Mutually exclusive events — addition rule: Visual walkthrough

2.7.6 · D2 · Maths › Statistics & Probability — Intermediate › Mutually exclusive events — addition rule

Hum sab kuch ek honest idea par banate hain: probability dots ki fair counting hai.

Step 1 — Poori duniya ko dots ki tarah draw karo

KYA. Socho ki ek experiment ke har possible outcome ko ek single dot ke roop mein ek bade rectangle ke andar imagine karo. Rectangle woh sample space hai, jise $S$ likha jaata hai. Agar hum ek fair die roll karein, toh chhe dots hain: faces $1, 2, 3, 4, 5, 6$ .

KYUN. Pehle hum "is ya us ki chance" ki baat kar sakein, humein ek aisi jagah chahiye jahan sab kuch jo ho sakta hai rehta ho. Counting tab hi fair hoti hai jab hum poori pool dekh sakein jisme se hum count kar rahe hain — woh pool hai $S$ . Dekho Probability — basic definitions & sample space.

PICTURE. Chhe dots, har ek equally likely, box $S$ mein floating hain.

Figure — Mutually exclusive events — addition rule

Step 2 — Ek event dots ka ek fenced-off group hai

KYA. Ek event bas unhi dots mein se kuch hoti hai, circled off. Lo $A =$ "even number roll karo" $= {2, 4, 6}$ . Exactly unhi teen dots ke around ek blue loop draw karo.

KYUN. Hum bahut kam single outcome ki parwah karte hain; hum ek sawaal ki parwah karte hain jaise "kya yeh even hai?". Us sawaal ka jawab dots ka ek subset deta hai — woh jo answer "yes" banate hain. Unhein circle karna ek vague sawaal ko countable dots mein badal deta hai.

PICTURE. Blue loop $A$ mein $2, 4, 6$ capture hain; dots $1, 3, 5$ bahar rehte hain.

Step 3 — Do events, aur overlap jo tum literally dekh sakte ho

KYA. Ek doosra event add karo $B =$ "3 se bada number roll karo" $= {4, 5, 6}$ , orange loop ki tarah draw kiya gaya. Ab dekho jahan blue aur orange cross karte hain: dots $4$ aur $6$ dono loops ke andar baithe hain. Yeh shared region hai intersection $A \cap B = {4, 6}$ .

KYUN. Hum " $A$ ya $B$ ki chance" ki taraf ja rahe hain. Us sawaal ki poori mushkil is crossing region mein rehti hai — woh dots jo ek saath do clubs mein belong karti hain. Ise abhi naam dena ( $A \cap B$ ) hume baad mein handle karne deta hai. Yeh ek Venn diagram hai.

PICTURE. Blue loop aur orange loop overlap kar rahe hain; lens-shaped middle (green shading) mein $4$ aur $6$ hain.

Step 4 — Kyun naive adding over-count karti hai (double-count visible banaya gaya)

KYA. Union count karne ki koshish karo seedha add karke: $n (A) + n (B) = 3 + 3 = 6$ . Lekin true union ${2, 4, 5, 6}$ mein sirf 4 dots hain. Humein 6 mili. Extra 2 exactly $4$ aur $6$ hain.

KYUN. Jab tum "blue loop mein dots" count karte ho toh tumne already $4$ aur $6$ count kar liye. Jab tum phir "orange loop mein dots" count karte ho toh tum $4$ aur $6$ dobara count karte ho. Har overlap dot do baar tally ho gaya. Yahi woh jagah hai jahan error paida hoti hai — aur tum do guilty dots ko point kar sakte ho.

PICTURE. Do shared dots red mein lit, har ek blue ke liye ek baar aur orange ke liye ek baar tally hote dikhaye gaye — har ek par "counted twice" ka tag.

Step 5 — Overlap ko ek baar subtract karo: inclusion–exclusion

KYA. Humne shared dots ko exactly ek copy zyada add kar diya. Toh overlap ki ek copy hata do: $n (A \cup B) = n (A) + n (B) - n (A \cap B)$ Check karo: $3 + 3 - 2 = 4$ . ✓ Yeh un chaar dots se match karta hai jo hum dekh sakte hain.

KYUN. Humne jab loops add kiye toh har dot include ki, lekin overlap ko do baar include kiya; toh iska ek copy exclude karo. Sab kuch add karo, jo double-add hua usse wapas lo — inclusion–exclusion. Dekho General addition rule & inclusion–exclusion.

PICTURE. Wahi Venn, green lens ko "ek baar lift out" kiya ja raha hai — ek arrow ${4, 6}$ ki ek copy door kheench raha hai, 4 ka clean count chhodke.

Step 6 — Special case: loops ko alag kheeecho

KYA. Ab events change karo taaki woh koi dot share na karein. Lo $A = {2}$ (blue) aur $B = {5}$ (orange) die par. Loops ab touch nahi karte: $A \cap B = \emptyset$ (woh empty set, koi dots nahi).

KYUN. Agar crossing region mein kuch nahi hai, toh kuch bhi do baar count nahi hua — toh subtract karne ke liye kuch nahi hai. Correction term ignore nahi ki jaati; woh genuinely $0$ hai kyunki $n (A \cap B) = 0$ .

