replacement gives the higher probability ().
Looking back — why?	Removing a bean (of either colour) without replacing it changes the pool for draw two. If the first bean was yellow, green becomes relatively more common among the remaining ( vs the original ), which boosts the chance of getting green on draw two exactly when draw one failed to. This effect outweighs the (expected) reduction in from part (b).

Answers: (a) ; (b) ; (c) Without replacement is lower for “both green”, because after removing one green bean, only of the remaining are green (), a smaller proportion than the original ; (d) as scaffolded — without replacement gives the higher .

Checks for Understanding

(6 minutes — exit ticket, collected)

A bag has red and blue counters ( total).

1. Two draws with replacement. Find .
2. Two draws without replacement. Find .
3. Explain why your two answers differ.
4. A tree diagram shows on the first branch. Following , on the second branch. Find .

Answers: 1. ; 2. ; 3. Without replacement, removing a red counter after a red first draw reduces the proportion of red counters left (, compared with the unchanged used with replacement); 4. .

Common Misconceptions

Misconception	How to pre-empt it
Using the original probabilities on every branch, even without replacement.	Require students to re-count what remains before labelling any “second stage” branch.
Adding probabilities along a single path instead of multiplying.	Anchor the rule “along a path, multiply; between separate paths, add” with a worked contrast every time.
Forgetting to add multiple relevant paths for an “or” / “exactly one” outcome.	Model circling every path that satisfies the event before adding, as in “exactly one red”.
Believing branch probabilities from any single point should sum to something other than .	Check every set of sibling branches sums to before proceeding, as demonstrated in Activity 1.
Assuming without replacement always makes “success” less likely.	Activity 3(d) is a deliberate counterexample — “at least one green” is more likely without replacement.

Enrichment — Competition-Style Problems

E1 (AMC Junior style). A bag has red and blue counters. Two are drawn without replacement. Find .

Answer

E2 (Kangaroo style). A spinner has each spin, independently. It is spun times. Find .

Answer

Three paths give exactly two wins: , , , each with probability .

E3 (Challenge). The jelly bean jar from Activity 3 has green and yellow beans. Three beans are drawn without replacement. Find .

Answer

Homework

1. A bag has red and blue counters ( total). Two draws with replacement. Find .
2. Same bag, two draws without replacement. Find .
3. A tree diagram has on the first branch; ; . Find and .
4. Explain, using the idea of a changing denominator, why without-replacement probabilities differ from with-replacement probabilities.
5. Reasoning. A friend says: “Since we removed one counter, the total probability across all branches at the second stage no longer equals .” Explain why this is incorrect.
6. Challenge. A box has chocolates: dark, milk. Three are eaten one after another, without replacement. Find .

Answers: 1. . 2. . 3. ; . 4. Removing an item changes both the total remaining and the count of each colour remaining, so the fraction representing each later probability changes even though the rule “probability = favourable ÷ total” stays the same. 5. At each individual branching point, the probabilities of the branches leaving that point still sum to — the second-stage branches are just different numbers because they describe a smaller, updated pool; there is no missing probability anywhere in the tree. 6. Draws and must both be milk, then draw is dark: .