Conditional probability — P(A - B) = P(A∩B) - P(A): Worked examples

2.7.8 · D3 · Maths › Statistics & Probability — Intermediate › Conditional probability — P(A - B) = P(A∩B) - P(A)

The scenario matrix

Har conditional-probability question jo tum meet karte ho, in case classes mein se kisi ek mein aata hai. Hum har cell ko kam se kam ek baar hit karenge.

#	Case class	Kya cheez tricky hai	Worked in
C1	Basic overlap (equally likely, ek shrunken universe mein count karo)	$B$ tak restrict karna bhool jaana	Ex 1
C2	Flip the bar — $P(A	B) $v s$ P(B	A)$
C3	Chaining without replacement (multiplication rule)	universe har step pe shrink hoti hai	Ex 3
C4	Independence check — is $P(A	B)=P(A)$?	overlap dependent lag sakta hai
C5	Degenerate: $P (B) = 0$ (impossible condition)	formula zero se divide karta hai	Ex 5
C6	Degenerate: $A \subseteq B$ or $B \subseteq A$ (nested events)	answer $0$ ya $1$ hota hai	Ex 5
C7	Condition on a complement $P(A	B^c)$	numerator ke $A$ ko touch mat karo
C8	Real-world word problem (base rates / false positives)	condition reverse karne se galat answer aata hai	Ex 7
C9	Limiting behaviour — condition ek single outcome ki taraf shrink hoti hai	probabilities $0$ ya $1$ pe snap karti hain	Ex 8
C10	Condition on one cell of a partition (disjoint $B_{1}, \dots, B_{n}$ )	parts disjoint hone chahiye aur sab cover karne chahiye	Ex 9
C11	Exam twist — condition on a union	overlap ko double-count mat karo	Ex 10

Prerequisite pictures jinpe hum lean karte hain: Sample space and events (the universe), Tree diagrams (chaining), Multiplication rule of probability, Independence of events, Bayes' Theorem, Law of total probability.

C1 — Basic overlap

Worked example Ex 1 — Fair die, even world tak restrict karo

Ek fair six-sided die roll karo. Maano $A = {$ score is a prime $} = {2, 3, 5}$ aur $B = {$ score is even $} = {2, 4, 6}$ . $P (A ∣ B)$ nikalo.

Forecast: abhi ek fraction guess karo — even numbers mein se kitne prime hain? Apna guess likho.

Step 1 — universe draw karo aur $B$ shade karo.

Yeh step kyun? Conditional probability matlab hai "sab kuch jo

B

ke bahar hai use throw away karo." Picture naye (chhote) universe ko literal banati hai: sirf

{2, 4, 6}

bachte hain.

Step 2 — overlap $A \cap B$ nikalo. Woh primes jo even bhi hain: sirf $2$ . To $A \cap B = {2}$ aur $P (A \cap B) = \frac{1}{6}$ . Yeh step kyun? Numerator un outcomes ko count karta hai jo dono mein hain — figure mein amber cell.

Step 3 — $P (B)$ nikalo. $B = {2, 4, 6}$ mein $3$ outcomes hain, to $P (B) = \frac{3}{6} = \frac{1}{2}$ . Yeh step kyun? $B$ denominator hai: yeh us duniya ka size hai jismein ab hum rehte hain.

Step 4 — divide karo. $P (A ∣ B) = \frac{P ( A \cap B )}{P ( B )} = \frac{1/6}{3/6} = \frac{1}{3} .$ Yeh step kyun? Yeh definition hi hai — shrunken universe par overlap. Divide karna chhote amber slice ko us duniya ke size se rescale karta hai jismein hum ab rehte hain, ek raw count ko ek probability mein badalta hai jo $B$ ke andar $1$ tak sum hoti hai.

Verify: ${2, 4, 6}$ tak directly restrict karo aur andar primes count karo: sirf ${2}$ , yaani $3$ mein se $1$ . Same $\frac{1}{3}$ . ✔

C2 — Flip the bar

Worked example Ex 2 — Same events, opposite condition

Bilkul same $A = {2, 3, 5}$ , $B = {2, 4, 6}$ use karo Ex 1 se, ab $P (B ∣ A)$ nikalo aur compare karo.

