Sample spaces, events, and axioms of probability

1.3.1 · AI-ML › Probability & Statistics

Sample space kya hota hai?

Yeh kyun zaroori hai? Agar "saari possibilities" define nahi ki, toh hum measure nahi kar sakte ki kuch "kitna likely" hai. Yeh aise hai jaise "fraction kya hai?" poochha jaye bina yeh bataye ki "kis cheez ka fraction?"

Sample spaces ke examples

Event kya hota hai?

Subsets kyun? Kyunki hum aksar outcomes ke groups ki parwah karte hain. "Even number roll karna" event {2, 4, 6} hai, koi ek single outcome nahi.

Events ke types

Simple event: single outcome, jaise {H} ya {(3,5)}
Compound event: multiple outcomes, jaise {sum≥ 10} = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}
Certain event: Ω khud (probability = 1)
Impossible event: ∅ (probability = 0)

Event A define karo = "sum 7 hai" $A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}$ $∣ A ∣ = 6$

Event B define karo = "kam se kam ek die par 6 aaya" $B = {(6, 1), (6, 2), .., (6, 6), (1, 6), (2, 6), .., (5, 6)}$ $∣ B ∣ = 11$ 11 kyun, 12 kyun nahi? (6,6) ko ek baar count kiya jaata hai, do baar nahi (inclusion-exclusion principle)

Event operations:

A ∪ B = "sum 7 hai YA kam se kam ek 6 aaya" (union)
A ∩ B = {(1,6), (6,1)} = "sum 7 hai AUR kam se kam ek 6 aaya" (intersection)
A^c = "sum 7 NAHI hai" (complement, A ke alawa saare outcomes)

Probability ke teen axioms

Axiom 1 (Non-negativity): $P (E) \geq 0 \forall E \subseteq Ω$

Axiom 2 (Normalization): $P (Ω) = 1$

Axiom 3 (Countable Additivity): Mutually exclusive events $E_{1}, E_{2}, ..$ ke liye: $P (E_{1} \cup E_{2} \cup ..) = P (E_{1}) + P (E_{2}) + ...$ (agar $E_{i} \cap E_{j} = \emptyset$ for $i \neq = j$ )

Yeh teeno kyun?

Axiom 1 kyun: Probabilities certainty ke fractions hain. Negative probability ka koi physical meaning nahi. Tum "−20% sure" nahi ho sakte.

Axiom 2 kyun: KUCH NA KUCH hona hi hai. Agar tum coin flip karo, heads ya tails aayega (yeh assume karte hue ki coin edge par nahi girti, jo hamare model se bahar hai). Saare possible outcomes ki total probability = certainty = 100% = 1.

Axiom 3 kyun: Agar events ek saath nahi ho sakte (mutually exclusive), toh "A ya B" ki probability bas "unhe jodo." Agar (1,1) aur (2,3) ek saath roll nahi ho sakte, toh P({(1,1)} ∪ {(2,3)}) = P({(1,1)}) + P({(2,3)}).

Non-mutually-exclusive events ka kya? Woh Axiom 3 nahi hai. Hum isse derive karte hain:

Derivation:

Likho $A \cup B = A \cup (B \cap A^{c})$ (B ko "sirf-B" wale part mein split karo)
Yeh disjoint hain: $A$ aur $B \cap A^{c}$ overlap nahi karte
Axiom 3 se: $P (A \cup B) = P (A) + P (B \cap A^{c})$
Lekin $B = (B \cap A) \cup (B \cap A^{c})$ , yeh bhi disjoint hain
Toh $P (B) = P (B \cap A) + P (B \cap A^{c})$
Rearrange karo: $P (B \cap A^{c}) = P (B) - P (A \cap B)$
Step 3 mein substitute karo: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ ∎

Yeh step kyun? Humein overlap $A \cap B$ ko double-count hone se bachana tha.

Axioms se useful properties derive karna

Proof:

$A$ aur $A^{c}$ mutually exclusive hain: $A \cap A^{c} = \emptyset$
$A \cup A^{c} = Ω$ (har outcome ya toh A mein hai ya nahi)
Axiom 2 se: $P (Ω) = 1$
Axiom 3 se: $P (A \cup A^{c}) = P (A) + P (A^{c})$
Combine karo: $1 = P (A) + P (A^{c})$
Rearrange karo: $P (A^{c}) = 1 - P (A)$ ∎

ML intuition: Agar tumhara spam classifier kehta hai P(spam) = 0.7, toh P(not-spam) = 0.3. Unka sum 1 hona chahiye.

