The Tilted Board — Teacher Answers

1

Explore the Board

4.5 · Experimental probability · Simulation · Widget interaction

Task 1a Observe the level board (p = 0.5, n = 6)

Expected histogram shape: Symmetric, bell-shaped. Peak at bin 3. Roughly reflects the 1:6:15:20:15:6:1 ratio (Pascal row n = 6).

Key observation to draw out: The distribution is symmetric about the middle bin. Every peg decision is a 50/50 coin flip.

▸ Let groups run 50 and then 500 balls. Ask: "does 50 look symmetric? Does 500?" — this motivates why we need many trials to see the theoretical shape emerge.

Task 1b Tilt to p = 0.7 — sketch and compare

Expected change: The histogram skews right — peak shifts toward the higher-numbered bins. The distribution becomes asymmetric.

Peak location (n = 6, p = 0.7): Most balls land around bin 4 or 5. The theoretical mode is at \(k = 4\).

Key question to ask each group: "Where does the peak sit? Can you predict it before dropping?" This is the hook for Phase 4 (mean = np).

▸ If groups say "around 4 or 5," accept both — the distribution is spread. The exact mode for n=6, p=0.7 is k=4 with P(X=4) ≈ 0.3241. The mean is np = 4.2, which falls between bins 4 and 5.

Task 1c Predict the peak — before dropping

Accept any reasonable conjecture at this stage. Students may guess "the tilt percentage × number of rows" — this is intuitive and essentially correct (np). Validate all approaches; don't confirm the formula yet.

▸ Do NOT reveal E(X) = np here. This is the Phase 4 payoff. Acknowledge predictions and say "we'll return to this."

Consolidation prompt after Phase 1: "Two things seem to control the shape of the histogram — what are they?" Target answers: the number of rows (n) and the tilt (p). "What does tilting do to the centre of the pile? What does it do to the spread?" These observations set up Phases 3 and 4.

2

Count the Paths

4.8 · Tree diagrams · Pascal's triangle · Deriving P(X = k)

Task 2a n = 3 tree, p = 0.7 — write every path probability

All 8 paths and their probabilities:

Path	Bin (# rights)	Probability
LLL	0	\(0.3 \times 0.3 \times 0.3 = 0.3^3\)
LLR	1	\(0.3 \times 0.3 \times 0.7 = 0.3^2 \times 0.7\)
LRL	1	\(0.3 \times 0.7 \times 0.3 = 0.3^2 \times 0.7\)
RLL	1	\(0.7 \times 0.3 \times 0.3 = 0.3^2 \times 0.7\)
LRR	2	\(0.3 \times 0.7 \times 0.7 = 0.3 \times 0.7^2\)
RLR	2	\(0.7 \times 0.3 \times 0.7 = 0.3 \times 0.7^2\)
RRL	2	\(0.7 \times 0.7 \times 0.3 = 0.3 \times 0.7^2\)
RRR	3	\(0.7 \times 0.7 \times 0.7 = 0.7^3\)

▸ Key moment: because p ≠ 0.5, students can clearly see that LLR and RLL have the same probability (the factors just reorder), but LLR ≠ LLR-with-different-p. The exponent structure — k factors of 0.7, (n−k) factors of 0.3 — is unmistakeable here in a way that p=0.5 hides.

Task 2b Group by bin — exponent form and path counts

Grouped by bin:

Bin k	Paths	Count	Prob of one path
0	LLL	1	\(p^0(1-p)^3 = 0.3^3 = 0.027\)
1	LLR, LRL, RLL	3	\(p^1(1-p)^2 = 0.7 \times 0.09 = 0.063\)
2	LRR, RLR, RRL	3	\(p^2(1-p)^1 = 0.49 \times 0.3 = 0.147\)
3	RRR	1	\(p^3(1-p)^0 = 0.7^3 = 0.343\)

One path to bin k has probability: \(p^k(1-p)^{n-k}\)

Path counts: 1 · 3 · 3 · 1 — the 3rd row of Pascal's triangle.

Task 2c Write P(X = k) and name the pieces

\[ P(X = k) = \underbrace{\binom{n}{k}}_{\text{count paths}} \times \underbrace{p^k(1-p)^{n-k}}_{\text{prob one path}} \]

k	Paths	\(p^k(1-p)^{3-k}\)	\(P(X=k)\)
0	1	\(0.027\)	0.0270
1	3	\(0.063\)	0.1890
2	3	\(0.147\)	0.4410
3	1	\(0.343\)	0.3430

Sum = 1.0000 ✓

Name: The path count \(\binom{n}{k}\) is the binomial coefficient — the number of ways to choose k rights from n decisions.

