Mathematics • Year 9 • Unit 4 • Lesson 14

Two-Stage Events and Tree Diagrams

Build fluency with the multiplication rule, tree diagrams, and the difference between "with replacement" (independent) and "without replacement" (dependent), from a worked example through guided practice to eight independent problems.

Build · I Do / We Do / You Do

1. I do, fully worked example

Read every step. The reason on the right tells you why, not just what.

Problem. A bag has 3 red and 2 blue marbles. Two marbles are drawn one after the other. Find P(both red) (a) with replacement and (b) without replacement.

3/5 2/5 R B R B R B 1st draw → 2nd draw
Multiply along the branches; the second-draw fractions change if you don't replace.

Step 1, Identify the two stages.

Stage 1: first draw. Stage 2: second draw.

Step 2, With replacement: probabilities don't change.

P(1st red) = 3/5. After replacement the bag is unchanged, so P(2nd red) = 3/5.

Reason: "with replacement" makes the two events independent.

Step 3, Multiplication rule along the branches.

P(R then R) = 3/5 × 3/5 = 9/25.

Reason: AND → multiply along a tree branch.

Step 4, Without replacement: the bag changes.

P(1st red) = 3/5. After keeping the red out, only 2 red and 2 blue are left → P(2nd red, given 1st was red) = 2/4.

Step 5, Multiply along the branch.

P(R then R) = 3/5 × 2/4 = 6/20 = 3/10.

Reason: without replacement = dependent events; denominator drops by 1.

Step 6, Tree sketch (with replacement).

            3/5  R --- 3/5 R   (RR  9/25)
           /           \--- 2/5 B   (RB  6/25)
   start
           \
            2/5  B --- 3/5 R   (BR  6/25)
                       \--- 2/5 B   (BB  4/25)

Answer: (a) 9/25 with replacement. (b) 3/10 without replacement.

Stuck? Revisit lesson § "With Replacement" vs "Without Replacement" for the same kind of example.

2. We do, fill in the missing steps

Same structure as Section 1, but with the working faded. Fill in each blank line. 5 marks

Problem. A bag has 5 red and 5 blue marbles. Two are drawn (a) with and (b) without replacement. Find P(both red) for each.

Step 1, Total marbles = ______ .

Step 2, With replacement:

P(1st red) = ______ ÷ ______ . After replacement, P(2nd red) = ______ ÷ ______ .

Step 3, Multiply along branch:

P(RR with replacement) = ______ × ______ = ______ (simplified).

Step 4, Without replacement:

P(1st red) = ______ . After removing one red, P(2nd red, given 1st was red) = ______ ÷ ______ .

Step 5, Multiply along branch:

P(RR without replacement) = ______ × ______ = ______ (simplified).

Step 6, Which is bigger and why?

Stuck? Revisit lesson § "Misconceptions", without replacement makes "both red" slightly less likely because the bag has one fewer red after the first draw.

3. You do, independent practice

Show working under each problem. Foundation = one branch; Standard = combine outcomes; Extension = with vs without replacement.

Foundation, single branch

3.1 Two fair coins are tossed. Find P(HH). 1 mark

3.2 A fair die is rolled and a fair coin tossed. Find P(rolling a 6 AND tossing a head). 1 mark

3.3 Bag has 2 red and 3 blue marbles. With replacement, find P(red then red). 1 mark

3.4 Same bag as 3.3 but without replacement, find P(red then red). 1 mark

Standard, combine outcomes

3.5 Two fair coins are tossed. Find P(exactly one head) by listing all four sample-space outcomes. 2 marks

3.6 A bag has 3 green and 2 yellow marbles. Two are drawn with replacement. Find P(both green) and P(one of each colour). 3 marks

Extension, with vs without replacement

3.7 Same 3-green / 2-yellow bag as 3.6 but without replacement. Find P(both green) and P(one of each colour). 3 marks

3.8 Compare your answers to 3.6 and 3.7. State which probability for "both green" is larger and explain in one sentence why. 2 marks

Stuck on 3.7? After removing one green, the bag has 2 green and 2 yellow. "One of each" means GY OR YG, find both and add.

How did this worksheet feel?

What I'll revisit before next class:

Answers, Do not peek before attempting

Section 2, We do (5R / 5B bag)

Total = 10.
With replacement: P(1st R) = 5/10. P(2nd R) = 5/10. P(RR) = 5/10 × 5/10 = 25/100 = 1/4.
Without replacement: P(1st R) = 5/10. P(2nd R, given 1st was R) = 4/9. P(RR) = 5/10 × 4/9 = 20/90 = 2/9.
With replacement is bigger (1/4 = 0.25 vs 2/9 ≈ 0.222) because without replacement leaves one fewer red marble in the bag, lowering the second-draw probability.

3.1, P(HH)

Independent coins. P(HH) = 1/2 × 1/2 = 1/4.

3.2, P(6 AND H)

Independent. P(6 and H) = 1/6 × 1/2 = 1/12.

3.3, P(red then red) with replacement, 2R/3B bag

P(RR) = 2/5 × 2/5 = 4/25.

3.4, P(red then red) without replacement, 2R/3B bag

P(RR) = 2/5 × 1/4 = 2/20 = 1/10.

3.5, P(exactly one head), two coins

Sample space = {HH, HT, TH, TT}. Exactly one head = {HT, TH}. P = 2/4 = 1/2.

3.6-3G/2Y with replacement

P(GG) = 3/5 × 3/5 = 9/25.
P(one of each) = P(GY) + P(YG) = 3/5 × 2/5 + 2/5 × 3/5 = 6/25 + 6/25 = 12/25.

3.7-3G/2Y without replacement

P(GG) = 3/5 × 2/4 = 6/20 = 3/10.
P(one of each) = P(GY) + P(YG) = 3/5 × 2/4 + 2/5 × 3/4 = 6/20 + 6/20 = 12/20 = 3/5.

3.8, Compare

P(both green) with replacement = 9/25 = 0.36; without replacement = 3/10 = 0.30. With replacement is larger after one green is taken out without replacement, only 2 of the remaining 4 marbles are green, so the second-draw probability drops from 3/5 to 2/4.

HSCScience • Mathematics • Year 9 • Unit 4 • Lesson 14 • Worksheet 1 (Build)