Probability space — sample space Ω, sigma-algebra F, measure P — Kolmogorov axioms: Worked examples

4.9.1 · D3 · Maths › Probability Theory & Statistics › Probability space — sample space Ω, sigma-algebra F, measure

Formula Do consequences jo hum baar baar use karenge (A1–A3 se prove kiye hain yahan)

Yeh extra axioms nahi hain — yeh A1–A3 se follow karte hain, aur hum inhe ek baar prove karte hain taaki page akela khada reh sake.

(T0) Impossible event ki probability $0$ hai: $P (\emptyset) = 0$ . Disjoint sequence $A_{1} = Ω, A_{2} = A_{3} = \dots = \emptyset$ lo; inki union $Ω$ hai. A3 se, $P (Ω) = P (Ω) + \sum_{i \geq 2} P (\emptyset)$ , toh $\sum_{i \geq 2} P (\emptyset) = 0$ . Har term $\geq 0$ hai by A1, toh har term $0$ honi chahiye. Isliye $P (\emptyset) = 0$ . $■$

(T1) Continuity from above: agar $A_{1} \supseteq A_{2} \supseteq \dots$ shrink karke $A = ⋂_{n} A_{n}$ pe aate hain, toh $P (A_{n}) \to P (A)$ . (T2) Continuity from below: agar $B_{1} \subseteq B_{2} \subseteq \dots$ grow karke $B = ⋃_{n} B_{n}$ tak jaate hain, toh $P (B_{n}) \to P (B)$ . Yeh kyun hold karte hain: dono A3 ke packaged forms hain. (T2) ke liye, growing sequence ko disjoint "rings" $B_{1}, (B_{2} ∖ B_{1}), (B_{3} ∖ B_{2}), \dots$ ki tarah likho; A3 $P (B)$ ko ring-probabilities ki infinite sum mein badal deta hai, aur pehle $n$ rings ka partial sum exactly $P (B_{n})$ hai — toh partial sums (jo infinite sum ki definition se total tak converge karte hain) hi hain $P (B_{n}) \to P (B)$ . (T1), (T2) se complements lekar follow karta hai. Isliye A3 mein "countable" matter karta hai — yahi toh probability ko infinite limits ke through pass hone deta hai.

Scenario matrix

Har probability-space question inhi cells mein se kisi ek mein aata hai. Har worked example neeche uss cell ko tag karta hai jo woh hit karta hai.

Cell	Scenario class	Tricky kya hai	Example
C1	Finite space, disjoint events	Plain additivity (A3)	Ex 1
C2	Finite space, overlapping events	Overlap subtract karna padega (inclusion–exclusion)	Ex 2
C3	Complement shortcut	"At least one" $1 - P (none)$ se aasaan	Ex 3
C4	Degenerate events: $\emptyset$ aur $Ω$	Certain aur impossible ke edges	Ex 4
C4′	Finite space, non-uniform atoms	Atomic weights $\sum_{ω \in A} P ({ω})$ add karo, counts nahi	Ex 4b
C5	Continuous space, intervals	Length = probability, single points ki length $0$ hai	Ex 5
C6	Limiting sequence (shrinking aur growing)	Continuity of $P$ from above aur below	Ex 6
C7	Real-world word problem	English → $Ω, F, P$ mein translate karo	Ex 7
C8	Exam twist: teen overlapping events	General inclusion–exclusion, sign bookkeeping	Ex 8

Shuru karne se pehle, ek symbol jo hum bar bar use karte hain:

Ex 1 — Finite space, disjoint events [C1]

Worked example Do cards, koi overlap nahi

Ek standard $52$ -card deck se ek card draw ki jaati hai; har card equally likely hai, toh $Ω$ mein $52$ outcomes hain aur $P (A) = ∣ A ∣/52$ . Maan lo $A = {card ek King hai}$ aur $B = {card ek Queen hai}$ . $P (A \cup B)$ nikalo = probability ki card ek King ya Queen hai.

Forecast: Kya aapko lagta hai yahan kuch subtract karna padega? Ek fraction guess karo.