Forecast: kya yeh $\frac{1}{3}$ phir se hoga? Guess karo haan/nahi.

Step 1 — numerator rakho, yeh symmetric hai. $A \cap B = {2}$ abhi bhi, to $P (A \cap B) = \frac{1}{6}$ . Yeh step kyun? " $A$ aur $B$ dono" ko order ki parwah nahi — isliye numerator Ex 1 se same hai. Yahi exactly trap hai.

Step 2 — denominator ko new condition $A$ mein change karo. Ab given event $A = {2, 3, 5}$ hai, to hum $P (A) = \frac{3}{6} = \frac{1}{2}$ se divide karte hain. Yeh step kyun? "Bottom = the Bar's Buddy." Condition bar ke right mein baith ta hai; yahan woh $A$ hai.

Step 3 — divide karo. $P (B ∣ A) = \frac{1/6}{3/6} = \frac{1}{3} .$ Yeh step kyun? Ex 1 jaisi same definition, lekin duniya ab $A$ hai, $B$ nahi. Hum overlap ko $P (A)$ se rescale karte hain taaki new universe $A$ ke andar probabilities $1$ tak sum hon — alag rescaling alag meaning deta hai.

Step 4 — coincidence ko dhyan se padho. Yahan $P (A ∣ B) = P (B ∣ A) = \frac{1}{3}$ sirf isliye kyunki $P (A) = P (B) = \frac{1}{2}$ . Kisi ek event ka size change karo aur dono alag ho jaate hain — Ex 7 dekho. Yeh step kyun? Yeh galat lesson "woh hamesha equal hote hain" ko khatam karne ke liye. Woh equal hote hain iff $P (A) = P (B)$ .

Verify: Bayes' Theorem kehta hai $P (B ∣ A) = P (A ∣ B) \frac{P ( B )}{P ( A )} = \frac{1}{3} \cdot \frac{1/2}{1/2} = \frac{1}{3}$ . ✔

C3 — Chaining without replacement

Worked example Ex 3 — Teen aces ek row mein

Standard $52$ -card deck se $3$ cards draw karo, no replacement. $P (all three are aces)$ nikalo.

Forecast: $(\frac{4}{52})^{3}$ se bada ya chhota? Guess karo.

Step 1 — chain ko tree se banao.

Yeh step kyun? Tree diagrams dikhate hain ki universe har node par shrink hoti hai — aces bachi aur cards bachi dono ek success ke baad ek kam ho jaati hain.

Step 2 — pehla draw. $P (A_{1}) = \frac{4}{52}$ . Yeh step kyun? Abhi koi information nahi; $52$ cards mein $4$ aces raw base rate hai.

Step 3 — second draw given pehla ace tha. $P (A_{2} ∣ A_{1}) = \frac{3}{51}$ . Yeh step kyun? Universe shrink ho gayi: "pehla ace tha" par conditioning karne se $51$ cards mein $3$ aces bachte hain.

Step 4 — third draw given pehle do aces the. $P (A_{3} ∣ A_{1} \cap A_{2}) = \frac{2}{50}$ . Yeh step kyun? Do aces ja chuke hain, $50$ cards bachti hain — condition universe ko phir se tighten karti hai.

Step 5 — branch ke along multiply karo. $P = \frac{4}{52} \cdot \frac{3}{51} \cdot \frac{2}{50} = \frac{24}{132600} = \frac{1}{5525} .$ Yeh step kyun? General Multiplication rule of probability conditionals ko chain karta hai: $P (A_{1} \cap A_{2} \cap A_{3}) = P (A_{1}) P (A_{2} ∣ A_{1}) P (A_{3} ∣ A_{1} \cap A_{2})$ .

Verify: $\frac{1}{5525} \approx 0.000181$ , aur $(\frac{4}{52})^{3} \approx 0.000455$ — sahi value chhoti hai kyunki aces hatane se baad wala har ace rarer hota jaata hai. ✔

C4 — Independence check

Worked example Ex 4 — Kya

$B$ jaanna $A$ ko change karta hai? $52$ mein se ek card draw karo. $A = {$ the card is a King $}$ , $B = {$ the card is a Heart $}$ . Kya $A$ , $B$ se independent hai? $P (A ∣ B)$ nikalo.

