Mutually exclusive events — addition rule: Question bank

2.7.6 · D5 · Maths › Statistics & Probability — Intermediate › Mutually exclusive events — addition rule

Shuru karne se pehle, woh exact symbols jinhe hum har jagah use karte hain (pehle samjhe, tabhi use kiye):

Neeche diya figure teeno ek nazar mein dikhata hai — jab bhi koi symbol aaye, ise dobara dekho.

Figure — Mutually exclusive events — addition rule

Mutually exclusive = do events jinke circles kabhi touch nahi karte, isliye ek hi trial mein dono nahi ho sakte ( $A \cap B = \emptyset$ ). Overlap = woh lens $A \cap B$ ; addition rule ka poora drama isi baat par hai ki woh lens empty hai ya nahi.

True or false — justify

Do events ek saath mutually exclusive bhi ho sakte hain aur unki probabilities ka sum 1 se upar bhi ja sakta hai.

False — agar

A \cap B = \emptyset

hai toh

P (A \cup B) = P (A) + P (B)

, jo 1 se zyada nahi ho sakta, isliye

P (A) + P (B) \leq 1

zaroor hoga.

Agar

P (A) + P (B) = 1

hai, toh events mutually exclusive zaroor honge.

False — sum ka 1 hona overlap ke baare mein kuch nahi batata; jaise

P (A) = 0.6, P (B) = 0.4

phir bhi

0.3

tak overlap kar sakte hain agar

P (A \cup B) = 0.7

ho. Sum-to-1 fact exclusivity ke liye na zaroori hai, na kaafi.

Mutually exclusive events hamesha independent hote hain.

False — yeh toh almost ulte hain: exclusivity ka matlab hai

P (A \cap B) = 0

, lekin independence ke liye chahiye

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

, jo

0

tabhi hoga jab kisi ek event ki probability zero ho.

Jo do events positive probability wale hain aur mutually exclusive hain, woh kabhi independent nahi hote.

True — independence ke liye zaroori hoga

P (A) P (B) = P (A \cap B) = 0

, jo dono probabilities positive hone par impossible hai, isliye

A

ka hona

B

ko force karta hai nahi hone ke liye (strong dependence).

General addition rule

P (A \cup B) = P (A) + P (B) - P (A \cap B)

sirf tab kaam karta hai jab events overlap karte hain.

False — yeh hamesha valid hai; disjoint events ke liye subtract hone wala term bas

0

hota hai, toh yeh bina kisi special case ke simple addition mein collapse ho jaata hai.

Ek event aur uska complement

A^{'}

(jo kuch bhi

A

mein nahi hai) mutually exclusive hote hain.

True — koi bhi outcome ek saath "

A

mein" aur "

A

mein nahi" nahi ho sakta, isliye

A \cap A^{'} = \emptyset

; yeh milkar poore sample space

S

ko bhi cover karte hain.

Agar

A

aur

B

mutually exclusive hain, toh unke complements

A^{'}

aur

B^{'}

bhi mutually exclusive hote hain.

False — disjoint

A, B

dono ke bahar bhi jagah hoti hai; woh bahar wale outcomes dono

A^{'}

aur

B^{'}

mein aate hain, isliye

A^{'} \cap B^{'}

aksar non-empty hota hai.

Teen events ka pairwise mutually exclusive hona matlab ek trial mein zyada se zyada ek hi ho sakta hai.

True — pairwise disjoint ka matlab hai koi bhi do events koi outcome share nahi karte, isliye koi bhi single outcome teen mein se zyada se zyada ek mein hoga (ya phir teeno se bahar bhi ho sakta hai).

Spot the error

"Ek die par

P (even OR > 3)

= \frac{3}{6} + \frac{3}{6} = 1

hai, toh yeh certain hai."

Error hai overlap

{4, 6}

ko ignore karna; sahi answer hai

\frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{2}{3}

, kyunki 4 aur 6 do baar count ho gaye the.

P (A) = 0.6

P (B) = 0.7

, toh

P (A \cup B) = 1.3

Probability kabhi 1 se zyada nahi ho sakti, toh yeh signal hai ki events overlap karte hain; true union zyada se zyada 1 hai, aur general rule rearrange karne par overlap milta hai

P (A \cap B) = P (A) + P (B) - P (A \cup B) \geq 0.6 + 0.7 - 1 = 0.3

"Dono unrelated events ke baare mein hain, toh mutually exclusive aur independent ek hi baat hai."

"Unrelated" ek feeling hai, definition nahi; exclusivity unions ke baare mein hai (

P (A \cap B) = 0

) jabki independence products ke baare mein hai (

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

) — bilkul alag equations hain.

A

aur

B

dono ek saath nahi ho sakte, toh

P (A \cap B) = P (A) \cdot P (B)

Exclusive events ke liye

P (A \cap B) = 0

hota hai, product nahi; multiplication rule independent events ke liye hai, jo bilkul alag situation hai.

"Kisi bhi do events ke liye,

P (A \cup B) = P (A) + P (B)

Yeh overlap term ko bina check kiye drop kar deta hai; yeh sirf tab valid hai jab confirm ho ki

A \cap B = \emptyset

hai, warna shared outcomes double-count ho jaate hain.

"Kyunki

A \cap B = \emptyset

hai, main safe rehne ke liye

P (A \cap B)

subtract kar lunga — lekin maine ise

0.1

estimate kiya."

Agar woh sach mein disjoint hain toh overlap exactly

0

hai, toh

0.1

guess karna ek aisa overlap invent karna hai jo hai hi nahi aur ek galat, chhota answer deta hai.