Proof:

Likho $B = A \cup (B \cap A^{c})$ (B matlab A aur kuch extra)
Yeh disjoint hain
Axiom 3 se: $P (B) = P (A) + P (B \cap A^{c})$
Axiom 1 se: $P (B \cap A^{c}) \geq 0$
Isliye: $P (B) \geq P (A)$ ∎

Intuition: "Bade events ki probability chhoti nahi ho sakti." Agar B mein A ke saath aur bhi cheezein hain, toh B kam se kam utna hi likely hai.

Finite sample spaces mein probabilities compute karna

Finite Ω ke liye jahan equally likely outcomes hon (jaise fair dice, random draws):

Yeh kyun kaam karta hai:

Har outcome ω ki same probability hoti hai: $P ({ω}) = \frac{1}{∣Ω∣}$
Event E disjoint outcomes ka union hai: $E = {ω_{1}} \cup {ω_{2}} \cup ...$
Axiom 3 se: $P (E) = \sum_{ω \in E} P ({ω}) = ∣ E ∣ \cdot \frac{1}{∣Ω∣}$

Yeh step kyun? 36 mein se har outcome equally likely hai (fair dice). Event "sum=7" mein unme se 6 hain.

Estimated: $P (spam) \approx \frac{300}{1000} = 0.3$ $P (not-spam) \approx \frac{700}{1000} = 0.7$

Check: $0.3 + 0.7 = 1$ ✓ (Axiom 2)

ML connection: Yeh Bernoulli distribution ke liye maximum likelihood estimation hai.

Common mistakes

Yeh sahi kyun lagta hai: Disjoint events ke liye, yeh sach mein sahi hai (Axiom 3). Beginners isko overgeneralize kar dete hain.

Fix: Check karo ki events overlap karte hain ya nahi. Agar $A \cap B \neq = \emptyset$ , toh inclusion-exclusion use karo: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$

Example: Do dice, A = "pehla die 6 hai", B = "sum 10 hai"

$P (A) = 6/36 = 1/6$
$P (B) = 3/36 = 1/12$ (pairs: (4,6), (5,5), (6,4))
$A \cap B = {(6, 4)}$ , toh $P (A \cap B) = 1/36$
$P (A \cup B) = 1/6 + 1/12 - 1/36 = 6/36 + 3/36 - 1/36 = 8/36 = 2/9$

Agar hum bas add karte: $1/6 + 1/12 = 1/4 \neq = 2/9$ ✗

Yeh sahi kyun lagta hai: Mathematically, |Ω| = 2. Classical probability 1/2 deta hai.

Fix: Equally likely outcomes ke liye symmetry chahiye (fair coin, fair die, random draw). Weather symmetric NAHI hai. Axioms non-uniform probabilities allow karte hain. Humein data ya model chahiye.

Fix: Axiom 1 aur 2 bound karte hain P(E) ∈ [0, 1]. Agar tum kuch is range se bahar compute karo, toh ya toh algebra mein galti hui ya tumhara model incoherent hai.

ML/AI se connections

Conditional Probability events par build karta hai: P(A|B) ke liye events A, B ka matlab samajhna zaroori hai
Bayes' Theorem in axioms par tika hai; inke bina Bayesian inference undefined hai
Random Variables Ω se ℝ tak ke functions hain; sample space domain hai
Probability Distributions events ko ek structured tarike se probabilities assign karte hain
Maximum Likelihood Estimation frequentist probability use karta hai: P(data|model) as |favorable|/|total| limit mein
Classification Metrics (precision, recall) confusion matrix events par conditional probabilities hain
Entropy and Information un probability distributions par uncertainty measure karta hai jo inhi axioms se bani hain

Recall Ek 12-saal ke bachche ko samjhao

Socho tumhare paas ek bag mein marbles hain: 3 red, 2 blue, 5 green. Tum aankhein band karke ek nikalte ho.