Forecast: kya "yeh Heart hai" jaanna King hone ki chance change karta hai?

Step 1 — overlap. $A \cap B = {$ King of Hearts $}$ , ek card, to $P (A \cap B) = \frac{1}{52}$ . Yeh step kyun? Exactly ek card dono King aur Heart hai — woh single card numerator hai.

Step 2 — condition. $P (B) = \frac{13}{52} = \frac{1}{4}$ ( $13$ Hearts). Yeh step kyun? $B$ given event hai, to iska size denominator hai.

Step 3 — divide karo. $P (A ∣ B) = \frac{1/52}{13/52} = \frac{1}{13} .$ Yeh step kyun? Definition — hum single overlap card ko $13$ -card Heart world se rescale karte hain taaki Hearts mein se Kings ka fraction mile.

Step 4 — $P (A)$ se compare karo. $P (A) = \frac{4}{52} = \frac{1}{13}$ . Kyunki $P (A ∣ B) = P (A)$ , events independent hain — Heart hona Kinghood ke baare mein kuch nahi batata. Yeh step kyun? Independence of events precisely woh case hai jab $P (A ∣ B) = P (A)$ .

Verify: independence ka matlab yeh bhi hai $P (A \cap B) = P (A) P (B)$ : check karo $\frac{1}{4} \cdot \frac{1}{13} = \frac{1}{52}$ . ✔

C5 & C6 — Degenerate inputs

Worked example Ex 5 — Jab machine break ho ya trivialise ho

Same fair die. Teen edge questions ka jawab do.

(a) $P (B) = 0$ — impossible condition. Maano $B = {$ score is $7}$ . Kisi bhi $A$ ke liye $P (A ∣ B)$ nikalo. Step 1. $P (B) = 0$ kyunki $7$ koi face nahi hai. Yeh step kyun? $B$ empty hai — koi outcome ise realise nahi karta. Step 2. $P (A ∣ B) = \frac{P ( A \cap B )}{0}$ — undefined. Yeh step kyun: formula ka fine print hai $P (B) > 0$ ; zero se divide karna koi value nahi deta, to "given the impossible" ka koi jawab nahi hai.

(b) Nested $A \subseteq B$ . Maano $A = {2}$ , $B = {2, 4, 6}$ . $P (A ∣ B)$ nikalo. Step 1. $A \cap B = A = {2}$ , to $P (A \cap B) = \frac{1}{6}$ . Yeh step kyun? Kyunki $A$ poori tarah $B$ ke andar rehta hai, overlap poora $A$ hai. Step 2. $P (A ∣ B) = \frac{1/6}{3/6} = \frac{1}{3}$ . Yeh step kyun? Hum overlap ko $P (B)$ se divide karte hain bilkul waise hi; kyunki $A \subseteq B$ hai to numerator $P (A)$ ke barabar hai, isliye formula clean shape $P (A ∣ B) = \frac{P ( A )}{P ( B )}$ mein reduce ho jaata hai.

(c) Reverse nesting $B \subseteq A$ . Maano $A = {2, 4, 6}$ , $B = {4}$ . $P (A ∣ B)$ nikalo. Step 1. $A \cap B = B = {4}$ , $P (A \cap B) = \frac{1}{6}$ . Yeh step kyun? $B$ , $A$ ke andar baitha hai, to overlap poora $B$ hai. Step 2. $P (A ∣ B) = \frac{1/6}{1/6} = 1$ . Yeh step kyun? Hum phir bhi $P (B)$ se divide karte hain, lekin ab condition ka har outcome pehle se $A$ mein hai; overlap poori condition ke barabar hai, to ratio $1$ hone par majboor hai — certainty.

Verify: (b) $\frac{1}{3}$ aur (c) $1$ ; aur $P (B ∣ B) = 1$ hamesha, parent mein rescaling logic se match karta hai. ✔

C7 — Complement par condition

Worked example Ex 6 — "NOT even" given, prime kitna likely hai?

