Probability Distributions — Applied Maths 12

1. Random Variables aur Probability Distribution

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

Discrete RV: finite ya countable values (jaise dice roll = 1,2,3,4,5,6) Continuous RV: uncountable values (jaise height, weight, time)

Probability Distribution Table dikhata hai har value ka probability:

X	0	1	2	3	Total
P(X)	1/8	3/8	3/8	1/8	1

Mean (Expected Value): E[X] = Σ xᵢP(xᵢ) Variance: Var(X) = E[X²] − (E[X])² = Σ xᵢ²P(xᵢ) − μ² Standard Deviation: σ = √Var(X)

Udaharan: Mean aur Variance

X = {0,1,2,3} with P = {1/8, 3/8, 3/8, 1/8}
E[X] = 0(1/8)+1(3/8)+2(3/8)+3(1/8) = 0+0.375+0.75+0.375 = 1.5
E[X²] = 0+0.375+1.5+1.125 = 3
Var(X) = 3 − (1.5)² = 3 − 2.25 = 0.75
σ = √0.75 ≈ 0.866

2. Binomial Distribution

Jab n independent trials hote hain, har trial mein sirf 2 outcomes (success/failure) aur success ki probability p constant ho.

P(X = r) = ⁿCᵣ · pʳ · qⁿ⁻ʳ where q = 1−p, r = 0,1,2,...,n Mean: μ = E[X] = np Variance: σ² = npq SD: σ = √(npq)

Udaharan: Binomial

Ek fair coin 6 baar throw ki. Exactly 4 heads aane ki probability?
n=6, p=1/2, q=1/2, r=4
P(X=4) = ⁶C₄ × (1/2)⁴ × (1/2)²
= 15 × 1/16 × 1/4 = 15/64 ≈ 0.234

Udaharan: Quality Control

Factory mein 10% defective items hain. 5 items mein exactly 2 defective?
n=5, p=0.1, q=0.9, r=2
P(X=2) = ⁵C₂ × (0.1)² × (0.9)³
= 10 × 0.01 × 0.729 = 0.0729

3. Poisson Distribution

Rare events count karne ke liye use hota hai — ek fixed interval mein average λ events hote hain.

P(X = x) = (e⁻λ × λˣ) / x! x = 0,1,2,... Mean: E[X] = λ Variance: Var(X) = λ (mean = variance!) SD: σ = √λ

Binomial to Poisson Approximation: Jab n bahut bada ho aur p bahut chhota ho (n→∞, p→0), toh Binomial ≈ Poisson with λ=np

Udaharan: Traffic Accidents

Ek highway par average 2 accidents/day hote hain. P(exactly 3 accidents)?
λ=2, x=3
P(X=3) = e⁻² × 2³ / 3! = 0.1353 × 8 / 6 = 0.1804

Udaharan: Call Center

Average 5 calls/minute aate hain. P(0 calls in a minute)?
λ=5, x=0
P(X=0) = e⁻⁵ × 5⁰ / 0! = e⁻⁵ ≈ 0.0067

4. Normal Distribution

Sabse important continuous distribution. Bell-shaped, symmetric about mean. Heights, weights, IQ — sab normal follow karte hain.

f(x) = (1/σ√2π) × e^[−(x−μ)²/2σ²] Parameters: μ (mean), σ (standard deviation) Standard Normal: Z = (X − μ)/σ → N(0,1)

Empirical Rule (68-95-99.7):
P(μ−σ < X < μ+σ) = 68.27%
P(μ−2σ < X < μ+2σ) = 95.45%
P(μ−3σ < X < μ+3σ) = 99.73%

Udaharan: Exam Scores

Exam scores: μ=70, σ=10. P(score > 80)?
Z = (80−70)/10 = 1
P(X>80) = P(Z>1) = 1−Φ(1) = 1−0.8413 = 0.1587 ≈ 15.87%

Udaharan: Height

Men ki height: μ=170cm, σ=6cm. P(165 < X < 175)?
Z₁=(165−170)/6=−0.83, Z₂=(175−170)/6=0.83
P = Φ(0.83)−Φ(−0.83) = 0.7967−0.2033 = 0.5934 ≈ 59.3%

4.1 Standard Normal Table Values

Z	Φ(Z) = P(X≤Z)	Z	Φ(Z)
0.0	0.5000	1.5	0.9332
0.5	0.6915	1.96	0.9750
1.0	0.8413	2.0	0.9772
1.28	0.8997	2.5	0.9938
1.645	0.9500	3.0	0.9987

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