Conditional probability — P(A - B) = P(A∩B) - P(A): Visual walkthrough

2.7.8 · D2 · Maths › Statistics & Probability — Intermediate › Conditional probability — P(A - B) = P(A∩B) - P(A)

Yeh the parent note ka picture-first companion hai. Agar koi word naya lage, uske saath ek picture hogi.

Step 1 — Draw the whole world of outcomes

KYA HAI. Ek experiment (die roll karo, card draw karo) kai tarike se khatam ho sakta hai. Har possible ending ek outcome hai. Unhe saare ek box mein collect karo — woh box sample space hai, likha jaata hai $S$ . Poori idea ke liye Sample space and events dekho; yahan sirf itna chahiye: " $S$ = saare dots ka box."

KYUN. "Kisi cheez ke chance" ki baat karne se pehle, humein ek fixed universe chahiye jiske andar measure kar sakein. "Aadha" tab tak nahi bol sakte jab tak pata na ho "aadha kiska". $S$ wahi "kiska" hai.

PICTURE. Har chhota chalk dot ek outcome hai. Abhi ke liye maano ki har dot equally likely hai — winner banne ka same chance. Total dots ki sankhya ko $N$ likhte hain.

Figure — Conditional probability — P(A - B) = P(A∩B) - P(A)

Step 2 — Mark the two events $A$ and $B$

KYA HAI. Ek event bas kuch dots ke around ek chosen loop hai — $S$ ka ek subset. Hum do loops draw karte hain: $A$ aur $B$ . Yeh usually overlap karte hain.

KYUN. Conditional probability do events ke beech ek relationship hai. Yeh poochhne se pehle ki ek doosre ko kaise affect karta hai, dono ko stage par laana zaroori hai.

PICTURE. Blue loop $A$ , pink loop $B$ . Lens-shaped overlap — dots jo dono loops ke andar hain — ko $A \cap B$ likha jaata hai aur " $A$ and $B$ " padha jaata hai. Symbol $\cap$ ek chhota cup hai jo sirf wahi pakadta hai jo dono share karte hain.

$A \cap B$ ::: overlap region; dots jo dono events ko ek saath satisfy karte hain.
$n (A \cap B)$ ::: us overlap mein kitne dots hain.

Step 3 — Someone tells you " $B$ happened" → the world shrinks

KYA HAI. Ab sabse important move. Ek dost result dekh ke bolta hai: "Main promise karta hoon outcome $B$ ke andar hai." $B$ ke bahar har dot ab impossible hai — use board se mita do.

KYUN. Yeh ek tathya — " $B$ hua" — information hai. Information possibilities ko khatam karti hai. $B$ ke bahar wale dots ab jeet nahi sakte, toh unhe count bhi nahi karna chahiye. Pink loop $B$ tumhara naya, chhota box ban jaata hai.

PICTURE. Greyed-out dots ko fade hote dekho. Jo bachta hai woh exactly $B$ hai: ek chhota sample space jisme $n (B)$ dots hain.

Step 4 — Ask the question inside the new box

KYA HAI. Hum $B$ ke andar hain. Ab poochho: jo dots bache, unme se kaunse $A$ mein bhi hain? Woh exactly overlap dots $A \cap B$ hain — wahi $A$ -dots hain jo mite nahi.

KYUN. Koi bhi $A$ -dot jo $B$ ke bahar tha woh Step 3 mein mar gaya. Toh hamare chhote world ke andar, " $A$ hota hai" ka matlab sirf " $A \cap B$ dot" ho sakta hai. Isliye overlap, na ki poora $A$ , formula ka star hai.

PICTURE. Yellow, overlap dots ko highlight karta hai — survivors ke andar winners.

Step 1 ka count rule lagate hain, lekin box $B$ ke andar (toh total ab $n (B)$ hai, $N$ nahi): $P (A ∣ B) = \frac{n ( A \cap B )}{n ( B )}$

numerator $n (A \cap B)$ ::: yellow dots — survivors jo $A$ mein bhi hain.
denominator $n (B)$ ::: saare surviving dots — naya "total".
the fraction ::: chhote box ka woh slice jo $A$ leta hai.