Same fair die, $A = {$ prime $} = {2, 3, 5}$ , $B = {$ even $} = {2, 4, 6}$ . Ab $B^{c}$ par condition karo (die odd hai). $P (A ∣ B^{c})$ nikalo.

Forecast: kya $P (A ∣ B) + P (A ∣ B^{c})$ $1$ tak add hoga? Compute karne se pehle haan/nahi guess karo.

Step 1 — complement banao. $B^{c} = {1, 3, 5}$ (odd faces), to $P (B^{c}) = \frac{3}{6} = \frac{1}{2}$ . Yeh step kyun? $B^{c}$ par conditioning ka matlab hai new universe "sab kuch jo $B$ mein nahi hai." Hum condition ko uske complement mein change kar sakte hain, lekin bar ke left mein event $A$ unchanged rehta hai.

Step 2 — $A$ ka new universe ke saath overlap. $A \cap B^{c} = {2, 3, 5} \cap {1, 3, 5} = {3, 5}$ , to $P (A \cap B^{c}) = \frac{2}{6}$ . Yeh step kyun? Numerator un primes ko count karta hai jo odd bhi hain — woh primes jo odd world mein survive karte hain.

Step 3 — divide karo. $P (A ∣ B^{c}) = \frac{2/6}{3/6} = \frac{2}{3} .$ Yeh step kyun? Definition — shrunken (odd) universe par overlap.

Step 4 — do false "complement" laws check karo. Ex 1 se, $P (A ∣ B) = \frac{1}{3}$ . To $P (A ∣ B) + P (A ∣ B^{c}) = \frac{1}{3} + \frac{2}{3} = 1$ — yahan yeh $1$ ke barabar hota hai, lekin sirf coincidence se (dono conditions ka size $3$ hai aur primes evenly split hote hain). Jo law hamesha true hai woh event par hai: $P (A ∣ B) + P (A^{c} ∣ B) = 1$ , $A$ ko change karo, kabhi condition ko nahi. Yeh step kyun? Parent ka mistake box warn karta hai ki $P (A ∣ B) + P (A ∣ B^{c}) = 1$ ek law nahi hai; yeh example dikhata hai ki yeh accident se hold ho sakta hai, to tum ise rely nahi kar sakte.

Verify: hamesha-true law: $P (A^{c} ∣ B) = \frac{P ( A ^{c} \cap B )}{P ( B )}$ jahan $A^{c} \cap B = {4, 6}$ hai, $\frac{2/6}{3/6} = \frac{2}{3}$ deta hai, aur $\frac{1}{3} + \frac{2}{3} = 1$ . ✔

C8 — Real-world word problem

Worked example Ex 7 — Rare disease, achha test, surprising answer

Ek disease $2%$ population ko affect karti hai. Ek test beemar logon mein se $95%$ ke liye positive result deta hai (true positive) aur healthy logon mein se $5%$ ke liye (false positive). Tum test positive aate ho. $P (sick ∣ +)$ nikalo.

Forecast: ek number guess karo. Zyaadatar log "lagbhag $95%$ " guess karte hain. Dekho kya hota hai.

Step 1 — $10000$ ka ek concrete crowd lo. Sick $= 2%$ of $10000 = 200$ . Healthy $= 9800$ . Yeh step kyun? Whole numbers shrunken universe ko visible banate hain — yeh Law of total probability disguise mein hai.

Step 2 — har group ko test result ke hisaab se split karo ( $2 \times 2$ tree banao).

Sick aur

+

95%

200 = 190

. Healthy aur

+

5%

9800 = 490

. Yeh step kyun? Hume jo numerator chahiye woh hai "sick aur positive"; denominator hai "sabhi positive."

Step 3 — new universe tak restrict karo: sabhi positive. Total positives $= 190 + 490 = 680$ . Yeh step kyun? Hum $+$ par condition kar rahe hain, to $+$ ab duniya hai.

Step 4 — us duniya mein sick ka fraction. $P (sick ∣ +) = \frac{190}{680} = \frac{19}{68} \approx 0.279.$ Yeh step kyun? Definition — favourable (sick $+$ ) over total (sab $+$ ).

