Mathematics • Year 8 • Unit 4 • Lesson 18

Tree Diagrams

Build fluency with drawing tree diagrams, multiplying along branches, adding path probabilities, and adjusting for with/without replacement.

Build · I Do / We Do / You Do

1. I do — fully worked example

Watch how we build a tree diagram, multiply branch probabilities, and add paths.

Problem. A bag has 3 Red and 2 Blue marbles. Two marbles are drawn with replacement. Find P(both Red) and P(one of each colour).

Step 1 — Find branch probabilities (same for every draw with replacement).

P(R) = 3/5 P(B) = 2/5

Reason: with replacement, the bag is restored to 5 marbles before the second draw.

Step 2 — Draw the tree (Stage 1: R or B; Stage 2: R or B from each).

Four paths: RR, RB, BR, BB

Step 3 — Multiply along each path.

P(RR) = 3/5 × 3/5 = 9/25
P(RB) = 3/5 × 2/5 = 6/25
P(BR) = 2/5 × 3/5 = 6/25
P(BB) = 2/5 × 2/5 = 4/25

Reason: along a single path the events are sequential — we multiply because both must occur.

Step 4 — Check all endpoints sum to 1.

9/25 + 6/25 + 6/25 + 4/25 = 25/25 = 1 ✓

Step 5 — Add paths for combined events.

P(both R) = P(RR) = 9/25
P(one of each) = P(RB) + P(BR) = 6/25 + 6/25 = 12/25

Answer: P(both Red) = 9/25 and P(one of each colour) = 12/25.

Stuck? Revisit lesson § "Tree Diagram Structure" — multiply ALONG a path, add ACROSS paths.

2. We do — fill in the missing steps

Same shape as Section 1, but with the working faded. Fill in each blank. 4 marks

Problem. A bag has 4 Red and 1 Blue marble. Two marbles are drawn with replacement. Find P(both Red), P(both Blue), and P(one of each).

Step 1 — Branch probabilities (same at every draw):

P(R) = ______ / ______ P(B) = ______ / ______

Step 2 — List the four paths:

______, ______, ______, ______

Step 3 — Multiply along paths:

P(RR) = 4/5 × 4/5 = ______ / 25
P(RB) = ______ × ______ = ______ / 25
P(BR) = ______ × ______ = ______ / 25
P(BB) = 1/5 × 1/5 = ______ / 25

Step 4 — Check sum = 1:

______ + ______ + ______ + ______ = ______ / 25 = ______ ✓

Step 5 — Combined events:

P(both R) = ______ P(both B) = ______ P(one of each) = ______ + ______ = ______

Stuck? With replacement, P(R) = 4/5 and P(B) = 1/5 at BOTH stages. Multiply along each path to find each path's probability.

3. You do — independent practice

Show your working under each problem. Foundation tests recall, standard uses the tree method, and extension uses the without-replacement rule.

Foundation — branch values and recall

3.1 Two fair coins are flipped. State the four endpoint probabilities on a tree diagram. 1 mark

3.2 Why do all endpoint probabilities on a complete tree diagram sum to 1? Answer in one sentence. 1 mark

3.3 When do you ADD probabilities and when do you MULTIPLY when using a tree diagram? Use the words "path" and "branches". 1 mark

3.4 A bag has 6 marbles. After drawing one WITHOUT replacement, how many remain? Why does this matter for the second-stage branch probabilities? 1 mark

Standard — apply the tree method

3.5 A coin is flipped three times. Find P(exactly 2 heads) by multiplying along the favourable paths and adding. List the paths used. 2 marks

3.6 A bag has 2 Yellow and 3 Green marbles. Two are drawn with replacement. Find P(both Yellow) and P(at least one Green). 2 marks

Extension — without replacement

3.7 A bag has 3 Yellow and 2 Green marbles. Two are drawn without replacement. Show your tree branch probabilities at Stage 2 (which depend on Stage 1) and find P(both Yellow). 2 marks

3.8 Same bag (3 Y, 2 G, no replacement). Use the complement to find P(at least one Green). Show working. 2 marks

Stuck on 3.8? P(at least one G) = 1 − P(no G) = 1 − P(YY). You found P(YY) in 3.7.

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What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (4 Red, 1 Blue, with replacement)

Step 1: P(R) = 4/5, P(B) = 1/5.
Step 2: Paths = RR, RB, BR, BB.
Step 3: P(RR) = 16/25; P(RB) = 4/5 × 1/5 = 4/25; P(BR) = 1/5 × 4/5 = 4/25; P(BB) = 1/25.
Step 4: 16/25 + 4/25 + 4/25 + 1/25 = 25/25 = 1 ✓.
Step 5: P(both R) = 16/25, P(both B) = 1/25, P(one of each) = 4/25 + 4/25 = 8/25.

3.1 — Two coins, endpoint probabilities

P(HH) = P(HT) = P(TH) = P(TT) = 1/4 each.

3.2 — Why endpoints sum to 1

Because the endpoints represent every possible outcome of the experiment, and the total probability of "something happens" is always 1 (certainty).

3.3 — Add vs multiply

Multiply the branch probabilities ALONG a single path (sequential events both must occur). Add the path probabilities ACROSS different paths (when an outcome can be reached by more than one route).

3.4 — Without replacement

5 marbles remain after drawing one from 6. This matters because the second-stage branch probabilities use 5 as the denominator (not 6), and the numerator depends on which marble was drawn first.

3.5 — Three coins, P(exactly 2 H)

Each path has probability 1/2 × 1/2 × 1/2 = 1/8. Favourable paths: HHT, HTH, THH (3 paths). P(exactly 2 H) = 3 × 1/8 = 3/8.

3.6 — 2 Y, 3 G with replacement

P(Y) = 2/5, P(G) = 3/5 at every draw.
P(both Y) = 2/5 × 2/5 = 4/25.
P(at least one G) = 1 − P(no G) = 1 − P(YY) = 1 − 4/25 = 21/25.

3.7 — 3 Y, 2 G without replacement, P(both Y)

Stage 1: P(Y) = 3/5. Stage 2 if Y first: only 2 Y left out of 4 marbles, so P(Y | Y) = 2/4. Stage 2 if G first: P(Y | G) = 3/4.
P(YY) = 3/5 × 2/4 = 6/20 = 3/10.

3.8 — At least one G (complement)

P(at least one G) = 1 − P(both Y) = 1 − 3/10 = 7/10.