Stand Up & Math
Stand Up & Math
The Flagged Essay
AHL 4.13 · Bayes' Theorem · HL Only · Groups of 3–4 · ~50–55 min
1
Phase 1
Reading the Evidence
Priors · Likelihoods · Conditional probability from a table
Scenario A school's integrity committee has reviewed 1 000 essays using an AI-detection tool. The table below summarises the results from last semester.
AI-assistedHuman-writtenTotal
Flagged by detector15382235
Not flagged27738765
Total1808201000
Task 1a — Prior probabilities
Find P(AI-assisted) and P(Human-written) from the table.
These are the prior probabilities — our belief before seeing the detector result.
Task 1b — Likelihoods
Find all four conditional probabilities: P(Flagged | AI), P(Flagged | Human), P(Not flagged | AI), P(Not flagged | Human).
These are likelihoods — how probable each detector outcome is given the essay's true origin.
Task 1c — First instinct
The detector flags an essay. Estimate P(AI | Flagged) as a group and write it on the board — you'll revisit after Phase 2.
2
Phase 2
The Bayes Flip
Draw a tree from scratch · Read the posterior · Meet the formula
Known from Phase 1 P(AI) = 0.18 · P(Human) = 0.82 · P(Flagged | AI) = 0.85 · P(Flagged | Human) = 0.10 · P(Not flagged | AI) = 0.15 · P(Not flagged | Human) = 0.90
Task 2a — Draw the tree on your board
Using the Phase 1 values, construct a fully labelled probability tree from scratch. First branches: origin of essay. Second branches: detector result. Calculate all four end-node products.
Task 2b — Read the posterior from your tree
An essay is flagged. Find P(AI | Flagged) from your tree: AI flagged path ÷ sum of all flagged paths. Compare to Task 1c — surprised?
Reference tree below — only look after you've tried it on the board.
Phase 2 reference tree (check after attempting)
0.18 0.82 AI Human 0.85 0.15 0.10 0.90 AI ∩ Flagged 0.85 × 0.18 = 0.153 ✦ AI ∩ Not Flagged 0.15 × 0.18 = 0.027 Human ∩ Flagged 0.10 × 0.82 = 0.082 ✦ Human ∩ Not Flagged 0.90 × 0.82 = 0.738
✦ Flagged paths: 0.153 + 0.082 = 0.235  ·  Posterior = 0.153 ÷ 0.235 ≈ 0.651
Task 2c — Verify from the table
Of all flagged essays in the Phase 1 table, what fraction were AI-assisted? Does this match your tree?
Task 2d — Someone already wrote this down
This method is called Bayes' theorem. Label every part of the formula using your tree — which part is which path?
\( P(H_1 \mid F) = \dfrac{P(F \mid H_1)\cdot P(H_1)}{P(F \mid H_1)\cdot P(H_1) + P(F \mid H_2)\cdot P(H_2)} \)
3
Phase 3
Sequential Update
Posterior becomes prior · Draw a new tree · Vocabulary diversity signal
New evidence The essay also scores low on vocabulary diversity. P(Low vocab | AI) = 0.76 · P(Low vocab | Human) = 0.24. Your Phase 2 posterior P(AI | Flagged) ≈ 0.6511 is now the new prior. Erase the old root — replace it.
Task 3a — Draw a new tree on your board
Same two-branch structure as Phase 2, but new root probabilities (0.6511 / 0.3489) and vocabulary diversity as the second-level branches. Calculate all four end-node products.
Task 3b — Read the updated posterior
Find P(AI | Flagged ∩ Low vocab): AI low-vocab path ÷ sum of all low-vocab paths. Confirm with the Bayes formula from Task 2d.
Reference tree below — only look after you've tried it on the board.
Phase 3 reference tree — updated prior (check after attempting)
Given: Flagged 0.6511 0.3489 AI Human 0.76 0.24 0.24 0.76 AI ∩ Low vocab 0.76 × 0.6511 = 0.4948 ✦ AI ∩ Not low vocab 0.24 × 0.6511 = 0.1563 Human ∩ Low vocab 0.24 × 0.3489 = 0.0837 ✦ Human ∩ Not low vocab 0.76 × 0.3489 = 0.2652
✦ Low vocab paths: 0.4948 + 0.0837 = 0.5786  ·  Posterior = 0.4948 ÷ 0.5786 ≈ 0.855
Task 3c — Interpret
Compare Phase 3 posterior (≈ 85.5%) to Phase 2 (≈ 65.1%). How much did the vocabulary signal move the needle? What does this mean for the committee?
4
Phase 4
The Threshold
Two chained trees · Critical prior · GDC required
Committee rule The committee acts only if P(AI | Flagged ∩ Low vocab) > 0.90. With p = 0.18 the posterior was ≈ 85.5% — just below. Find the minimum p (to 2 d.p.) that pushes the posterior above 90%.
Task 4a — Two-tree chain structure
Draw two chained trees on your board with p as the unknown root. Tree 1 (detector signal) gives intermediate posterior r(p). Tree 2 (vocab signal) uses r(p) as its root. Write the general expressions before substituting numbers.
\( r(p) = \dfrac{0.85p}{0.85p + 0.10(1-p)} \quad \rightarrow \quad \text{use } r(p) \text{ as root of Tree 2} \)
Chain structure — substitute p = 0.25 then p = 0.26 to bracket the answer
TREE 1 — Detector p p 1−p AI H 0.85 0.15 0.10 0.90 0.85p ✦ 0.15p 0.10(1−p) ✦ 0.90(1−p) r(p) = 0.85p/denom → TREE 2 — Vocab r(p) r(p) 1−r AI H 0.76 0.24 0.24 0.76 0.76·r(p) ✦ → numerator 0.24·r(p) 0.24·(1−r) ✦ → denom 0.76·(1−r(p))
p = 0.25 → r ≈ 0.7391 → final posterior ≈ 89.97% ✗  ·  p = 0.26 → r ≈ 0.7492 → final posterior ≈ 90.44% ✓  ·  Answer: p = 0.26
Task 4b — Test p = 0.25 and p = 0.26
Substitute each value into your two-tree chain and run the full calculation. Find the minimum p (to 2 d.p.) such that the final posterior exceeds 90%.
Task 4c — Reflect
Currently p = 0.18. How much must it increase before the committee can act? What does this tell you about the role of the prior in a Bayesian system?