Worked examples — Bayes' theorem — derivation and applications

2.7.9 · D3 · Maths › Statistics & Probability — Intermediate › Bayes' theorem — derivation and applications

Yeh page Bayes' theorem ki practice floor hai. Parent note ne formula build kiya; yahan hum use har tarah ke input ke against drill karte hain jo ek problem de sakti hai — normal cases, extreme priors, extreme tests, ties, degenerate zeros, aur exam traps. Agar tum poora matrix survive kar sako, toh koi bhi Bayes question tumhe surprise nahi kar sakta.

Sab kuch ek hi formula pe tika hai. Hum ise poore time view mein rakhenge:

Kuch bhi compute karne se pehle, aao ek baar phir in pieces ko seedhe saadon mein samjhein, kyunki har worked example inhe naam se refer karta hai.

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The scenario matrix

Har Bayes problem is table ki ek row hai. Jo examples aate hain wo Cell ke saath label hain jo wo cover karti hai, taaki tum har ek ko tick kar sako.

Cell	Scenario class	Inputs mein kya unusual hai	Example
C1	Standard rare-cause, 2 hypotheses	small prior, good test	Example 1
C2	High prior (common cause)	prior ab tiny nahi raha	Example 2
C3	Multi-way partition ($n>2$)	teen ya zyada mutually exclusive causes	Example 3
C4	Degenerate: perfect test ($P(B\mid \bar{A})=0$)	ek likelihood exactly $0$ hai	Example 4
C5	Degenerate: prior $=0$ ya $=1$	belief pehle se certain	Example 5
C6	Useless evidence (independence)	likelihoods equal → $B$ tumhe kuch nahi batata	Example 6
C7	Real-world word problem	tumhe numbers khud extract karne padenge	Example 7
C8	Sequential updating (limiting behaviour)	posterior next prior ban jaata hai; limit dekho	Example 8
C9	Exam twist — complement ya odds form maangta hai	answer plain posterior nahi hai	Example 9

Har cell input ki ek alag shape exercise karti hai: "rarity" ke signs (small vs large prior), degenerate corners (ek probability $0$ ya $1$ ke barabar), ek tie (independence), ek limiting process (kaafi updates), aur do tarike jisse exams question ko disguise karte hain. Mil ke yeh poora space fill karte hain.

Do mental tools jinka hum sahara lete hain:

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Example 1 — C1 · Standard rare-cause case

Step 1 — prior aur likelihoods list karo. $P(A)=0.02,P(\bar{A})=0.98,P(+\mid A)=0.90,P(+\mid \bar{A})=\mathrm{0.10.}$ Yeh step kyun? Hume allergy kitni rare hai (prior) aur test kaisa behave karta hai (likelihood) ko alag karna hai — parent note ki poori warning yahi hai ki yeh alag cheezein hain.

Step 2 — evidence $P(+)$ ko head-count ki tarah build karo. $10,000$ bacche lo (figure dekho):

• Allergic: $0.02\times 10,000=200$. Inme se, $0.90\times 200=180$ positive test karte hain (blue).
• Not allergic: $9,800$. Inme se, $0.10\times 9,800=980$ positive test karte hain (orange false alarms).

$P(+)=\frac{180+980}{10,000}=\frac{1160}{10,000}=\mathrm{0.116.}$ Yeh step kyun? Ek positive do tareekon se aa sakta hai (real ya false alarm); evidence mein dono include hone chahiye, warna posterior normalise nahi hoga — parent note ki teesri galti.

Step 3 — Bayes apply karo (saare flagged logon mein se, real ka fraction kya hai?). $P(A\mid +)=\frac{180}{180+980}=\frac{180}{1160}\approx 0.1552=15.5\%.$ Yeh step kyun? Bayes literally "true positives $÷$ all positives" hai — conditional probability ki definition positive column tak restrict ki gayi.

Verify: dono posteriors ka sum $1$ hona chahiye: $P(\bar{A}\mid +)=980/1160\approx 0.845$, aur $0.1552+0.8448=1$. ✓ Answer $90\%$ se kaafi neeche hai — rarity ne test ko overwhelm kar diya, bilkul waise jaise forecast tha agar tumne "under $20\%$" guess kiya tha.

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Example 2 — C2 · Ek common cause (large prior)

Step 1 — naya prior, same likelihoods. $P(A)=0.40,P(\bar{A})=0.60$. Yeh step kyun? Hum sirf prior ke effect ko isolate kar rahe hain — test untouched hai.

