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.
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.
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 | 1st 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.
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 | 1st red) = ______ ÷ ______ .
Step 5 — Multiply along branch:
P(RR without replacement) = ______ × ______ = ______ (simplified).
Step 6 — Which is bigger and why?
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
How did this worksheet feel?
What I'll revisit before next class:
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 | 1st 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.