Probability 🎲

Class 11 se aage — Bayes' theorem, random variables, aur distributions!

🏥 Doctor ka Test

Ek rare disease ka test 99% accurate hai (agar bimari hai toh 99% chance positive aata hai). Lekin disease sirf 0.1% population mein hai. Agar kisi ka test positive aaya — toh kya actually bimari hai? Answer shocking hai — sirf 9% chance! Kyun? Kyunki bahut kam log bimari se afflicted hain. Yeh Bayes' Theorem ka most famous example hai — conditional probability ka updated estimate bayesian reasoning se.

1. Conditional Probability

P(A|B) = P(A ∩ B) / P(B) — "B given, A hone ki probability"

Multiplication Theorem: P(A ∩ B) = P(A) × P(B|A) = P(B) × P(A|B)

P(A|B) = P(A ∩ B) / P(B) [P(B) ≠ 0] /* Independent Events */ A and B independent ⟺ P(A|B) = P(A) ⟺ P(A∩B) = P(A)×P(B) /* Multiplication Theorem for n events */ P(A∩B∩C) = P(A) × P(B|A) × P(C|A∩B)

✅ Example: Two dice roll. P(sum=8 | first die=3)?

Given first die = 3. For sum = 8: need second die = 5. P = 1/6

2. Total Probability Theorem

B₁, B₂, ..., Bₙ ek partition of S (mutually exclusive, exhaustive) hain. Tab:

P(A) = Σᵢ P(Bᵢ) × P(A|Bᵢ)

Total Probability — Tree Diagram S P(B₁) P(B₂) B₁ B₂ P(A|B₁) P(A|B₂) A A' A A' P(A) = P(B₁)P(A|B₁) + P(B₂)P(A|B₂)

3. Bayes' Theorem

Bayes' theorem: diya hua ki A hua — B ke hua hone ki probability kya hai?

P(Bᵢ|A) = P(Bᵢ) × P(A|Bᵢ) —————————————————— Σⱼ P(Bⱼ) × P(A|Bⱼ) Yaad karo: "Posterior = Prior × Likelihood / Normalizer"

✅ Worked Example: Bayes' Theorem

Bag I mein 3 red, 4 black balls. Bag II mein 5 red, 6 black balls. Ek bag randomly choose kiya aur ek ball nikali — red hai. Probability ki Bag I se nikali?

P(Bag I) = P(Bag II) = 1/2
P(Red|Bag I) = 3/7, P(Red|Bag II) = 5/11
P(Red) = (1/2)(3/7) + (1/2)(5/11) = 3/14 + 5/22 = 33/154 + 35/154 = 68/154
P(Bag I|Red) = (1/2 × 3/7) / (68/154) = (3/14) × (154/68) = 33/68

P(Bag I | Red ball) = 33/68 ≈ 0.485

4. Random Variables aur Probability Distribution

Random variable X: sample space ke har outcome ko ek real number assign karta hai.

Discrete RV: Countable values (jaise number of heads)

Expected Value: E(X) = Σ x·P(x)

Variance: Var(X) = E(X²) - [E(X)]²

✅ Expected Value Example

X0123
P(X)1/83/83/81/8

E(X) = 0×(1/8)+1×(3/8)+2×(3/8)+3×(1/8) = 0+3/8+6/8+3/8 = 12/8 = 3/2

5. Binomial Distribution

X ~ B(n, p) — n trials, p = success probability each trial P(X=r) = ⁿCᵣ × pʳ × (1-p)^(n-r) Mean = np Variance = np(1-p) = npq (q=1-p) SD = √(npq)

✅ Binomial Example

Fair coin 5 baar uchhaali. P(exactly 3 heads)?

n=5, p=1/2, r=3

P(X=3) = ⁵C₃ × (1/2)³ × (1/2)² = 10 × 1/8 × 1/4 = 10/32 = 5/16

6. Practice Questions

Q1. Two cards are drawn successively without replacement from 52-card deck. P(both kings)?

Solution

P(first king) = 4/52. P(second king|first king) = 3/51. P(both kings) = (4/52)(3/51) = 12/2652 = 1/221

Q2. A die is thrown twice. Events A={odd on first}, B={total≥8}. Are A,B independent?

Solution

P(A)=1/2. P(B)=favorable/36. Pairs summing ≥8: (2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)=15. P(B)=15/36. P(A∩B): A∩B means odd first AND sum≥8: (3,5),(3,6),(5,3),(5,4),(5,5),(5,6)=6/36. P(A)×P(B)=1/2×15/36=15/72≠6/36=1/6. So NOT independent.

Q3. In Binomial distribution with n=10, p=1/3. Find mean and variance.

Solution

Mean = np = 10×1/3 = 10/3. Variance = npq = 10×(1/3)×(2/3) = 20/9.