Step 1 — Har event count karo. $4$ Kings hain aur $4$ Queens hain, toh $∣ A ∣ = 4$ , $∣ B ∣ = 4$ , jisse $P (A) = \frac{4}{52}$ aur $P (B) = \frac{4}{52}$ milte hain. Yeh step kyun? Equally-likely finite space pe, probability sirf counting hai.

Step 2 — Disjointness check karo. Ek single card ek saath King aur Queen dono nahi ho sakti, toh $A \cap B = \emptyset$ . Yeh cell C1 hai: disjoint. Yeh step kyun? Sirf jab $A \cap B = \emptyset$ ho tabhi hum plain additivity use kar sakte hain — warna double-counting ho jaati hai.

Step 3 — Add karo (finite additivity, A3 se derived). $P (A \cup B) = P (A) + P (B) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13} .$ Yeh step kyun? Disjoint pieces ki probabilities add hoti hain — yeh A3 ka empties-se-padded version hai.

Verify: $A \cup B = {8 face-cards jo K ya Q hain}$ , toh $∣ A \cup B ∣ = 8$ aur $8/52 = 2/13 \approx 0.1538$ . ✓ Kuch subtract nahi hua, jo hamari disjoint forecast se match karta hai.

Ex 2 — Finite space, overlapping events [C2]

Worked example Die pe Even YA greater-than-3 (classic overlap)

Ek fair die roll karo: $Ω = {1, 2, 3, 4, 5, 6}$ , $P (A) = ∣ A ∣/6$ . Maan lo $A = {even} = {2, 4, 6}$ aur $B = {> 3} = {4, 5, 6}$ . $P (A \cup B)$ nikalo.

Forecast: Agar aapne naively $\frac{3}{6} + \frac{3}{6} = 1$ kiya, toh aap kehte ki event certain hai. Kya yeh sach hai? (Outcome $1$ na even hai na $> 3$ ...) Toh naive answer galat hona chahiye. Sahi value guess karo.

Step 1 — Overlap nikalo. $A \cap B = {4, 6}$ , toh $P (A \cap B) = \frac{2}{6}$ . Yeh step kyun? $4$ aur $6$ ek baar $A$ mein count hote hain aur ek baar $B$ mein — inclusion–exclusion exactly isi double count ko remove karne ke liye hai.

Step 2 — Inclusion–exclusion apply karo. $P (A \cup B) = P (A) + P (B) - P (A \cap B) = \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} = \frac{2}{3} .$ Yeh step kyun? $P (A \cap B)$ subtract karna overcount ko correct karta hai.

Verify: $A \cup B = {2, 4, 5, 6}$ directly, $∣ A \cup B ∣ = 4$ , toh $4/6 = 2/3 \approx 0.667$ . ✓ $1$ nahi — outcome $1$ correctly exclude hua.

Ex 8 mein extend hue general pattern ke liye Inclusion–exclusion principle dekho.

Ex 3 — Complement shortcut [C3]

Worked example Chaar rolls mein "at least one six"

Ek fair die chaar baar independently roll karo. Maan lo $E = {kam se kam ek roll mein 6 aaya}$ . $P (E)$ nikalo.

Forecast: Directly, "at least one" exactly-one, exactly-two, exactly-three, exactly-four mein split hoti hai — bahut messy. Guess karo ki answer $\frac{1}{2}$ se upar hai ya neeche.

Step 1 — Event ka complement lo. "At least one six" iska ulta hai "koi six nahi kisi bhi roll mein". Toh $E^{c} = {chaaon rolls mein koi six nahi}$ . Yeh step kyun? Complement rule $P (E) = 1 - P (E^{c})$ (A2 + additivity se) ek 4-way sum ko ek single easy product mein badal deta hai.

Step 2 — Ek roll mein no six ki probability. Paanch faces "not six" hain, toh $P (no six, one roll) = \frac{5}{6}$ .

Step 3 — Chaaon rolls mein six se bacho. Rolls independent hain (upar define kiya tha: independent events ki joint probability multiply hoti hai), toh $P (E^{c}) = \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} = (\frac{5}{6})^{4} = \frac{625}{1296} .$ Yeh step kyun? Kyunki $P (chaaon rolls mein no six) = P (roll 1 no six) \dots P (roll 4 no six)$ mutual independence se — isse hum outcomes enumerate karne se bach jaate hain.