Step 5 — moral. Sirf lagbhag $28%$ , $95%$ nahi! Kyunki healthy group itna bada hai, uske $5%$ false positives ( $490$ ) sick group ke true positives ( $190$ ) ko swamp kar dete hain. Yeh Bayes' Theorem flip hai: $P (sick ∣ +) \neq = P (+ ∣ sick)$ . Yeh step kyun? Memory mein fix karne ke liye ki bar reverse karna (base-rate neglect) classic real-world error hai.

Verify: Bayes directly: $\frac{0.02 \cdot 0.95}{0.02 \cdot 0.95 + 0.98 \cdot 0.05} = \frac{0.019}{0.068} \approx 0.279$ . ✔

C9 — Limiting behaviour

Worked example Ex 8 — Condition ko ek single point tak squeeze karna

Fair die roll karo. $A = {6}$ fix karo. Dekho $P (A ∣ B_{k})$ kaise change hota hai jab condition $B_{k}$ shrink hoti hai: $B_{1} = {4, 5, 6}$ , phir $B_{2} = {5, 6}$ , phir $B_{3} = {6}$ .

Forecast: jaise $B$ , ${6}$ ke paas jaata hai, answer kahan jaata hai?

Step 1 — $B_{1} = {4, 5, 6}$ . $A \cap B_{1} = {6}$ , to $P (A ∣ B_{1}) = \frac{1/6}{3/6} = \frac{1}{3}$ . Yeh step kyun? Universe mein teen mein se ek favourable outcome.

Step 2 — $B_{2} = {5, 6}$ . $P (A ∣ B_{2}) = \frac{1/6}{2/6} = \frac{1}{2}$ . Yeh step kyun? Universe half ho gayi, lekin favourable outcome $6$ abhi bhi survive karta hai — fraction badhta hai.

Step 3 — $B_{3} = {6}$ . $P (A ∣ B_{3}) = \frac{1/6}{1/6} = 1$ . Yeh step kyun? Condition ab wahi outcome hai — certainty.

Step 4 — limit padho. Jaise condition us single outcome ke around tighten hoti hai jo $A$ ko true banata hai, $P (A ∣ B_{k}) \to 1$ . Agar hum ${4, 5}$ ki taraf squeeze karte ( $6$ miss karte), to overlap vanish ho jaata aur $P (A ∣ B_{k}) \to 0$ . Yeh step kyun? Ek condition limit karna ya to $A$ ko guarantee karta hai ya forbid — answer $1$ ya $0$ par snap karta hai, endpoint par kabhi beech mein nahi rukta.

Verify: sequence hai $\frac{1}{3}, \frac{1}{2}, 1$ — strictly increasing to $1$ . ✔

C10 — Partition ke ek cell par condition

Worked example Ex 9 — Teen coloured urns mein balls (ek sachi partition)

Ek factory parts ko teen disjoint machines $M_{1}, M_{2}, M_{3}$ se bhejti hai jo milke sab parts handle karti hain: $M_{1}$ $50%$ banati hai, $M_{2}$ $30%$ , $M_{3}$ $20%$ . Ek part defective hone ki probability $M_{1}, M_{2}, M_{3}$ par respectively $0.02, 0.04, 0.10$ hai. $P (defective ∣ M_{2})$ nikalo, aur phir overall $P (defective)$ .

Forecast: kya $P (defective ∣ M_{2})$ sirf $0.04$ hai, ya machine shares matter karte hain?

Step 1 — check karo ki $M_{1}, M_{2}, M_{3}$ ek partition banate hain. Woh disjoint hain (ek part exactly ek machine se aata hai) aur sab cover karte hain ( $0.5 + 0.3 + 0.2 = 1$ ). Yeh step kyun? Law of total probability tabhi kaam karta hai jab conditions disjoint parts hon jo poore sample space ko tile karin — yahi "partition" ka matlab hai.

Step 2 — single cell $M_{2}$ par condition karo. $P (defective ∣ M_{2}) = 0.04$ directly given hai — yeh already $M_{2}$ world ke andar defectives ka fraction hai. Yeh step kyun? Ek cell par conditioning ka matlab hai universe sirf woh cell hai; per-cell defect rate exactly conditional probability hai, koi rescaling nahi chahiye.