Step 2 — evidence. $P(+)=(0.90)(0.40)+(0.10)(0.60)=0.36+0.06=\mathrm{0.42.}$ Yeh step kyun? Same two-channel sum; sirf group sizes badli hain.

Step 3 — posterior. $P(A\mid +)=\frac{0.36}{0.42}\approx 0.8571=85.7\%.$ Yeh step kyun? Bayes engine, unchanged.

Verify: complement $0.06/0.42\approx 0.1429$; $0.8571+0.1429=1$. ✓ Ek identical test $15.5\%$ trustworthy se $85.7\%$ trustworthy ho gaya, purely isliye kyunki prior badha. Yeh parent note ka "prior aur likelihood dono matter karte hain" wala point hai.

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Example 3 — C3 · Three-way partition

Step 1 — teeno causes ke liye priors aur likelihoods. $P(F)=0.5,P(W)=0.3,P(S)=0.2,$ $P(\text{FREE}\mid F)=0.02,P(\text{FREE}\mid W)=0.05,P(\text{FREE}\mid S)=\mathrm{0.60.}$ Yeh step kyun? $n=3$ mutually exclusive aur exhaustive sources ke saath, partition mein teen cells hain; evidence sum mein har ek appear hona chahiye.

Step 2 — evidence (teeno ke upar sum). $P(\text{FREE})=(0.02)(0.5)+(0.05)(0.3)+(0.60)(0.2)=0.010+0.015+0.120=\mathrm{0.145.}$ Yeh step kyun? Law of Total Probability teen terms ke saath — ek bhi term bhoolna denominator ko corrupt kar deta hai.

Step 3 — spam ke liye posterior. $P(S\mid \text{FREE})=\frac{0.120}{0.145}\approx 0.8276=82.8\%.$ Yeh step kyun? Spam contribution $0.120$, friend/work ke tiny contributions pe dominate karta hai.

Verify: teeno posteriors ka sum $1$ hai: $\frac{0.010+0.015+0.120}{0.145}=\frac{0.145}{0.145}=1$. ✓ Yeh Naive Bayes Classifier ka seed hai.

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Example 4 — C4 · Ek perfect (degenerate) test

Step 1 — inputs, note karo ki ek likelihood exactly $0$ hai. $P(C)=0.05,P(+\mid C)=0.80,P(+\mid \bar{C})=0.$ Yeh step kyun? Likelihood mein zero ek degenerate corner hai — yeh ek pura channel annihilate kar dega.

Step 2 — evidence. $P(+)=(0.80)(0.05)+(0)(0.95)=0.04+0=\mathrm{0.04.}$ Yeh step kyun? False-alarm channel vanish ho jaata hai kyunki uski likelihood $0$ hai — koi bhi non-carrier kabhi flag nahi ho sakta.

Step 3 — posterior. $P(C\mid +)=\frac{0.04}{0.04}=1.$ Yeh step kyun? Jab sirf carriers hi kabhi positive test kar sakte hain, toh ek positive proof hai. Dhyan do $P(C\mid +)=1$ even though $P(+\mid C)=0.80\ne 1$ — dono conditionals genuinely alag hain (prosecutor's-fallacy lesson).

Verify: $P(\bar{C}\mid +)=0/0.04=0$; sum $1$ hota hai. ✓ Degenerate lekin consistent — jab tak $P(+)>0$ hai tab tak hum zero se divide nahi kar rahe.

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Example 5 — C5 · Ek certain prior ($P(A)=0$)

Step 1 — inputs, prior ek corner pe pinned hai. $P(D)=0,P(\bar{D})=1,P(+\mid D)=0.99,P(+\mid \bar{D})=\mathrm{0.10.}$ Yeh step kyun? $P(A)=0$ doosra degenerate corner hai — belief pehle se certain hai. Dono likelihoods problem mein stated hain, toh koi bhi symbol use karne se pehle define kiya gaya hai.

Step 2 — evidence. $P(+)=(0.99)(0)+(0.10)(1)=0+0.10=\mathrm{0.10.}$ Yeh step kyun? Disease channel ek zero prior se multiply hota hai, toh kuch contribute nahi karta.