Dhyan do: yeh hamara pehla asli formula hai, aur yeh seedha picture se nikla.

Step 5 — Trade dot-counts for probabilities

KYA HAI. Dots count karna sirf tab kaam karta hai jab saare dots equally likely hon. Us assumption se azaad hone ke liye, fraction ke upar aur neeche dono ko $N$ se divide karo (poore board ke total dots).

KYUN. Kisi fraction ke upar aur neeche ko same number se divide karna uski value kabhi nahi badalta — lekin yeh har count $n (\cdot)$ ko Step 1 ki probability $P (\cdot) = n (\cdot) / N$ mein convert kar deta hai. Yahi swap formula ko un duniyaon mein bhi kaam karne deta hai jahan outcomes equally likely nahi hote.

PICTURE. Same fraction, relabelled: baayein counts daayin probabilities mein badal jaate hain.

$P (A ∣ B) = \frac{n ( A \cap B )}{n ( B )} = \frac{n ( A \cap B ) / N}{n ( B ) / N} = \frac{P ( A \cap B )}{P ( B )}$

$\frac{n ( A \cap B )}{N} = P (A \cap B)$ ::: poore board ka overlap slice.
$\frac{n ( B )}{N} = P (B)$ ::: poore board ka pink slice.
their ratio ::: "saari cheez mein se overlap" ko "sirf $B$ mein se overlap" mein rescale karta hai.

Step 6 — $P (B)$ se kyun divide karo, $P (A)$ se kyun nahi?

KYA HAI. Formula ko uss ek case par test karo jiska answer hum pehle se pakka jaante hain: $P (B ∣ B)$ kya hai? Agar tumhe pata hai $B$ hua, toh " $B$ hone ka chance" $1$ hona chahiye — bilkul pakka.

KYUN. Ek sahi rescaling mein naya box poori certainty tak add up karna chahiye. $P (B)$ se divide karo toh hota hai; $P (A)$ se divide karo toh nahi — bas yahi cheez choice force karti hai.

PICTURE. Pink box, poori certainty ke saath, meter $1$ par pinned.

$P (B ∣ B) = \frac{P ( B \cap B )}{P ( B )} = \frac{P ( B )}{P ( B )} = 1 ✓$

$B \cap B = B$ ::: koi set khud ke saath overlap karo toh wahi milta hai.
result $= 1$ ::: exactly woh certainty jo humne maangi thi. $P (A)$ se divide karne par $P (B) / P (A)$ milta — $1$ barabar hone ki koi wajah nahi, toh yeh galat hai.

Step 7 — The forbidden case: $P (B) = 0$

KYA HAI. Agar $B$ impossible ho — ek khali loop, $n (B) = 0$ , toh $P (B) = 0$ ?

KYUN. Tab denominator $0$ hai, aur $\frac{kuch bhi}{0}$ undefined hai. Lekin "zero se divide mat karo" se bhi gehri wajah hai: tumhe bataya nahi ja sakta ki ek impossible event hua. Andar khade hone ke liye koi chhota box nahi hai — box khali hai. Sawaal hi meaningless hai.

PICTURE. Ek khali pink loop jiske upar chalk ka cross — condition karne ke liye koi dots nahi.

Step 8 — Multiplication rule mein rearrange karo (bonus picture)

KYA HAI. Boxed formula ke dono sides ko $P (B)$ se multiply karo.

KYUN. Kabhi kabhi tumhe $P (A ∣ B)$ nahi chahiye; tumhe $P (A \cap B)$ chahiye — chance ki dono hoon. Yahi woh cheez hai jo ek tree diagram chain karta hai, aur jo multiplication rule kehta hai.

