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09-quiz.html
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/* ===== Shared helpers for KP18: DAG drawing, parent sets, CPT parameter counting ===== */
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/* 09 — Three-question self-check */
const Q=[
{q:'六個二元特徵在完全不作獨立性假設時，完整聯合機率分佈需要多少個獨立機率值？',
o:['A) 64 個','B) 63 個','C) 14 個','D) 6 個'],a:1,
ok:'答對了！2⁶ = 64 種狀態的機率和為 1，所以只有 63 個自由參數。',
no:'注意：64 種狀態的機率總和必須是 1，因此最後一個可以由其餘算出，獨立參數是 2⁶ − 1 = 63。'},
{q:'講義標示貝氏網路為 13 個參數，而依指定父集合逐節點相加的二元 CPT 槽位是 14，下列何者為正確理解？',
o:['A) 來源教材完全算錯了，不能參考','B) 兩者只是計數口徑的微小差異，重點是兩者都約 14，相較 63 達成數量級的大幅縮減','C) 13 代表包含了所有的補數機率，14 則沒有','D) 14 是包含了測試集樣本後的修正數量'],a:1,
ok:'正確！第 5 頁逐節點相加得 1+1+2+4+2+4 = 14；無論 13 或 14，相較 63 都省了約八成。',
no:'關鍵不在 13 或 14：兩者只是計數口徑差一點，重點是它們與 63 相比的數量級差距。'},
{q:'給定條件獨立式 p(x₃ | x₂, x₁) = p(x₃ | x₂)，若尚未觀察中介變數 x₂，則 x₁ 與 x₃ 的關係為何？',
o:['A) 依然嚴格獨立，彼此機率毫無關聯','B) 因資訊不足，數學上完全無法計算','C) 透過中介 x₂ 間接相依，數值上 P(x₃=1 | x₁=1) = 0.74 ≠ P(x₃=1 | x₁=0) = 0.26','D) 兩者必然形成完全互斥關係'],a:2,
ok:'太棒了！第 6 頁可以拖曳滑桿驗證：未觀察 x₂ 時差距 0.48，已知 x₂ 後差距為 0。',
no:'條件獨立不是「無關」：沒觀察 x₂ 時，x₁ 仍透過 x₂ 影響 x₃（0.74 對 0.26）；只有知道 x₂ 之後，x₁ 才不再提供新情報。'}];
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