PICTURE. Do alag loops ke beech mein gap; crossing region empty hai, $\emptyset$ se mark ki gayi.

Step 7 — Degenerate & edge cases (koi bhi scene jo humne nahi dikhaya)

KYA & KYUN & PICTURE — chaar boundary situations, har ek apne Venn look ke saath:

Ek loop doosre ko swallow kar leti hai ( $A \subseteq B$ , jaise $A = {4}$ , $B = {4, 5, 6}$ ). Overlap $= A$ khud, toh $P (A \cup B) = P (B)$ . Picture: chhoti blue loop poori tarah orange ke andar.
Loops identical hain ( $A = B$ ). Toh $P (A \cup B) = P (A)$ — do baar add karke ek baar subtract karne se ek copy bachti hai.
Empty event ( $A = \emptyset$ , $P (A) = 0$ ). Koi dots wala event add karna kuch nahi badalta: $P (\emptyset \cup B) = P (B)$ .
Loops box bhar deti hain — complement case: $A$ aur $A^{'}$ kuch share nahi karte aur saath mein pure $S$ cover karte hain. Toh $P (A) + P (A^{'}) = \frac{n ( S )}{n ( S )} = 1$ , jisse $P (A^{'}) = 1 - P (A)$ milta hai.

Recall Kyun "mutually exclusive" "independent" ka

opposite hai Question ::: Agar $A$ aur $B$ mutually exclusive hain aur $P (A), P (B) > 0$ , kya woh independent hain? Answer ::: Nahi. Exclusive ka matlab hai "agar $A$ hota hai, toh $B$ nahi ho sakta" — $A$ ko jaanna tumhe $B$ ke baare mein sab kuch bata deta hai (woh impossible hai). Yeh strong dependence hai. Independence (Independent events — multiplication rule) ka matlab hai $A$ tumhe $B$ ke baare mein kuch nahi batata. Exclusive union add karne ki duniya mein rehta hai; independent intersection multiply karne mein rehta hai.

Step 8 — Sanity check: probabilities 1 se zyada nahi ho sakti

KYA. Maano $P (A) = 0.6$ , $P (B) = 0.7$ . Kya woh mutually exclusive ho sakte hain?

KYUN. Agar hote, toh $P (A \cup B) = 0.6 + 0.7 = 1.3$ — pure box se bada, jo impossible hai ( $P \leq 1$ ). Picture force karti hai loops ko overlap karne ke liye: unka shared share kam se kam $0.6 + 0.7 - 1 = 0.3$ hai.

PICTURE. Ek box jo sirf "size 1" ka hai, do loops ke saath jo itne bade hain ki unhe zaroor intersect karna hai, forced overlap $\geq 0.3$ red mein shaded.

Ek-picture summary

Upar sab kuch, compress kiya gaya: do loops, overlap ek baar subtract ki (general rule), aur wahi picture jisme loops alag khichi gayi hain toh overlap gayab ho jaati hai (mutually exclusive rule).

Recall Feynman retelling — ek 12-saal ke bachche ko batao

Ek table imagine karo jo dots se bhari hai — har woh possible cheez jo ho sakti hai. Blue lasso ko "yes" dots ke around loop karo question A ke liye, aur orange lasso ko "yes" dots ke around question B ke liye. "A ya B" ki chance chahiye? Blue dots gino, orange dots gino, unhe add karo. Lekin dono lassos ne jo bhi dot pakdi woh abhi do baar count ho gayi — toh use ek baar cross out karo. Yeh single "overlap ko ek baar cross out karna" hi poora general rule hai: $P (A) + P (B) - P (A \cap B)$ . Ab dono lassos ko alag kheeecho jab tak woh touch na karein — koi dot dono mein nahi, cross out karne ke liye kuch nahi, toh bas add karo. Woh alag khichi lassos hi "mutually exclusive" ka matlab hain. Same rule, ek mein zero. Aur agar lassos kabhi itni badi ho jayein ki unke shares 1 se aage add ho jayein, woh alag nahi khichi ja sakti — woh overlap karne ke liye force hain.

Active-recall

Union dot-language mein

kisi bhi loop mein dots;

n (A \cup B)

Double-count kahan se aata hai?

overlap dots (

A \cap B

) har loop ke liye ek baar tally ho jaate hain.

P (A \cap B)

exactly ek baar kyun subtract karte hain?

humne use do baar add kiya, toh ek extra copy hataate hain.

Picture mein mutually exclusive

loops touch nahi karte;

A \cap B = \emptyset

, toh subtracted term

0

hai.

Kya

P (A) = 0.6, P (B) = 0.7

exclusive ho sakte hain?

Nahi —

1.3 > 1

; unhe

\geq 0.3

overlap karna hi hoga.

Picture se complement

A

aur

A^{'}

kuch share nahi karte aur

S

bhar dete hain, toh

P (A^{'}) = 1 - P (A)

Connections

Mutually exclusive events — addition rule
Probability — basic definitions & sample space
General addition rule & inclusion–exclusion
Venn diagrams in probability
Independent events — multiplication rule
Complementary events