Mathematics • Year 9 • Unit 4 • Lesson 14
Tree Diagrams — Mixed Challenge
Combine ideas from Lessons 13 and 14 — single probabilities, the complement rule, tree diagrams, and the multiplication rule with and without replacement. Choose the right approach for each problem, fix a classmate's error, and tackle an open-ended challenge.
1. Mixed problems — choose the right tool
Decide which rule or tree to use before you start writing. Show working. 3 marks each
1.1 A bag has 6 red and 4 blue marbles. Two are drawn with replacement. Find P(both blue) and P(at least one red).
1.2 Same 6R / 4B bag, but without replacement. Find P(both blue) and P(one of each colour).
1.3 Two fair coins are tossed. Find P(exactly one head) by drawing the tree and adding the relevant branches.
1.4 A box has 4 red and 4 blue marbles. Two are drawn without replacement. Find P(different colours).
1.5 Independent events A and B have P(A) = 0.3 and P(B) = 0.5. Find P(A and B), P(A or B) and P(neither A nor B).
1.6 A bag has 3 red, 2 blue and 1 green marble. Two are drawn without replacement. Find P(at least one red).
2. Find the mistake
Another student tried to find P(both red) when two marbles are drawn from a bag of 4 red and 6 blue without replacement. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — 4R/6B, two draws without replacement:
Line 1: P(1st red) = 4/10 = 2/5.
Line 2: After taking out one red, 4 red and 6 blue remain, so P(2nd red) = 4/10.
Line 3: Multiplication rule: P(both red) = 2/5 × 4/10 = 8/50.
Line 4: Simplify: P(both red) = 4/25.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer for P(both red).
Stuck? Without replacement: after one red is taken, only 3 red are left and the total drops to 9. The student forgot to subtract from both.3. Open-ended challenge — design a fair game
This question has more than one valid answer. 4 marks
3.1 Design a "two-draw" game from a bag of only red and blue marbles (whole numbers of each, at least one of each colour) that is exactly fair:
- You draw two marbles without replacement.
- You win if you draw two of the same colour (RR or BB).
- Your friend wins if you draw one of each colour (RB or BR).
- For the game to be fair, you both must have P(win) = 1/2.
For your bag:
(i) Write the number of red and blue marbles.
(ii) Compute P(same colour) and P(different colours) and confirm both equal 1/2.
(iii) Explain in one sentence why a bag with very different numbers of red and blue marbles (e.g. 1 red and 9 blue) would not give a fair game.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — 6R/4B with replacement
P(BB) = (4/10)² = 16/100 = 4/25. P(at least one red) = 1 − P(BB) = 1 − 4/25 = 21/25.
1.2 — 6R/4B without replacement
P(BB) = 4/10 × 3/9 = 12/90 = 2/15.
P(one of each) = P(RB) + P(BR) = 6/10 × 4/9 + 4/10 × 6/9 = 24/90 + 24/90 = 48/90 = 8/15.
1.3 — Two coins, exactly one head
Branches: HH (1/4), HT (1/4), TH (1/4), TT (1/4). Exactly one H = HT or TH. P = 1/4 + 1/4 = 1/2.
1.4 — 4R/4B without replacement, different colours
P(RB) = 4/8 × 4/7 = 16/56 = 2/7. P(BR) = 4/8 × 4/7 = 2/7. P(different) = 2/7 + 2/7 = 4/7.
1.5 — Independent A and B
P(A and B) = 0.3 × 0.5 = 0.15.
P(A or B) = P(A) + P(B) − P(A and B) = 0.3 + 0.5 − 0.15 = 0.65.
P(neither) = 1 − P(A or B) = 1 − 0.65 = 0.35.
1.6 — 3R/2B/1G, two draws without replacement, at least one red
Use the complement. P(no red) = P(both from the 3 non-red) = 3/6 × 2/5 = 6/30 = 1/5. P(at least one red) = 1 − 1/5 = 4/5.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) Without replacement, after a red is taken out there are only 3 red left (not 4) and the total is now 9 (not 10). The student forgot to update both numerator and denominator.
(c) Corrected working: P(1st red) = 4/10 = 2/5. P(2nd red | 1st red) = 3/9 = 1/3. P(both red) = 2/5 × 1/3 = 2/15.
3 — Open-ended challenge (sample solution)
Try a bag with 1 red and 3 blue (total 4). P(same colour) without replacement = P(RR) + P(BB) = 1/4 × 0/3 + 3/4 × 2/3 = 0 + 6/12 = 1/2. P(different) = 1/2 ✓. This bag works.
Another working bag: 3 red and 1 blue (by symmetry — same calculation).
One more: 6 red and 3 blue (total 9). P(same) = 6/9 × 5/8 + 3/9 × 2/8 = 30/72 + 6/72 = 36/72 = 1/2 ✓.
(iii) If the bag has 1 red and 9 blue, almost every two-draw combination will be a pair of blues, so P(same colour) is far above 1/2 and the game is heavily biased toward you — not fair.
Marking: 2 marks for a valid bag (and totals stated); 1 mark for clear working showing both probabilities equal 1/2; 1 mark for a coherent explanation of why an unbalanced bag fails. Accept any bag with R red and B blue such that R(R−1) + B(B−1) = RB + BR = R × B (i.e., it works out to fair).