Step 3 — posterior. $P(D\mid +)=\frac{0}{0.10}=0.$ Yeh step kyun? Bayes aisi hypothesis ko resurrect nahi kar sakta jise tumne probability $0$ assign ki ho. Zero prior ek permanent verdict hai; yeh kabhi bhi kisi cause ko literally impossible declare karne ka danger hai.

Verify: numerator $0$ hai, toh kisi bhi positive test ke liye posterior $0$ hai. ✓ Lesson: prior ko exactly $0$ ya $1$ kabhi set mat karo jab tak tumhara sach mein matlab yeh na ho ki "koi bhi evidence mere mind ko badal nahi sakta."

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Example 6 — C6 · Useless evidence (ek tie → independence)

Step 1 — inputs, equal likelihoods spot karo. $P(F)=0.30,P(B)=0.70,P(H\mid F)=P(H\mid B)=\mathrm{0.50.}$ Yeh step kyun? Equal likelihoods aisi evidence ka signature hai jo kuch bhi discriminate nahi karti.

Step 2 — evidence. $P(H)=(0.50)(0.30)+(0.50)(0.70)=0.15+0.35=\mathrm{0.50.}$

Step 3 — posterior. $P(F\mid H)=\frac{(0.50)(0.30)}{0.50}=\frac{0.15}{0.50}=\mathrm{0.30.}$ Yeh step kyun? Posterior prior ke barabar hai, $0.30$. Jab $P(H\mid F)=P(H)$, events independent hain aur $P(F\mid H)=P(F)$ — Bayes sahi tarike se irrelevant data pe update karne se mana kar deta hai.

Verify: $0.30=P(F)$ exactly. ✓ Yeh parent note ka "special case jahan $P(A\mid B)=P(A)$" hai.

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Example 7 — C7 · Real-world word problem (numbers extract karo)

Step 1 — words ko symbols mein translate karo. Maan lo $B$ = cab Blue hai, $W_B$ = witness "Blue" kehta hai. $P(B)=0.30,P(G)=0.70,P(W_B\mid B)=0.80,P(W_B\mid G)=\mathrm{0.20.}$ Yeh step kyun? Word problem ka sabse mushkil hissa yeh decide karna hai ki kaunsa number prior hai aur kaunsa likelihood. "$80\%$ accurate" ek likelihood hai (true colour diya gaya); "$30\%$ Blue" prior hai.

Step 2 — evidence: witness kitni baar "Blue" kehta hai? $1000$ cabs use karo (figure):

• Blue cabs: $300$; witness $0.80\times 300=240$ ke liye "Blue" kehta hai.
• Green cabs: $700$; witness galti se $0.20\times 700=140$ ke liye "Blue" kehta hai.

$P(W_B)=\frac{240+140}{1000}=\mathrm{0.38.}$ Yeh step kyun? Kaafi saari green cabs, ek low error rate pe bhi, kaafi saare false "Blue" calls produce karti hain.

Step 3 — posterior. $P(B\mid W_B)=\frac{240}{240+140}=\frac{240}{380}\approx 0.6316=63.2\%.$ Yeh step kyun? Bahut saari green cabs answer ko witness ki $80\%$ accuracy se kaafi neeche khich leti hain.

Verify: complement $140/380\approx 0.3684$; $0.6316+0.3684=1$. ✓ $80\%$ se neeche — green cabs ke base rates matter karte hain, classic base-rate-neglect trap.

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Example 8 — C8 · Sequential updating & uski limit

Step 1 — pehla red. Prior $P(L)=0.5$. $P(\text{red})=(0.9)(0.5)+(0.5)(0.5)=0.45+0.25=0.70,$ $P(L\mid \text{red})=\frac{0.45}{0.70}\approx \mathrm{0.6429.}$ Yeh step kyun? Ek draw belief update karta hai; posterior ab next draw ke liye prior ban jaata hai — yeh belief updating in action hai.

Step 2 — doosra red. New prior $P(L)=0.6429$. $P(\text{red})=(0.9)(0.6429)+(0.5)(0.3571)=0.5786+0.1786=0.7571,$ $P(L\mid \text{red})=\frac{0.5786}{0.7571}\approx \mathrm{0.7642.}$ Yeh step kyun? Bayes chain karna; kyunki draws independent hain, hum equivalently likelihoods $0.9^n$ vs $0.5^n$ multiply kar sakte hain.