Step 4 — Complement wapas lo. $P (E) = 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.518.$

Verify: $625 + 671 = 1296$ ✓ (complementary events ki probabilities $1$ tak add hoti hain). Value $\approx 0.518 > \frac{1}{2}$ , intuition se match karta hai ki chaar chances ek six ko more-likely-than-not banate hain.

Ex 4 — Degenerate events: $\emptyset$ aur $Ω$ [C4]

Worked example Impossible aur certain

Die space $Ω = {1, \dots, 6}$ pe evaluate karo: (a) $P (roll a 7)$ , (b) $P (roll \leq 6)$ , (c) $P (roll is 2 and 5)$ .

Forecast: Inme se kaun sa $0$ hai, kaun sa $1$ ? Padhne se pehle teeno guess karo.

Step 1 — (a) 7 roll karo. Koi face $7$ ke barabar nahi, toh yeh event $\emptyset$ hai. Upar prove kiye theorem (T0) se (A1 + A3 se), $P (\emptyset) = 0$ . Yeh step kyun? $\emptyset$ mein koi outcome nahi; (T0) iske probability ko $0$ force karta hai.

Step 2 — (b) $\leq 6$ roll karo. Har face $\leq 6$ satisfy karta hai, toh yeh event poora space $Ω$ hai. Axiom A2 se, $P (Ω) = 1$ . Yeh step kyun? A2 total mass ko $1$ define karta hai: kuch na kuch toh hoga.

Step 3 — (c) Roll $2$ AND $5$ hai. Ek single roll do numbers ek saath nahi ho sakti, toh ${2} \cap {5} = \emptyset$ , isliye probability $0$ by (T0). Yeh step kyun? "AND" intersection hai; disjoint singletons ka intersection impossible event hai.

Verify: $P (\emptyset) = 0/6 = 0$ ✓, $P (Ω) = 6/6 = 1$ ✓, aur ${2} \cap {5} = \emptyset \Rightarrow 0$ ✓. Yeh range $[0, 1]$ ke do edges hain.

Ex 4b — Finite space with non-uniform atoms [C4′]

Worked example Ek loaded (biased) die

Ek die aise weighted hai ki uske chhah faces equally likely nahi hain. Atomic probabilities — har single outcome $ω$ ki probability, likhi jaati hai $P ({ω})$ — yeh hain: $P ({1}) = 0.1, P ({2}) = 0.1, P ({3}) = 0.1, P ({4}) = 0.2, P ({5}) = 0.2, P ({6}) = 0.3.$ Nikalo (a) $P (Ω)$ sanity check ke liye, (b) $P (A)$ jahan $A = {even} = {2, 4, 6}$ , (c) $P (B)$ jahan $B = {> 3} = {4, 5, 6}$ , (d) $P (A \cup B)$ .

Forecast: Counting formula $∣ A ∣/6$ ab galat hai — kyun? Guess karo ki $P (even)$ $\frac{1}{2}$ se bada hai ya chhota, given ki face $6$ heavy hai.

Step 1 — Check karo ki atoms 1 tak add hote hain. $0.1 + 0.1 + 0.1 + 0.2 + 0.2 + 0.3 = 1.0 = P (Ω)$ . Yeh step kyun? Valid measure ke liye A2 chahiye: $Ω$ pe total mass exactly $1$ honi chahiye. Agar atoms $1$ tak add nahi karte, toh yeh probability measure nahi hai.

Step 2 — Ek event ki probability = uske atoms ka sum. Kisi bhi $A$ ke liye, kyunki singletons ${ω}$ disjoint hain aur unki union $A$ hai, additivity (A3) deta hai $P (A) = \sum_{ω \in A} P ({ω}) .$ Yeh step kyun? Yeh general discrete rule hai; uniform formula $∣ A ∣/∣Ω∣$ iska special case hai jab har atom $1/∣Ω∣$ ke barabar ho.