Sample space hai "bag ke saare marbles" — har woh marble jo tum nikal sakte ho. Total 10 marbles hain.

Ek event woh hai jo tumhe chahiye, jaise "maine red marble nikala" ya "maine red ya blue nikala." Events bas marbles ke groups hain.

Probability ke teen rules (axioms) yeh hain:

Chances negative nahi ho sakte. Red nikalne ka "minus 30% chance" nahi ho sakta — yeh bakwaas hai.
KOI NA KOI marble nikala jaayega (maano tum nikalte ho). Toh "koi bhi marble" ka chance 100% hai, jise hum 1 likhte hain.
Agar do events ek saath nahi ho sakte (jaise ek marble nikalte ho jo red BHI ho aur blue BHI — yeh impossible hai), toh "red YA blue" ka chance bas hai: red ka chance + blue ka chance.

Bas itna hai! Yeh teen simple rules hume AI aur ML mein har cheez ke liye probabilities calculate karne dete hain.

Summary

Sample space Ω: saare possible outcomes, exhaustive aur mutually exclusive
Event E: Ω ka ek subset jise hum probability assign karte hain
Teen axioms: Non-negativity, normalization, additivity (disjoint events ke liye)
Derived properties: complement rule, inclusion-exclusion, monotonicity
Classical probability: P(E) = |E|/|Ω| jab outcomes equally likely hon
ML foundation: saare probabilistic models (Bayes, logistic regression, generative models) inhi axioms par tike hain

#flashcards/ai-ml

Sample space (Ω) kya hota hai?

Ek random experiment ke saare possible outcomes ka set, jo exhaustive (har possible outcome contain kare) aur mutually exclusive (outcomes overlap na karein) hona chahiye.

Probability theory mein event kya hota hai?

Sample space ka ek subset (E ⊆ Ω) jo un outcomes ka collection represent karta hai jismein hum interested hain.

Probability ka Axiom 1 (non-negativity) batao.

P(E) ≥ 0 saare events E ke liye. Probabilities negative nahi ho sakti.

Probability ka Axiom 2 (normalization) batao.

P(Ω) = 1. Pure sample space ki probability (kuch na kuch hona hi hai) 1 hoti hai.

Probability ka Axiom 3 (countable additivity) batao.

Mutually exclusive events E₁, E₂, ... ke liye: P(E₁ ∪ E₂ ∪ ...) = P(E₁) + P(E₂) + ... Disjoint events ke union ki probability unki individual probabilities ka sum hai.

Axioms se complement rule P(A^c) = 1 - P(A) derive karo.

(1) A aur A^c mutually exclusive hain: A ∩ A^c = ∅. (2) Yeh Ω ko partition karte hain: A ∪ A^c = Ω. (3) Axiom 2 se: P(Ω) = 1. (4) Axiom 3 se: P(A ∪ A^c) = P(A) + P(A^c). (5) Isliye: 1 = P(A) + P(A^c), toh P(A^c) = 1 - P(A).

Do events ke liye inclusion-exclusion principle kya hai?

P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Hum intersection subtract karte hain taaki un outcomes ko double-count na karein jo A aur B dono mein hain.

P(E) = |E|/|Ω| formula kab use kar sakte hain?

Sirf tab jab sample space finite HO AUR saare outcomes equally likely hon (uniform distribution), jaise fair dice ya uniform population se random draws.

Sirf isliye P(rain) = 1/2 kyun nahi keh sakte ki do outcomes hain (rain/no rain)?

Formula P(E) = |E|/|Ω| sirf equally likely outcomes ke liye kaam karta hai. Rain aur no-rain symmetric nahi hain; probabilities assign karne ke liye humein data ya weather model chahiye.

Do dice ke liye P(sum ≥ 10) kya hai?

Favorable outcomes hain {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}, toh |E| = 6. Total outcomes |Ω| = 36. Isliye P(E) = 6/36 = 1/6.

Agar A ⊆ B, toh P(A) aur P(B) mein kya relationship hai?

P(A) ≤ P(B). Yeh monotonicity property hai: bade events (zyada outcomes) ki probability chhoti nahi ho sakti.

Figure — Sample spaces, events, and axioms of probability