Probability — classical, empirical, axiomatic (Kolmogorov axioms): Worked examples

2.7.5 · D3 · Maths › Statistics & Probability — Intermediate › Probability — classical, empirical, axiomatic (Kolmogorov ax

Neeche sab kuch sirf parent mein already bani hui ideas se hai:

Outcome, sample space $S$ , event $A$ — ek possible result, sabhi results ka set, aur us set ka ek subset.
$P (A) = \frac{m}{n}$ (classical): equally-likely total mein se favourable count.
$P (A) \approx \frac{f}{N}$ (empirical): $N$ tries mein kitni baar hua.
Teen axioms (No–Sure–Split) aur unse nikale gaye rules: complement $P (A^{c}) = 1 - P (A)$ , addition $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ .

Agar koi symbol naya hai, main use aate hi define kar deta hoon.

Scenario matrix

Is table ko ek checklist ki tarah socho. Har probability question jo tumhe milega in cells mein se kisi ek mein aata hai. Is page ka goal inhe sabhi hit karna hai.

Cell	Kya distinctive hai	Kaun sa tool answer deta hai	Example
C1 — disjoint	events saath nahi ho sakte	axiom 3: bas add karo	Ex 1
C2 — overlapping	events saath ho sakte hain	addition rule (overlap subtract karo)	Ex 2
C3 — complement / "at least one"	bahut saare favourable cases, ek aasaan opposite	$P (A^{c}) = 1 - P (A)$	Ex 3
C4 — no symmetry	outcomes equally likely nahi hain	empirical $f / N$	Ex 4
C5 — degenerate: certain / impossible	$P = 1$ ya $P = 0$ boundary	axioms 1–2 + derivations	Ex 5
C6 — counting-heavy	"favourable" ke liye combinations chahiye	Permutations & Combinations + classical	Ex 6
C7 — real-world word problem	pehle English → sets mein translate karo	set algebra + addition rule	Ex 7
C8 — limiting behaviour / continuous	$P = 0$ phir bhi possible hai; $N \to \infty$	axioms in infinite spaces + Law of Large Numbers	Ex 8
C9 — exam twist	ek cell jaisa lagta hai, actually doosra hai	sahi cell deliberately choose karo	Ex 9

Hum puri tarah Set Theory — Union, Intersection, Complement language mein events banate hain, isliye "and/or/not" ban jaata hai $\cap / \cup /^{c}$ .

Example 1 — Cell C1 (disjoint events)

Worked example Ek die: "2 aao

ya 5 aao" Ek fair six-sided die ek baar roll kiya jaata hai. $P (2 or 5)$ nikalo.

Forecast: six mein se do outcomes — aage padhne se pehle ek number guess karo. Kya yeh aadhe se zyada hai ya kam?

Sample space likho. $S = {1, 2, 3, 4, 5, 6}$ , sab equally likely, toh $n = 6$ . Yeh step kyun? Classical probability tabhi valid hai jab $n$ outcomes equally likely hon — ek fair die yeh guarantee karta hai (cube ki symmetry).
Events name karo. $A = {2}$ , $B = {5}$ . Overlap check karo: $A \cap B = \emptyset$ (ek roll ek saath 2 aur 5 nahi ho sakta). Yeh step kyun? Kyunki yeh disjoint hain, axiom 3 hume sirf add karne deta hai — koi subtraction nahi chahiye. Yeh Cell C1 hai.
Add karo. $P (A \cup B) = P (A) + P (B) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$ . Yeh step kyun? Axiom 3 (Split): disjoint events exactly add hote hain.

Verify: favourable outcomes hain ${2, 5}$ , toh directly $m / n = 2/6 = 1/3$ . ✓ Match karta hai. Aur $1/3 < 1/2$ , sensible hai.

Example 2 — Cell C2 (overlapping events)

Worked example Ek card: "face card

ya spade" Standard 52-card deck se ek card draw karo. Face card matlab Jack, Queen, ya King. $P (face \cup spade)$ nikalo.

