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).

P(both Red) = 3/5 × 3/5 = 9/25; P(one of each) = RB + BR = 6/25 + 6/25 = 12/25.

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.

How did this worksheet feel?

Got it Partly Lost

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.