Question bank — Probability — classical, empirical, axiomatic (Kolmogorov axioms)

False — jahan $n$ total outcomes hain aur $m$ favourable ones, formula ko yeh bhi chahiye ki $n$ outcomes equally likely hon; ek bent coin mein $n=2$ hota hai lekin dono halves symmetric nahi hain, isliye $m/n$ wahan meaningless hai. (Set Theory — Union, Intersection, Complement mein, $A$ sirf box $S$ ka ek subset hai; uske elements count karna tabhi probability deta hai jab har element equally weighted ho.)

True — disjoint events ke liye axiom 3 deta hai $P(A\cup B)=P(A)+P(B)\le P(S)=1$, isliye 1 se zyada hona force karta hai $A\cap B\ne \varnothing$ (double-counted overlap hi sum ko 1 se upar push karta hai).

False — $P=0$ nonempty events se attach ho sakta hai. Yeh kisi bhi continuous dartboard par exact point ke liye hota hai, lekin saath hi ek countably-infinite space mein non-uniform assignment ke saath bhi: kisi specific outcome ko weight 0 diya ja sakta hai jabki woh ek possible result bana rehta hai. Axioms $P(A)=0$ ko bina $A=\varnothing$ ke allow karte hain.

False — short runs noisy hote hain; Law of Large Numbers (relative frequency $f/N$ ka true value par converge karna) sirf $N\to \infty$ par settling ka promise karta hai, isliye zyada trials ka matlab hai tighter, zyada reliable estimate.

True — $A$ aur uska complement $A^c$ ($A$ ke blob ke bahar ki white region) disjoint hain aur $A\cup A^c=S$, isliye axiom 3 deta hai $P(A)+P(A^c)=P(S)$, aur axiom 2 right side ko 1 bana deta hai.

True — $A$ ka blob puri tarah $B$ ke andar baitha hai; $B=A\cup (B\cap A^c)$ ko disjoint pieces mein likho, isliye $P(B)=P(A)+P(B\cap A^c)$, aur kyunki extra piece ki probability $\ge 0$ hai (axiom 1), $P(B)\ge P(A)$.

False — yeh derived hai $A\cup B$ ko disjoint pieces $A$ aur $B\cap A^c$ mein split karke aur axiom 3 apply karke; kuch naya assume nahi kiya gaya (neeche "Why questions" mein poore steps hain).

False infinite spaces ke liye — jaise $P=0$ case mein, ek event ki $P=1$ ho sakti hai aur phir bhi occasionally miss ho sakta hai (uska complement ek nonempty $P=0$ set hai); finite sample spaces mein hi yeh har trial par guaranteed hai.

True — axiom poori sequence $A_1,A_2,\dots$ ke liye stated hai, yahi wajah hai ki $\sigma$-algebra ko countable unions ke under closed rehna zaroori hai: union ${\bigcup }_i{A}_i$ khud ek event hona chahiye taaki $P$ uska define ho sake.

King of Hearts dono mein count hota hai — 4 kings mein bhi aur 13 hearts mein bhi (yeh $A\cap B$ overlap mein rehta hai), isliye tumhe $P(K\cap H)=\frac{1}{52}$ subtract karna hoga: answer hai $\frac{16}{52}=\frac{4}{13}$.

Yeh gambler's fallacy hai; independent tosses ki koi memory nahi hoti, isliye $P(H)=0.5$ history se regardless. Law of Large Numbers cheezein balance karta hai early results ko kaafi naye results se swamp karke, compensate karke nahi.

Tabhi valid hai jab teeno pairwise disjoint hon. Overlaps ke saath, teen singles add karne par har pairwise overlap do baar count hota hai, isliye $P(A\cap B)$, $P(A\cap C)$, $P(B\cap C)$ subtract karo; lekin yeh subtraction triple region $A\cap B\cap C$ ko teen baar remove karta hai teen baar add karne ke baad (net zero), isliye use ek baar add back karo: $P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C)$.

11 sums equally likely nahi hain, isliye unpar classical counting invalid hai. Inki jagah 36 equally-likely ordered pairs count karo: $P(7)=\frac{6}{36}=\frac{1}{6}$.

Yeh silently classical symmetry apply kar raha hai; ek tack mein apne do states ke beech koi symmetry nahi hoti, isliye tumhe frequency measure karni padegi (jaise $290/500=0.58$), equal halves assume nahi kar sakte.

Non-negativity positive intersection force nahi karta; disjoint events (blobs jo touch nahi karte) ka $P(A\cap B)=0$ hota hai even jab dono events individually likely hon.