Forecast: 12 face cards hain aur 13 spades — kya answer $\frac{12 + 13}{52} = \frac{25}{52}$ hai? Guess karo, phir trap dekho.

Individual counts. $P (F) = \frac{12}{52}$ (4 suits × 3 faces), $P (S_{♠}) = \frac{13}{52}$ . Yeh step kyun? 52 equally likely cards → classical counting valid hai.
Overlap nikalo. Cards jo dono face aur spade hain: J♠, Q♠, K♠ → $P (F \cap S_{♠}) = \frac{3}{52}$ . Yeh step kyun? Yeh Cell C2 hai — events share karte hain outcomes, toh naive addition double-count karta hai.
Addition rule lagao. $P (F \cup S_{♠}) = \frac{12}{52} + \frac{13}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26} .$ Subtract kyun? 3 spade-faces $P (F)$ ke andar ek baar aur $P (S)$ ke andar ek baar count hue the; ek copy hatao taaki har outcome exactly ek baar count ho.

Verify: union directly list karo: 12 faces + (13 spades − 3 already counted) = 22 distinct cards. $22/52 = 11/26$ . ✓ Naive $25/52$ exactly $3/52$ overlap ke barabar galat tha.

Figure — Probability — classical, empirical, axiomatic (Kolmogorov axioms)

Example 3 — Cell C3 (complement, "at least one")

Worked example Char coin tosses: "at least one head"

Ek fair coin 4 baar toss karo. $P (at least one head)$ nikalo.

Forecast: char tries mein se "at least one" lagbhag certain lagta hai. $0.9$ se upar ya neeche?

Many-cases problem identify karo. "At least one head" ka matlab hai 1, 2, 3, ya 4 heads — bahut saare cases. Yeh step kyun? Inhe sab count karna tedious hai; complement ek hi case hai. Yeh Cell C3 hai.
Complement lo. "At least one head" ka opposite hai "koi head nahi" = sab tails = $T T T T$ . Yeh step kyun? $P (A^{c}) = 1 - P (A)$ (axioms se prove hua) hume mushkil event ko aasaan wale se calculate karne deta hai.
Aasaan event ki probability. Har toss independent hai $P (T) = \frac{1}{2}$ ke saath, toh $P (T T T T) = (\frac{1}{2})^{4} = \frac{1}{16}$ . Yeh step kyun? Char independent tails multiply hote hain (dekho Independence of Events).
Subtract karo. $P (at least one head) = 1 - \frac{1}{16} = \frac{15}{16} = 0.9375$ .

Verify: total outcomes $= 2^{4} = 16$ , sab equally likely; sirf $T T T T$ mein zero heads hain, toh $16$ mein se $15$ mein at least ek head hai → $15/16$ . ✓ Aur $0.9375 > 0.9$ , forecast se match karta hai.

Example 4 — Cell C4 (no symmetry, empirical)

Worked example Biased spinner

Ek spinner mein koi symmetry nahi hai (uneven weighting). Use $N = 400$ baar spin kiya jaata hai aur yeh $f = 148$ baar red par aata hai. $P (red)$ estimate karo, aur batao ki kitna off ho sakte ho.

Forecast: roughly kya fraction? Aur kya tum yahan $m / n$ use kar sakte ho?

Classical reject karo. Koi symmetry argument nahi hai, toh hum claim nahi kar sakte ki outcomes equally likely hain. Yeh step kyun? Classical $m / n$ ke liye equally-likely outcomes zaroori hain; symmetry ke bina yeh illegal hai. Yeh Cell C4 hai.
Relative frequency use karo. $P (red) \approx \frac{f}{N} = \frac{148}{400} = 0.37.$ Yeh step kyun? Empirical probability: bahut saare trials mein fraction measure karo.
Isko trust kyun karein. Law of Large Numbers kehta hai ki $f / N$ true value ki taraf settle hota hai jab $N$ badhta hai; $400$ trials kaafi large hain, toh $0.37$ ek reasonable estimate hai — exact nahi.