Step 3 — teesra red. New prior $P(L)=0.7642$. $P(\text{red})=(0.9)(0.7642)+(0.5)(0.2358)=0.6878+0.1179=0.8057,$ $P(L\mid \text{red})=\frac{0.6878}{0.8057}\approx \mathrm{0.8537.}$ Yeh step kyun? Chain mein teesra aur final update; har red belief ko Loaded bag ki taraf push karta rehta hai.

Step 4 — limit. $n$ reds ke baad odds hain $\frac{P(L)}{P(F)}={\left(\frac{0.9}{0.5}\right)}^n=1.8^n$. Kyunki $1.8>1$, yeh $\to \infty$, isliye $P(L\mid n\text{ reds})=\frac{1.8^n}{1.8^n+1}\underset{n\to \infty }{\to }1.$ Yeh step kyun? Evidence jo consistently ek hypothesis ko favour karta hai woh posterior ko certainty ki taraf drive karta hai — lekin finite draws mein kabhi $1$ nahi pahuncha (jab tak ek likelihood exactly $0$ na ho, cf. Example 4).

Verify: compact odds formula Step 3 ko reproduce karna chahiye: $\frac{1.8^3}{1.8^3+1}=\frac{5.832}{6.832}\approx 0.8537$. ✓ Step-by-step chain se match karta hai.

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Example 9 — C9 · Exam twist: complement & odds form

Step 1 — Bayes ke odds form se posterior odds. $\underset{\text{posterior odds}}{\underbrace{\frac{P(A\mid +)}{P(\bar{A}\mid +)}}}=\underset{\text{prior odds}}{\underbrace{\frac{P(A)}{P(\bar{A})}}}\times \underset{\text{likelihood ratio}}{\underbrace{\frac{P(+\mid A)}{P(+\mid \bar{A})}}}=\frac{0.02}{0.98}\times \frac{0.90}{0.10}.$ Yeh step kyun? Odds form evidence $P(+)$ ko completely sidestep karta hai — dono Bayes equations ko divide karne se common denominator cancel ho jaata hai.

$=0.020408\times 9=0.18367\approx 0.184(\text{about }1:5.44).$

Step 2 — posterior recover karo check karne ke liye (optional bridge). $P(A\mid +)=\frac{\text{odds}}{1+\text{odds}}=\frac{0.18367}{1.18367}\approx \mathrm{0.1552.}$ Yeh step kyun? Confirm karta hai ki odds Example 1 ke $15.5\%$ se match karte hain — exam ka "odds" answer wahi sach hai alag costume mein.

Step 3 — false-alarm probability bas complement hai. $P(\bar{A}\mid +)=1-P(A\mid +)\approx 1-0.1552=0.8448=84.5\%.$ Yeh step kyun? "Ek positive diya gaya, chance kya hai ki woh galat hai?" exactly $1$ minus posterior hai — koi naya computation nahi chahiye.

Verify: posterior odds $\times$ kuch nahi fraction wapas dena chahiye: $\frac{0.18367}{1+0.18367}=0.1552$, aur $0.1552+0.8448=1$. ✓

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Scenario matrix — kya humne har cell cover ki?

Recall Poora matrix tick karo

C1 standard rare cause ::: Example 1 ($15.5\%$) C2 large prior ::: Example 2 ($85.7\%$) C3 multi-way partition ::: Example 3 ($82.8\%$ spam) C4 perfect test, ek likelihood $=0$ ::: Example 4 ($P=1$) C5 certain prior $P(A)=0$ ::: Example 5 ($P=0$, unmovable) C6 useless evidence / independence ::: Example 6 (posterior $=$ prior $=0.30$) C7 real-world word problem ::: Example 7 (taxis, $63.2\%$) C8 sequential updating & limit ::: Example 8 (teen reds → $85.4\%$, limit $\to 1$) C9 exam twist: odds & complement ::: Example 9 (odds $0.184$, false alarm $84.5\%$)

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Connections

• Bayes' theorem — derivation and applications — parent formula jis par har example run karta hai.
• Conditional Probability — "flagged column mein se, real ka fraction kya hai."
• Law of Total Probability — Examples 3, 8 mein multi-term evidence.
• Independent Events — Example 6 mein tie ($P(A\mid B)=P(A)$).
• Tree Diagrams — Examples 1 aur 7 mein crowd-splitting pictures.
• Prior and Posterior Distributions — Example 8 mein sequential updating.
• Naive Bayes Classifier — Example 3 ka spam filter, generalised.

Concept Map