▸ This is the natural landing point for naming C(n,k). Students have derived the need for it from the tree — they have already computed it by hand. The formula gives a name to something they already found.

Task 2d Extend to n = 4 — Pascal's triangle and the nCr connection

Path counts for n = 1 through 5:

n	Path counts per bin
1	1	1
2	1	2	1
3	1	3	3	1
4	1	4	6	4	1
5	1	5	10	10	5	1

The pattern: each entry is the sum of the two entries above it — this is Pascal's triangle. Students can predict the n = 5 row without drawing 32 paths.

The closed-form expression: the number of paths to bin \(k\) in an \(n\)-row board is

\[ \binom{n}{k} = \frac{n!}{k!\,(n-k)!} \]

read as "\(n\) choose \(k\)" — the number of ways to select \(k\) rights from \(n\) decisions, without caring about order. This is exactly what students computed by hand in Tasks 2a–2c.

▸ Target question to ask groups: "You got 6 paths to bin 2 when n = 4. Can you get to 6 using a formula with 4 and 2?" Push them toward \(\frac{4!}{2!\,2!}\) if needed. The goal is that \(\binom{n}{k}\) feels like a shortcut they needed, not a formula dropped from above. ▸ Strong groups: ask "why does each Pascal row entry equal the sum of the two above it?" — this connects to the tree structure: every path to bin k in an (n+1)-row board came from bin k−1 or bin k in the previous row.

Why p = 0.7 first? At p = 0.5, every path has the same probability (0.5ⁿ) regardless of how many R's and L's it contains — which obscures the exponent structure entirely. Starting with p = 0.7 makes the contrast between p and (1−p) visible in every product, so students discover pᵏ(1−p)ⁿ⁻ᵏ from the paths themselves, before the formula is named.

3

The Formula and Its Shape

4.8 · Binomial formula · Conditions · Notation X ~ B(n, p)

Task 3a Full probability table — n = 4, p = 0.7

k	\(\binom{4}{k}\)	\(p^k\)	\((1-p)^{4-k}\)	\(P(X=k)\)
0	1	\(0.7^0 = 1\)	\(0.3^4 = 0.0081\)	0.0081
1	4	\(0.7^1 = 0.7\)	\(0.3^3 = 0.027\)	0.0756
2	6	\(0.7^2 = 0.49\)	\(0.3^2 = 0.09\)	0.2646
3	4	\(0.7^3 = 0.343\)	\(0.3^1 = 0.3\)	0.4116
4	1	\(0.7^4 = 0.2401\)	\(0.3^0 = 1\)	0.2401

Sum = 1.0000 ✓ Peak at k = 3.

Task 3b Verify against the simulator

Students set n = 4, p = 0.7, drop 500 balls. The observed histogram should approximately match the probabilities above — bin 3 tallest, then bin 4, then bin 2. Slight variation is expected due to randomness.

▸ A useful question: "Does your histogram look like your table? How close is 'close enough'?" This is an informal introduction to the relationship between theoretical and experimental probability.

Task 3c Conditions for the binomial model

Students should identify all four conditions from the Plinko context:

1. Fixed number of trials n — the ball passes through exactly n rows of pegs.

2. Each trial has exactly two outcomes — right (success) or left (failure).

3. Fixed probability p of success — the tilt is constant; every peg is the same.

4. Trials are independent — how the ball bounced at peg 2 has no effect on peg 3.

Notation: X ~ B(n, p)

For our verified example: \(X \sim B(4, 0.7)\)

Common misconception: Students often think "independent" means "equally likely." Clarify: a biased coin (p = 0.7) still gives independent flips — each flip is unaffected by the previous one. Independence is about influence, not balance.

4

Mean and Variance

4.8 HL · E(X) = np · Var(X) = np(1−p) · Effect of parameters

Task 4a Observe E(X) from the simulator — three cases

Students record observed means for n = 10 with varying p. Theoretical values:

n	p	E(X) = np	Var(X) = np(1−p)
10	0.3	3.00	2.10
10	0.5	5.00	2.50
10	0.7	7.00	2.10

Pattern: E(X) = np. The mean landing bin is simply "n × tilt probability."