PICTURE. Do-branch tree: pehli branch weighted $P (B)$ , doosri (conditional) branch weighted $P (A ∣ B)$ ; path ke saath multiply karo.

$P (A \cap B) = P (B) P (A ∣ B) = P (A) P (B ∣ A)$

$P (B)$ ::: pehli cheez ki probability (root branch).
$P (A ∣ B)$ ::: pehli ke given doosri ki probability (child branch).
product ::: poora path walk karne ki probability, yaani dono ki.

Aur independence turant dikh jaati hai: agar $B$ seekhne se kuch nahi badalta, $P (A ∣ B) = P (A)$ , toh $P (A \cap B) = P (A) P (B)$ . (Condition ko sahi se flip karne ke liye Bayes' Theorem; ek partition par conditionals sum karne ke liye Law of total probability.)

Ek-picture summary

Sab kuch compress kiya: poora board $\to$ $B$ tak shrink karo $\to$ $B$ mein se overlap count karo $\to$ probabilities mein rescale karo.

Recall Feynman: poori walkthrough ek 12-saal ke bachhe ko batao

Socho ek chalkboard dots se bhara hai — har dot game ke khatam hone ka ek tarika hai (Step 1). Kuch dots ke around ek blue loop aur ek pink loop draw karo; woh chhota football-jaisa part jahan dono cross karte hain woh "dono" hai (Step 2). Ab ek dost dekhta hai aur kasam khata hai ki asli dot pink loop ke andar hai. Turant, pink ke bahar har dot impossible ho jaata hai — unhe mita do. Tumhara board chhota ho gaya; pink loop ab tumhari poori duniya hai (Step 3). Poochho: jo dots abhi bhi zinda hain, unme se kaunse blue mein bhi hain? Sirf woh jo us football-jaisi crossing mein hain. Toh chance hai (football dots) ÷ (saare pink dots) — yahi hai $P (A ∣ B) = n (A \cap B) / n (B)$ (Step 4). Jab dots equally likely na hon tab bhi kaam karne ke liye, upar aur neeche dono ko total dot count se divide karo; counts probabilities ban jaate hain aur milta hai $P (A \cap B) / P (B)$ (Step 5). Pink se kyun divide karo, blue se kyun nahi? Kyunki agar dost bole "woh pink mein hai" toh "woh pink mein hone ka chance" ek pakki $1$ honi chahiye — aur sirf $P (B)$ se divide karne par woh milta hai (Step 6). Agar pink loop khali hai, toh dekhne ke liye kuch hai hi nahi, toh sawaal ka koi matlab nahi — isliye zaroori hai $P (B) > 0$ (Step 7). Formula ulta karo aur tumhe woh "dono" rule milta hai jo trees use karte hain (Step 8). Ho gaya — tumne definition draw ki.

Recall Quick self-check

Denominator $P (B)$ kyun hai, $P (A)$ kyun nahi? ::: Kyunki $B$ woh naya box hai jiske andar hum khade hain; sirf $P (B)$ se divide karne par $P (B ∣ B) = 1$ milta hai. " $B$ hua" ke baad kaunse dots bachte hain? ::: Sirf woh jo $B$ ke andar hain; naya total $n (B)$ hai. Poora $A$ nahi, sirf $A \cap B$ kyun count hota hai? ::: $B$ ke bahar wale $A$ -dots wipe ho gaye, toh $B$ ke andar "in $A$ " ka matlab "overlap mein" hai. Jab $P (B) = 0$ ho tab kya toot jaata hai? ::: Zero se division hoti hai, aur condition karne ke liye koi non-empty box nahi hai — sawaal meaningless hai.

Connections

Sample space and events — dots ka woh board jahan se shuru kiya.
Multiplication rule of probability — Step 8 ki rearranged form.
Independence of events — jab $P (A ∣ B) = P (A)$ ho.
Tree diagrams — Step 8 ki picture.
Bayes' Theorem — condition ko safely flip karna.
Law of total probability — ek partition par conditionals sum karna.