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Distributions

Short Notes

A probability distribution describes how probabilities are distributed over the values of a random variable. Key distributions for GATE CSE include Uniform, Binomial, Poisson, Exponential, and Normal.

Key Theories & Formulas

1. Discrete Distributions

Uniform Distribution (Discrete)

\(X \sim Uniform\{1, 2, ..., n\}\)

  • PMF: \(P(X = k) = \frac{1}{n}\)
  • Mean: \(E[X] = \frac{n+1}{2}\)
  • Variance: \(Var(X) = \frac{n^2 - 1}{12}\)

Bernoulli Distribution

\(X \sim Bernoulli(p)\) — single trial with success probability p

  • PMF: \(P(X = 1) = p\), \(P(X = 0) = 1-p\)
  • Mean: \(E[X] = p\)
  • Variance: \(Var(X) = p(1-p)\)

Binomial Distribution

\(X \sim Binomial(n, p)\) — number of successes in n independent trials

  • PMF: \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)
  • Mean: \(E[X] = np\)
  • Variance: \(Var(X) = np(1-p)\)

Geometric Distribution

\(X \sim Geometric(p)\) — number of trials until first success

  • PMF: \(P(X = k) = (1-p)^{k-1} p\), \(k = 1, 2, ...\)
  • Mean: \(E[X] = \frac{1}{p}\)
  • Variance: \(Var(X) = \frac{1-p}{p^2}\)
  • Memoryless Property: \(P(X > m+n | X > m) = P(X > n)\)

Poisson Distribution

\(X \sim Poisson(\lambda)\) — number of events in fixed interval

  • PMF: \(P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}\)
  • Mean: \(E[X] = \lambda\)
  • Variance: \(Var(X) = \lambda\)
  • Sum of Poissons: \(Poisson(\lambda_1) + Poisson(\lambda_2) = Poisson(\lambda_1 + \lambda_2)\)

2. Continuous Distributions

Uniform Distribution (Continuous)

\(X \sim Uniform(a, b)\)

  • PDF: \(f(x) = \frac{1}{b-a}\) for \(a \leq x \leq b\)
  • Mean: \(E[X] = \frac{a+b}{2}\)
  • Variance: \(Var(X) = \frac{(b-a)^2}{12}\)

Exponential Distribution

\(X \sim Exponential(\lambda)\)

  • PDF: \(f(x) = \lambda e^{-\lambda x}\) for \(x \geq 0\)
  • CDF: \(F(x) = 1 - e^{-\lambda x}\)
  • Mean: \(E[X] = \frac{1}{\lambda}\)
  • Variance: \(Var(X) = \frac{1}{\lambda^2}\)
  • Memoryless Property: \(P(X > s+t | X > s) = P(X > t)\)

Normal (Gaussian) Distribution

\(X \sim N(\mu, \sigma^2)\)

  • PDF: \(f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}\)
  • Mean: \(E[X] = \mu\)
  • Variance: \(Var(X) = \sigma^2\)
  • Standard Normal: \(Z = \frac{X - \mu}{\sigma} \sim N(0, 1)\)

68-95-99.7 Rule:

  • P(μ - σ < X < μ + σ) ≈ 68%
  • P(μ - 2σ < X < μ + 2σ) ≈ 95%
  • P(μ - 3σ < X < μ + 3σ) ≈ 99.7%

3. Key Relationships

Approximation Condition
\(Binomial(n,p) \approx Poisson(np)\) n large, p small, np moderate
\(Binomial(n,p) \approx Normal(np, np(1-p))\) n large, np ≥ 5, n(1-p) ≥ 5
\(Poisson(\lambda) \approx Normal(\lambda, \lambda)\) λ large (≥ 20)

Example Problems

Problem 1: X ~ Binomial(10, 0.3). Find P(X = 3).

Solution: $\(P(X=3) = \binom{10}{3}(0.3)^3(0.7)^7 = 120 \times 0.027 \times 0.0824 = \mathbf{0.267}\)$

Problem 2: Cars arrive at a toll booth at rate 3 per minute (Poisson). Find P(no cars in 30 seconds).

Solution:

  1. Rate for 30 sec = 3/2 = 1.5
  2. \(P(X=0) = \frac{e^{-1.5}(1.5)^0}{0!} = e^{-1.5} = \mathbf{0.223}\)

Problem 3: X ~ Exponential(0.5). Find P(X > 3 | X > 1).

Solution (using memoryless property): $\(P(X > 3 | X > 1) = P(X > 2) = e^{-0.5 \times 2} = e^{-1} = \mathbf{0.368}\)$

Hardest GATE Questions

Topic: Poisson Distribution Applications

Question (GATE 2018 Style): A computer system crashes on average 2 times per week (Poisson). Find the probability of at least 3 crashes in 2 weeks.

Solution:

  1. Rate for 2 weeks: λ = 4
  2. P(X ≥ 3) = 1 - P(X ≤ 2)
  3. P(X ≤ 2) = \(e^{-4}(1 + 4 + 8) = 13e^{-4} \approx 0.238\)
  4. P(X ≥ 3) = 1 - 0.238 = 0.762

Topic: Normal Distribution with Z-scores

Question (GATE 2017 Variant): Exam scores are N(70, 100). What percentage of students score between 60 and 85? (Given: Φ(1.5) = 0.9332, Φ(1) = 0.8413)

Solution:

  1. Z₁ = (60-70)/10 = -1
  2. Z₂ = (85-70)/10 = 1.5
  3. P(60 < X < 85) = Φ(1.5) - Φ(-1) = 0.9332 - (1 - 0.8413)
  4. = 0.9332 - 0.1587 = 77.45%

Topic: Distribution Identification

Question: X has MGF \(M_X(t) = e^{3(e^t - 1)}\). Identify the distribution and find P(X ≥ 2).

Solution:

  1. MGF of Poisson(λ) is \(e^{\lambda(e^t - 1)}\)
  2. Comparing: λ = 3, so \(X \sim Poisson(3)\)
  3. P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)
  4. = 1 - e⁻³ - 3e⁻³ = 1 - 4e⁻³ ≈ 1 - 0.199 = 0.801

Topic: Sum of Random Variables

Question: X₁ ~ Poisson(2), X₂ ~ Poisson(3), X₃ ~ Poisson(5) independent. Find Var(X₁ + X₂ + X₃).

Solution:

  1. Sum of independent Poissons: X₁ + X₂ + X₃ ~ Poisson(2+3+5) = Poisson(10)
  2. For Poisson: Var = λ
  3. Var(X₁ + X₂ + X₃) = 10

References