Intuitive justification: At each of the n pegs, the expected rightward movement is p. After n pegs, expected position = np.

Task 4b Variance investigation — write the formula

For fixed n = 10, varying p:

p	0.1	0.3	0.5	0.7	0.9
Var(X) = np(1−p)	0.90	2.10	2.50	2.10	0.90

Variance is maximised at p = 0.5. At p = 0.5 each peg decision is maximally uncertain — the ball is equally likely to go either way, so where it ends up is hardest to predict. As p → 0 or p → 1, outcomes become nearly certain and the spread collapses.

Symmetry: Var(n, p) = Var(n, 1−p) — the p = 0.3 and p = 0.7 histograms are mirror images with the same spread.

Formulas students should write at the end of this task:

\[ \text{E}(X) = np \qquad \text{Var}(X) = np(1-p) \]

▸ Make sure every group has both formulas written clearly on their board before moving on. Ask: "What is the standard deviation?" — SD = \(\sqrt{np(1-p)}\). This will appear in exam questions.

Task 4c Dr. Nemo's clownfish — first application + GDC

Distribution: \(X \sim B(25,\; 0.15)\)

Conditions met: fixed n = 25 trials, each clownfish independently has p = 0.15 of the unusual fin pattern, binary outcome, "all from different families" signals independence.

(i) \(P(X = 3)\)

\[ P(X=3) = \binom{25}{3}(0.15)^3(0.85)^{22} = 2300 \times 0.003375 \times 0.02798\ldots \]

\(0.85^{22} = 0.02798\ldots\) → \(P(X=3) = 2300 \times 0.003375 \times 0.02798 = \) 0.2174 (4 s.f.)

(ii) \(P(X \leq 4)\)

\[ P(X \leq 4) = P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4) \]

\(P(X=0) = 0.85^{25} = 0.01720\ldots\)

\(P(X=1) = 25 \times 0.15 \times 0.85^{24} = 0.07587\ldots\)

\(P(X=2) = \binom{25}{2}(0.15)^2(0.85)^{23} = 0.16067\ldots\)

\(P(X=3) = 0.21738\ldots\)

\(P(X=4) = \binom{25}{4}(0.15)^4(0.85)^{21} = 0.21099\ldots\)

Sum = 0.6821 (4 s.f.)

(iii) \(P(X \geq 2)\)

\[ P(X \geq 2) = 1 - P(X=0) - P(X=1) = 1 - 0.01720 - 0.07587 = \mathbf{0.9069} \text{ (4 s.f.)} \]

(iv) E(X) and Var(X)

\[ \text{E}(X) = np = 25 \times 0.15 = \mathbf{3.75} \]

\[ \text{Var}(X) = np(1-p) = 25 \times 0.15 \times 0.85 = \mathbf{3.1875} \]

GDC keystrokes — Casio ClassPad / fx-CG

P(X = 3): Menu → Statistics → Dist → Binomial PD → x=3, N=25, p=0.15 → 0.2174
P(X ≤ 4): Menu → Statistics → Dist → Binomial CD → x=4, N=25, p=0.15 → 0.6821
P(X ≥ 2): 1 − Binomial CD → x=1, N=25, p=0.15 → 1 − 0.09307 = 0.9069

GDC keystrokes — TI-Nspire CX

P(X = 3): Menu → 5 → 5 → D: Binomial Pdf → n=25, p=0.15, k=3 → 0.2174
P(X ≤ 4): Menu → 5 → 5 → E: Binomial Cdf → n=25, p=0.15, 0, 4 → 0.6821
P(X ≥ 2): 1 − Binomial Cdf → n=25, p=0.15, 0, 1 → 1 − 0.09307 = 0.9069

▸ The point of doing (i) by hand first is to confirm the GDC agrees. Note for students: "all from different families" is the independence signal — worth pointing out explicitly. Emphasise that in an exam, binomial probabilities must be found using technology (as stated in the guide for SL 4.8).

Consolidation from the walls: Choose 2–3 boards that show different approaches to the E(X) pattern. Highlight: (1) a group that noticed np from the table, (2) a group that verified it numerically, (3) if any group attempted the algebraic argument. Close by writing \(X \sim B(n, p)\), \(\text{E}(X) = np\), \(\text{Var}(X) = np(1-p)\) and ask: "What would the mean and variance be if n = 20, p = 0.4?" — a quick formative check.