Step 3 — (b) Even. $P (A) = P ({2}) + P ({4}) + P ({6}) = 0.1 + 0.2 + 0.3 = 0.6.$

Step 4 — (c) 3 se bada. $P (B) = P ({4}) + P ({5}) + P ({6}) = 0.2 + 0.2 + 0.3 = 0.7.$

Step 5 — (d) Union via inclusion–exclusion. Overlap $A \cap B = {4, 6}$ , toh $P (A \cap B) = 0.2 + 0.3 = 0.5$ . Phir $P (A \cup B) = 0.6 + 0.7 - 0.5 = 0.8.$ Yeh step kyun? Inclusion–exclusion kisi bhi measure ke baare mein ek theorem hai, uniform ho ya na ho — isne equal weights assume nahi kiye the.

Verify: Directly, $A \cup B = {2, 4, 5, 6}$ , toh $P = 0.1 + 0.2 + 0.2 + 0.3 = 0.8$ . ✓ Aur $P (even) = 0.6 > \frac{1}{2}$ kyunki heavy faces $4, 6$ even hain — forecast se match karta hai.

Ex 5 — Continuous space, intervals [C5]

Figure — Probability space — sample space Ω, sigma-algebra F, measure P — Kolmogorov axioms

Figure s01 — kya dekh rahe ho: grey bar poori stick $Ω = [0, 1]$ hai. Do green blocks intervals $[0.2, 0.5]$ (width $0.3$ ) aur $[0.7, 0.9]$ (width $0.2$ ) hain; unki widths hi unki probabilities hain. $0.4$ pe single red dot point event ${0.4}$ mark karta hai — iska zero width hai, toh probability $0$ . Neeche ke tick labels endpoints $0, 0.2, 0.4, 0.5, 0.7, 0.9, 1$ dikhate hain.

Step 1 — (a) Ek interval ki length. $P ([0.2, 0.5]) = 0.5 - 0.2 = 0.3$ . Yeh step kyun? Uniform model mein, probability hi length hai — yahi is $P$ ki definition hai.

Step 2 — (b) Ek single point. $P ({0.4}) = 0$ : ek point $0.4 - 0.4 = 0$ length ka interval hai. Yeh step kyun? Likho ${0.4} = ⋂_{n} [0.4, 0.4 + \frac{1}{n}]$ . Yeh intervals shrink karte hain, toh continuity from above (T1) se jo page ke top pe prove ki thi, $P ({0.4}) = lim_{n} P ([0.4, 0.4 + \frac{1}{n}]) = lim_{n} \frac{1}{n} = 0$ . Phir bhi $0.4$ ho sakta hai — probability $0 \neq =$ impossible.

Step 3 — (c) Do disjoint intervals. Yeh overlap nahi karte ( $0.5 < 0.7$ ), toh lengths add karo: $P = 0.3 (0.5 - 0.2) + 0.2 (0.9 - 0.7) = 0.5.$ Yeh step kyun? Disjoint events add hote hain (finite additivity), bilkul finite case ki tarah — machinery identical hai.

Verify: $0.3 \leq 1$ ✓, single point $0$ ✓, aur $0.3 + 0.2 = 0.5 \leq 1$ ✓. Do green bars milke stick ka aadha span karte hain.

Ex 6 — Limiting sequences: shrinking aur growing [C6]

Figure s02 — kya dekh rahe ho: upar red nested boxes shrinking $A_{n} = [0, 1/ n]$ hain $n = 1, 2, 4, 8$ ke liye; har ek half-transparent hai, aur tum dekh sakte ho ki woh $0$ pe black dot ke upar band ho rahe hain (woh dot $⋂_{n} A_{n} = {0}$ hai). Neeche blue nested boxes growing $B_{n} = [0, 1 - 1/ n]$ hain; har ek pichle se lamba hai, $1$ ke open end ki taraf creep karta hua (unka union $[0, 1)$ hai). Har box apni probability $P (A_{n}) = 1/ n$ ya $P (B_{n}) = 1 - 1/ n$ label ke saath hai.

Step 1 — (a) shrinking, number limit. $P (A_{n}) = \frac{1}{n} \to 0$ . Yeh step kyun? Har box ki length; numbers $0$ ki taraf jaate hain.