Verify: $148/400 = 0.37$ exactly, aur $0 \leq 0.37 \leq 1$ toh yeh valid probability hai. ✓

Example 5 — Cell C5 (degenerate: certain aur impossible)

Example 6 — Cell C6 (counting-heavy, combinations)

Worked example Committee: 5 mein se 2 choose kiye, dono women

Ek group mein 3 men aur 2 women hain. At random do log choose kiye jaate hain (order matter nahi karta). $P (both women)$ nikalo.

Forecast: sirf 2 women hain aur tum 2 log pick kar rahe ho — kya yeh chhoti chance hai? Ek fraction guess karo.

5 mein se 2 choose karne ke total ways. Order-free selection ek combination hai, likha jaata hai $(2 5)$ = "5 choose 2" = 2-element subsets ki sankhya. $(2 5) = \frac{5 !}{2 ! 3 !} = 10.$ Yeh step kyun? Har 2-person committee equally likely hai, toh classical $m / n$ apply hota hai jab hum committees count karein, ordered pairs nahi. Dekho Permutations & Combinations. Yeh Cell C6 hai.
Favourable committees. Dono women: 2 women mein se 2 choose karo → $(2 2) = 1$ . Yeh step kyun? $m$ = un committees ki sankhya jo event satisfy karti hain.
Divide karo. $P (both women) = \frac{m}{n} = \frac{1}{10} = 0.1$ .

Verify: 10 committees enumerate karo; exactly ek ${W_{1}, W_{2}}$ hai, toh $1/10$ . ✓ Chhoti chance, jaise forecast tha.

Example 7 — Cell C7 (real-world word problem)

Worked example Survey: newspaper aur internet

Ek sheher mein, $60%$ log newspaper padhte hain, $50%$ online news padhte hain, aur $20%$ dono karte hain. Ek random person choose karo. Woh probability nikalo ki woh in dono mein se kam se kam ek tarike se news padhe, aur woh probability ki woh kisi bhi tarike se news na padhe.

Forecast: agar tum bas $60% + 50% = 110%$ add karo toh kuch galat hai — probabilities $1$ se zyada nahi ho sakti. Kya fix karta hai?

Sets mein translate karo. Maano $A$ = "newspaper padhta hai", $B$ = "online padhta hai". Diya hai $P (A) = 0.60$ , $P (B) = 0.50$ , $P (A \cap B) = 0.20$ . Yeh step kyun? English "both" = intersection; "at least one of the two" = union. Word → set algebra. Yeh Cell C7 hai.
Addition rule. $P (A \cup B) = 0.60 + 0.50 - 0.20 = 0.90$ . Subtract kyun? Jo $20%$ dono karte hain woh $60%$ aur $50%$ dono ke andar count hue the; ek copy hatao taaki total $\leq 1$ rahe. Ab $0.90$ valid hai.
"Neither" complement hai. $P ((A \cup B)^{c}) = 1 - P (A \cup B) = 1 - 0.90 = 0.10$ . Yeh step kyun? "Neither" "at least one" ka opposite hai; complement rule use karo.

Verify: chaar disjoint regions ka sum 1 hona chahiye: sirf-newspaper $= 0.60 - 0.20 = 0.40$ ; sirf-online $= 0.50 - 0.20 = 0.30$ ; dono $= 0.20$ ; neither $= 0.10$ . Sum $= 0.40 + 0.30 + 0.20 + 0.10 = 1.00$ . ✓

Example 8 — Cell C8 (limiting / continuous: $P = 0$ phir bhi possible)

Worked example Dartboard: exact point hit karna

Ek dart $10$ cm radius ke circular board par uniformly random point par girta hai. Nikalo (a) probability ki yeh $1$ cm radius ke inner "bullseye" region mein gire, aur (b) probability ki yeh ek exact mathematical point par gire.