Complement rule poore event $A\cup B$ ko use karta hai, isliye yeh hai $1-P(A\cup B)=1-P(A)-P(B)+P(A\cap B)$ — overlap term survive karta hai.

Classical ko ek symmetry argument chahiye aur empirical ko infinite trials; teen axioms ki jagah koi bhi minimum rules state karte hain jo kisi bhi probability ko follow karni chahiye, isliye dono purani methods ek rigorous framework ke andar numbers assign karne ke valid tarike ban jaate hain.

Derivation har $n$ outcomes ko weight $\frac{1}{n}$ assign karta hai taaki woh 1 mein sum ho jayein; agar woh equally likely nahi hain, to koi single value $\frac{1}{n}$ correct nahi hogi aur $P(A)=m\cdot \frac{1}{n}$ collapse ho jaata hai.

Kyunki "and/or/not" exactly intersection/union/complement se map hote hain — jaise "$A$ or $B$" Venn box mein union blob $A\cup B$ hai (dekho Set Theory — Union, Intersection, Complement) — logical reasoning ko computable set algebra mein convert karte hue.

"At least one" kaafi cases mein split ho jaata hai, lekin uska complement "none" usually ek single case hota hai; $1-P(\text{none})$ compute karna ek lambi sum ki jagah ek term se replace kar deta hai.

Union ko non-overlapping pieces mein split karo: $A\cup B=A\cup (B\cap A^c)$, disjoint, isliye axiom 3 deta hai $P(A\cup B)=P(A)+P(B\cap A^c)$. Alag se $B=(B\cap A)\cup (B\cap A^c)$, disjoint, isliye $P(B\cap A^c)=P(B)-P(A\cap B)$. Substitute karne par milta hai $P(A)+P(B)-P(A\cap B)$; geometrically, $P(A)+P(B)$ overlap ko do baar shade karta hai, isliye ek copy remove ki jaati hai.

Ek single relative frequency $f/N$ sirf ek estimate hai; LLN guarantee karta hai ki yeh estimate ek fixed true value par converge karta hai jab $N\to \infty$, yahi cheez humein use probability call karne deti hai.

Kyunki yeh follow karta hai: $S=S\cup \varnothing$ disjoint deta hai $P(S)=P(S)+P(\varnothing )$, isliye $P(\varnothing )=0$ — koi alag assumption nahi chahiye.

Agar do blobs overlap karte hain, unke outcomes share hote hain, aur $P(A)+P(B)$ add karna us shared region ko double-count karta hai; sirf disjoint blobs union ko cleanly partition karte hain taaki probabilities simply sum ho jayein.

$P(S)=1$ axiom 2 (normalisation) se — yeh statement ki "box mein se koi na koi outcome occur hoga" certain hai.

Haan — unka overlap hona zaroori hai, kyunki disjointness force karti $\frac{3}{4}+\frac{3}{4}=1.5>1$; unke intersection ki probability kam se kam $\frac{3}{4}+\frac{3}{4}-1=\frac{1}{2}$ hai.

Generally nahi — Independence of Events ka matlab hai $P(A\cap B)=P(A)P(B)$ (overlap product hai), jabki mutually exclusive ka matlab hai $P(A\cap B)=0$ (koi overlap nahi); dono saath tab hold hote hain jab kisi ek event ki probability already 0 ho.

Ek pointer spin karo jo circle par uniformly land karta hai. Event "pointer 12 o'clock mark par exactly land nahi karta" ki probability 1 hai (woh single mark ek $P=0$ point hai), phir bhi kisi spin par woh wahan land ho sakta hai — isliye ek $P=1$ event "occasionally miss" ho sakta hai kyunki uska exception ek nonempty $P=0$ set hai.

Primes hain $\{2,3,5\}$, isliye $P=\frac{3}{6}=\frac{1}{2}$; classical apply hoti hai kyunki ek symmetric die ke saare cheh faces equally likely hain, dete hain $n=6$, $m=3$.

Sirf empirical (frequentist) wali — tum $f/N$ kaafi trials mein measure karte ho, kyunki classical counting ke liye equal-likelihood chahiye jo tum justify nahi kar sakte.

Outcomes $\{H,TH,TTH,\dots \}$ ek infinite disjoint list banate hain; countable additivity humein unki individual probabilities $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots =1$ sum karne deti hai, $P(S)=1$ confirm karte hue — sirf do-event rule se yeh impossible hai.