Step 2 — (a) shrinking, set limit. Har $A_{n}$ mein $0$ hai; koi bhi $x > 0$ bahar nikal jaata hai jab $\frac{1}{n} < x$ . Toh $⋂_{n} A_{n} = {0}$ aur $P ({0}) = 0$ . Yeh step kyun? Hum set limit ko number limit se compare karte hain. Continuity from above (T1): kyunki $A_{n} \supseteq A_{n + 1}$ shrink karke ${0}$ tak aate hain, $P (A_{n}) \to P ({0})$ . Dono routes $0$ dete hain. ✓

Step 3 — (b) growing, number limit. $P (B_{n}) = 1 - \frac{1}{n} \to 1$ . Yeh step kyun? Har growing box ki length poori stick ki length ki taraf jaati hai jab $\frac{1}{n} \to 0$ .

Step 4 — (b) growing, set limit. Har point $x < 1$ eventually andar aa jaata hai jab $1 - \frac{1}{n} > x$ ; sirf $x = 1$ kabhi included nahi hota (koi $n$ $1 - \frac{1}{n}$ ko $1$ tak nahi pahunchata). Toh $⋃_{n} B_{n} = [0, 1)$ , aur $P ([0, 1)) = 1 - 0 = 1$ (half-open interval ki length). Yeh step kyun? Hum phir se set limit ko number limit se compare karte hain.

Step 5 — (b) continuity from below se reconcile karo. Kyunki $B_{n} \subseteq B_{n + 1}$ grow karke $B = [0, 1)$ tak jaate hain, continuity from below (T2) kehta hai $lim_{n} P (B_{n}) = P (B)$ . Left side hai $lim_{n} (1 - \frac{1}{n}) = 1$ ; right side hai $P ([0, 1)) = 1$ . Dono agree karte hain. ✓

Step 6 — dono directions kaam kyun karte hain. (T1) aur (T2) dono countable additivity (A3) se page ke top pe derive kiye gaye the — "countable" exactly wahi hai jo probability ko infinite limits ke through pass hone deta hai. Finite additivity akele dono nahi kar sakti.

Verify: (a) $\frac{1}{n} \to 0$ aur $P ({0}) = 0$ agree karte hain. (b) $1 - \frac{1}{n} \to 1$ aur $P ([0, 1)) = 1$ agree karte hain. ✓

Ex 7 — Real-world word problem [C7]

Worked example Newspaper subscriptions (

$Ω, F, P$ English se build karo) Ek town mein, $60%$ paper $A$ padhte hain, $40%$ paper $B$ padhte hain, aur $25%$ dono padhte hain. Ek resident uniformly choose karo. Probability nikalo ki resident (a) kam se kam ek paper padhe, (b) koi bhi paper na padhe, (c) exactly ek paper padhe.

Forecast: (b) compute karne se pehle guess karo — "neither" $\frac{1}{4}$ se bada hai ya chhota?

Step 1 — Space set up karo. $Ω =$ sab residents, $P (X) =$ residents ka fraction jo $X$ mein hain. Translate karo: $P (A) = 0.60$ , $P (B) = 0.40$ , $P (A \cap B) = 0.25$ . Yeh step kyun? Word problem disguise mein probability space hai; $Ω, F, P$ naam dena axioms ko usable banata hai. ( $Ω$ finite hai, toh $F = 2^{Ω}$ — residents ka har group ek event hai.)

Step 2 — (a) At least one = union (inclusion–exclusion). $P (A \cup B) = 0.60 + 0.40 - 0.25 = 0.75.$ Yeh step kyun? "Both" woh overlap hai jo subtract karna hai, cell C2 logic reused.

Step 3 — (b) Neither = complement of union. $P ((A \cup B)^{c}) = 1 - 0.75 = 0.25.$ Yeh step kyun? Complement rule; "neither" "at least one" ka opposite hai.