"King ya Queen draw karna — ek card dono ho sakta hai, toh overlap subtract karo."

Koi bhi ek card ek saath King aur Queen nahi ho sakta, isliye yeh events exclusive hain aur overlap

0

hai; seedha

\frac{4}{52} + \frac{4}{52}

add karna sahi hai.

Why questions

Hum general rule mein overlap ko subtract kyun karte hain, add kyun nahi?

Jab hum

P (A) + P (B)

calculate karte hain, toh lens

A \cap B

mein har outcome

A

ke liye ek baar aur

B

ke liye ek baar count hota hai — yaani do baar — isliye use ek baar subtract karna ek honest single count restore karta hai.

Mutual exclusivity se subtraction term kyun vanish ho jaata hai, na ki poora rule change ho jaata hai?

Rule kabhi nahi badlta; exclusivity bas

A \cap B = \emptyset

set karti hai toh

P (A \cap B) = 0

ho jaata hai, matlab literally kuch bhi double-counted nahi jo hatana ho.

Positive-probability wale do mutually exclusive events independent kyun nahi ho sakte, intuitively?

Agar

A

ka hona

B

ko forbid karta hai, toh

A

hone ka pata chalne par

B

ki chance

0

ho jaati hai — yeh bahut badi information hai, bilkul ulta "kuch nahi batata" ke, jo independence demand karta hai.

Complement rule

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

actually addition rule ka disguise kyun hai?

A

aur

A^{'}

exclusive hain aur milkar

S

ko cover karte hain, toh

P (A) + P (A^{'}) = P (S) = 1

, jo seedha rearrange hokar

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

deta hai.

"OR matlab add karo" ek dangerous default kyun hai, safe nahi?

Yeh quietly assume karta hai ki koi overlap nahi, jo clean single-die examples mein sach hai lekin jab events ek saath ho sakti hain tab false ho jaata hai, aur silently answer inflate ho jaata hai.

Addition rule kaafi saare events ke liye

\sum_{i} P (A_{i})

tab hi kyun extend hota hai jab woh pairwise disjoint hon?

Pairwise disjointness guarantee karti hai ki koi bhi outcome kisi bhi do events mein share nahi hai, isliye jab tum sab pieces add karte ho toh koi bhi outcome kabhi ek se zyada baar count nahi hota.

Teen overlapping events ke liye inclusion–exclusion subtract karne ke baad ek term dobara add kyun karta hai?

Tum pehle har pairwise overlap subtract karte ho, lekin teeno mein aane wala ek outcome (

A \cap B \cap C

) ek baar zyada subtract ho jaata hai, isliye

P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap B) - P (A \cap C) - P (B \cap C) + P (A \cap B \cap C)

us triple lens ko ek baar wapas add karta hai — neeche figure dekho.

Edge cases

Agar

B = \emptyset

(impossible event) hai, toh kya

A

aur

B

mutually exclusive hain?

Haan — empty set kisi bhi cheez ke saath koi outcome share nahi karta, isliye

A \cap B = \emptyset

trivially ho jaata hai, aur

P (A \cup B) = P (A) + 0 = P (A)

expected tarah se.

Agar

A \subseteq B

hai (

A

ke har outcome

B

mein bhi hain), toh kya woh mutually exclusive ho sakte hain?

Sirf tab agar

A = \emptyset

ho; warna

A

ke outcomes

B

ke andar hain, isliye

A \cap B = A \neq = \emptyset

hai aur woh completely overlap karte hain.

Kya koi event khud se mutually exclusive ho sakta hai?

Sirf impossible event

\emptyset

, kyunki

A \cap A = A

hai, jo empty tabhi hoga jab

A

khud empty ho.

Agar

P (A) = 0

hai, toh kya

A

har doosre event se mutually exclusive hai?

Set-theoretically zaroor nahi —

A

phir bhi

B

ke saath outcomes share kar sakta hai, isliye

A \cap B

empty nahi bhi ho sakta. Lekin probabilistically: kyunki

A \cap B \subseteq A

hai, hume milta hai

P (A \cap B) \leq P (A) = 0

, jo force karta hai

P (A \cap B) = 0

; isliye

P (A \cup B) = P (A) + P (B) - 0 = P (B)

, toh

A

exclusive event ki tarah add hota hai chahe sets overlap karein. Sabak yeh hai: "disjoint sets" aur "zero-overlap probability" alag claims hain, aur yahan sirf doosra valid hai.

Exhaustive mutually exclusive events (ek partition) ki probabilities ka sum kya hona chahiye?

Exactly 1, kyunki poore sample space ko cover karne wale disjoint pieces bina overlap ke

P (S) = 1

tak add ho jaate hain.

Agar do events sample space ko cover karte hain lekin overlap karte hain, toh kya unki probabilities phir bhi 1 sum hoti hain?

Nahi — woh exactly overlap

P (A \cap B)

se zyada sum hoti hain, kyunki

P (A) + P (B) = P (A \cup B) + P (A \cap B) = 1 + P (A \cap B)

Kya teen events pairwise mutually exclusive ho sakte hain phir bhi unka union

S

ko cover na kare?

Haan — disjointness sirf overlap forbid karta hai, coverage ke baare mein kuch nahi kehta; teeno ke bahar wale outcomes simply kisi mein nahi hote.

Connections

Mutually exclusive events — addition rule
General addition rule & inclusion–exclusion
Independent events — multiplication rule
Complementary events
Venn diagrams in probability
Conditional probability