Forecast: (b) ke liye, point "hittable" hai — dart kahin na kahin girega zaroor. Toh kya uski probability zero hai ya positive?

Continuous matlab area, counting nahi. Infinitely many landing points ke saath, "equally likely" ka matlab hai "probability $\propto$ area". Yeh step kyun? Hum $n$ outcomes count nahi kar sakte; axioms phir bhi hold karte hain, lekin favourable area se measure hota hai. Yeh Cell C8 hai.
Part (a): area ratio. $P (bullseye) = \frac{π ( 1 ) ^{2}}{π ( 10 ) ^{2}} = \frac{1}{100} = 0.01.$ Yeh step kyun? Uniform distribution → probability total area ka fraction hai.
Part (b): ek single point ka zero area hota hai. Uska area $0$ hai, toh $P (exact point) = \frac{0}{π ( 10 ) ^{2}} = 0$ . Yeh step kyun? Phir bhi dart kisi point par girega — toh yahan $P = 0$ ka matlab impossible nahi hai. Infinite/continuous sample spaces mein axioms yeh allow karte hain; sirf finite spaces mein $P = 0 \Leftrightarrow$ impossible hota hai.

Verify: $π (1)^{2} / π (10)^{2} = 1/100 = 0.01$ ✓, aur $0/ (π \cdot 100) = 0$ ✓. Limiting lesson: jab ek region shrink hokar point ban jaata hai, uski probability $\to 0$ continuously.

Example 9 — Cell C9 (exam twist: disjoint lagta hai, nahi hai)

Worked example Do dice: "sum 7 hai

ya dono dice equal hain" Do fair dice roll karo. Maano $A$ = "sum $= 7$ ", $B$ = "doubles (dono dice equal)". $P (A \cup B)$ nikalo.

Forecast: bahut saare students turant $P (A) + P (B)$ add kar dete hain. Kya yeh yahan safe hai? Pehle ek shared outcome dhundho.

Sample space. Ordered pairs, $n = 36$ equally likely outcomes. Yeh step kyun? Ordering outcomes ko equally likely rakhta hai (parent ka two-dice logic).
Har event count karo. $A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}$ → $P (A) = \frac{6}{36}$ . $B = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}$ → $P (B) = \frac{6}{36}$ .
Overlap check karo — the twist. Kya koi pair dono sum-of-7 aur doubles ho sakta hai? Doubles ke sums hain $2, 4, 6, 8, 10, 12$ — kabhi 7 nahi. Toh $A \cap B = \emptyset$ . Yeh step kyun? Yeh Cell C9 hai: yeh lagta hai overlap problem hai, lekin events disjoint nikle. Tumhe disjointness verify karni hogi, kisi bhi taraf assume nahi karna.
Add karo (ab justified hai). $P (A \cup B) = \frac{6}{36} + \frac{6}{36} = \frac{12}{36} = \frac{1}{3}$ . Yeh step kyun? Kyunki $A \cap B = \emptyset$ hai, axiom 3 plain addition deta hai — koi subtraction nahi kyunki overlap genuinely empty hai.

Verify: union mein $6 + 6 = 12$ distinct pairs hain (koi repeat nahi, kyunki disjoint), toh $12/36 = 1/3$ . ✓ Lesson: intersection hamesha check karo; yahan subtraction term legitimately $0$ hai.

Recall Kaun sa cell, kaun sa tool?

Disjoint → bas add karo ::: axiom 3 Overlapping → intersection subtract karo ::: addition rule "At least one" → complement ::: $1 - P (A^{c})$ No symmetry → measure karo ::: empirical $f / N$ Order-free selection → combinations ::: $(r n)$ phir $m / n$ Continuous space → area ratio, aur $P = 0$ phir bhi possible ho sakta hai ::: measure not count