Step 4 — (c) Exactly one — pehle set-theoretically derive karo. "Exactly ek paper padhe" matlab " $A$ mein hai but $B$ mein nahi, ya $B$ mein hai but $A$ mein nahi": ${exactly one} = (A ∖ B) ⊔ (B ∖ A),$ ek disjoint union (dono halves mein koi resident common nahi). Additivity se, $P (exactly one) = P (A ∖ B) + P (B ∖ A) .$ Ab $P (A ∖ B) = P (A) - P (A \cap B)$ aur $P (B ∖ A) = P (B) - P (A \cap B)$ use karo (har set se shared overlap minus). Add karne pe: $P (exactly one) = (P (A) - P (A \cap B)) + (P (B) - P (A \cap B)) = P (A) + P (B) - 2 P (A \cap B) .$ Kyunki $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ , yeh $P (A \cup B) - P (A \cap B)$ ke barabar hai. Numerically: $P (exactly one) = 0.75 - 0.25 = 0.50.$ Yeh step kyun? Disjoint pieces $A ∖ B$ aur $B ∖ A$ se build karna dikhata hai ki shortcut $P (A \cup B) - P (A \cap B)$ kyun kaam karta hai — overlap ek baar remove hota hai kyunki "exactly one" "both" region ko poori tarah exclude karta hai.

Verify: Chaar disjoint regions (both, only $A$ , only $B$ , neither) ka sum $1$ hona chahiye: $0.25 + (0.60 - 0.25) + (0.40 - 0.25) + 0.25 = 0.25 + 0.35 + 0.15 + 0.25 = 1.00. ✓$ "Neither" $= 0.25$ exactly. ✓

Ex 8 — Exam twist: teen overlapping events [C8]

Worked example Three-set inclusion–exclusion (sign bookkeeping)

Students mein: $P (Maths = M) = 0.5$ , $P (Physics = P h) = 0.4$ , $P (Chem = C) = 0.3$ ; pairwise $P (M \cap P h) = 0.2$ , $P (M \cap C) = 0.15$ , $P (P h \cap C) = 0.1$ ; teeno $P (M \cap P h \cap C) = 0.05$ . $P (M \cup P h \cup C)$ nikalo = probability ki student teeno mein se kam se kam ek padhe.

Forecast: Naively $0.5 + 0.4 + 0.3 = 1.2 > 1$ add karna — impossible! Toh subtract karna padega. Corrected value guess karo.

Step 1 — Three-set formula likho. $P (M \cup P h \cup C) = \sum P (singles) - \sum P (pairs) + P (triple) .$ Yeh step kyun? Pairs singles se double-count ho jaate hain, toh unhe subtract karo; lekin triple phir bahut zyada subtract ho jaata hai, toh ek baar wapas add karo. Yeh alternating-sign pattern general Inclusion–exclusion principle hai.

Step 2 — Sahi signs ke saath plug in karo. $= (0.5 + 0.4 + 0.3) - (0.2 + 0.15 + 0.1) + 0.05.$ Yeh step kyun? $+$ singles, $-$ pairs, $+$ triple — ek bhi sign galat kiya toh answer $[0, 1]$ range tod deta hai.

Step 3 — Arithmetic. $= 1.2 - 0.45 + 0.05 = 0.80.$

Verify: $0.80 \in [0, 1]$ ✓ (ab impossible $1.2$ nahi hai). Sanity: yeh $\geq$ sabse bade single event $P (M) = 0.5$ hona chahiye aur $\leq$ unke sum $1.2$ ; actually $0.5 \leq 0.80 \leq 1.2$ . ✓

Recall

Recall Kaun si cell, kaun sa tool?

Disjoint finite events — seedha add karo? ::: Haan: $P (A) + P (B)$ (C1). Overlapping events — kya karna hai? ::: $P (A \cap B)$ subtract karo: inclusion–exclusion (C2, C8). Finite space lekin outcomes equally likely nahi? ::: $P (A) = \sum_{ω \in A} P ({ω})$ , atoms add karo (C4′). "At least one of many" — sabse fast route? ::: Complement: $1 - P (none)$ (C3). $P (\emptyset)$ aur $P (Ω)$ ? ::: $0$ aur $1$ (C4). $[0, 1]$ pe ek single point ki probability? ::: $0$ , lekin impossible nahi (C5). Shrinking $A_{n} ↓ A$ — kaun si continuity? ::: From above (T1): $P (A_{n}) \to P (A)$ (C6). Growing $B_{n} ↑ B$ — kaun si continuity? ::: From below (T2): $P (B_{n}) \to P (B)$ (C6). Independent events — kaise combine karo? ::: Multiply karo: $P (A \cap B) = P (A) P (B)$ (C3). Three-set formula sign pattern? ::: $+$ singles $-$ pairs $+$ triple (C8).