Step 3 — overall rate ke liye cells ko recombine karo. $P (defective) = \sum_{i} P (M_{i}) P (defective ∣ M_{i}) = 0.5 (0.02) + 0.3 (0.04) + 0.2 (0.10) .$ Yeh step kyun? Law of total probability har cell ke conditional ko us cell ke size se weight karta hai, phir sum karta hai — kyunki cells space ko partition karte hain.

Step 4 — compute karo. $P (defective) = 0.010 + 0.012 + 0.020 = 0.042.$ Yeh step kyun? Arithmetic total-probability sum khatam karta hai.

Verify: $0.5 \cdot 0.02 + 0.3 \cdot 0.04 + 0.2 \cdot 0.10 = 0.042$ , aur sirf $M_{2}$ par conditioning $0.04$ return karta hai. ✔

C11 — Exam twist: union par condition

Worked example Ex 10 — "

$B$ ya $C$ " par condition karo Fair die. $A = {$ prime $} = {2, 3, 5}$ . $B \cup C$ par condition karo jahan $B = {$ even $} = {2, 4, 6}$ aur $C = {$ score $\leq 2} = {1, 2}$ . $P (A ∣ B \cup C)$ nikalo.

Forecast: condition ab ek union hai — kya denominator $P (B) + P (C)$ hoga? Careful.

Step 1 — union dhyan se banao (inclusion–exclusion). $B \cup C = {2, 4, 6} \cup {1, 2} = {1, 2, 4, 6}$ . Note karo ki $2$ ek baar count hota hai, do baar nahi. Yeh step kyun? $P (B \cup C) = P (B) + P (C) - P (B \cap C) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6}$ . Naively $P (B) + P (C) = \frac{5}{6}$ add karna galat hoga.

Step 2 — $A$ ka union ke saath overlap. $A \cap (B \cup C) = {2, 3, 5} \cap {1, 2, 4, 6} = {2}$ , to $P = \frac{1}{6}$ . Yeh step kyun? Sirf $2$ prime bhi hai aur conditioning set mein bhi hai.

Step 3 — divide karo. $P (A ∣ B \cup C) = \frac{1/6}{4/6} = \frac{1}{4} .$ Yeh step kyun? Definition, sahi union denominator ke saath — shrunken universe ${1, 2, 4, 6}$ hai.

Verify: ${1, 2, 4, 6}$ tak restrict karo aur primes count karo: sirf ${2}$ — yeh $4$ mein se $1$ hai. Same $\frac{1}{4}$ . ✔

Recall Matrix par quick self-test

"Do aces already drawn given, next ace ki chance" kaunsa cell hit karta hai? ::: C3 — chaining without replacement. Die par $P (A ∣ {7})$ undefined kyun hai? ::: The condition mein $P (B) = 0$ hai; zero se divide karna undefined hai (C5). Agar $A \subseteq B$ hai, to $P (A ∣ B)$ simplify karo. ::: $P (A) / P (B)$ (C6). $B^{c}$ par conditioning karte waqt, kaun si quantity tum NAHI badal sakte? ::: Bar ke left mein event $A$ ; sirf condition $B^{c}$ banti hai (C7). Disease problem mein answer $95%$ kyun nahi hai? ::: Huge healthy group se false positives dominate karte hain (C8 / Bayes flip). Kaunsi do properties $M_{1}, M_{2}, M_{3}$ ko partition banati hain? ::: Disjoint (koi overlap nahi) aur exhaustive (probabilities $1$ tak sum hoti hain) (C10). Denominator for conditioning on $B \cup C$ ? ::: $P (B \cup C) = P (B) + P (C) - P (B \cap C)$ , $P (B) + P (C)$ nahi (C11).

Connections

Multiplication rule of probability — Ex 3 ki chaining ke peeche engine.
Tree diagrams — Ex 3 aur Ex 7 ke liye picture.
Independence of events — Ex 4 mein test.
Bayes' Theorem — Ex 2 aur Ex 7 mein sahi flip.
Law of total probability — Ex 9 mein partition recombination.
Sample space and events — woh universe jo hum baar baar shrink karte rehte hain.