Binomial and Poisson distributions

Objectives and prerequisites

Recognize the assumptions behind binomial and Poisson models; compute exact and tail probabilities; use their means and variances; rescale a Poisson rate to the observation interval; and distinguish an approximation from an exact model. Prerequisites: random variables, PMFs, combinations, independence, expectation and variance. Both distributions are explicitly listed in GATE 2026 CS. Bernoulli variables provide the building block; a full stochastic-process treatment is not included.

1. Bernoulli success indicators

A Bernoulli(p) variable X takes one with probability p and zero with probability 1−p, for 0≤p≤1. “Success” is simply the event being counted; it need not be desirable. E[X]=p. Since X²=X, Var(X)=E[X²]−E[X]²=p−p²=p(1−p).

Examples include whether a bit is flipped, a component passes a test, or a randomly sampled request exceeds a threshold. These interpretations do not prove independence across multiple observations. A model must state or justify that assumption separately.

2. Binomial model and its derivation

If n trials are mutually independent, each has the same success probability p, and X counts successes, then X~Bin(n,p). Here n is a fixed nonnegative integer and X takes values 0,…,n. For k in that range,

P(X=k)=C(n,k)pᵏ(1−p)ⁿ⁻ᵏ.

Choose the k successful positions in C(n,k) ways. Independence gives probability pᵏ(1−p)ⁿ⁻ᵏ for each specified sequence. Distinct sequences are disjoint, so add their equal probabilities. Outside the support, probability is zero. For p=0 or p=1, interpret the model directly as a constant variable rather than relying on ambiguous 0⁰ notation.

Example: five independent trials each succeed with probability .2. Exactly two successes has probability C(5,2)(.2)²(.8)³=10(.04)(.512)=.2048. The factor C(5,2) matters: (.2)²(.8)³ describes only one specified ordering of two successes and three failures.

“At least one” is easiest by complement: P(X≥1)=1−P(X=0)=1−(.8)⁵=.67232. “At most one” is P(X=0)+P(X=1)=(.8)⁵+5(.2)(.8)⁴=.73728. A tail means summing all included counts, not inserting its endpoint into the PMF.

3. Binomial mean and variance, and model failures

Write X=I₁+…+Iₙ with independent Bernoulli indicators. Linearity gives E[X]=np. Variance additivity under independence gives Var(X)=np(1−p), and standard deviation √[np(1−p)]. For n=10,p=.3, mean=3, variance=2.1 and standard deviation √2.1. Mean and variance are not generally equal.

The formula requires fixed n, binary success/failure for each trial, constant p, and mutual independence. Sampling without replacement from a finite population generally violates independence and changes the conditional success probability; the exact count model is hypergeometric, which is mentioned here only as a contrast. A binomial approximation may be reasonable when sampling has negligible effect on the population, but it is still an approximation. Varying success probabilities also invalidate the simple Bin(n,p) PMF even if trials remain independent.

Some stopping experiments have random trial count, such as stopping at the first success. They are not binomial with a fixed n. Identify what is fixed and what is counted before selecting a distribution.

4. Poisson count distribution

A Poisson(λ) variable X has support k=0,1,2,… and PMF

P(X=k)=e^(−λ) λᵏ/k!, for λ>0.

The parameter λ is the expected count for the specified interval or region, not a probability, and it can exceed one. The degenerate λ=0 case has X=0 with probability one. Normalization follows from the exponential series: e^(−λ)Σₖλᵏ/k!=e^(−λ)e^λ=1.

The moments are E[X]=λ and Var(X)=λ. One derivation uses k/k!=1/(k−1)! to obtain E[X]=λ, and k(k−1)/k!=1/(k−2)! to obtain E[X(X−1)]=λ². Then E[X²]=λ²+λ and Var(X)=λ. Its standard deviation is √λ. Unlike a binomial variable, a nondegenerate Poisson variable has no fixed finite upper bound.

If counts follow a homogeneous Poisson process with rate r per time unit, the count over length t has λ=rt. This modeling statement includes stationary independent increments; merely knowing an average rate does not prove a Poisson model. A traffic process with bursts or regular scheduled arrivals may not satisfy it.

5. Worked Poisson calculations and interval scaling

Suppose requests follow a homogeneous Poisson process at two per minute. Over three minutes, λ=2×3=6. P(no requests)=e^(−6); P(exactly two)=e^(−6)6²/2!=18e^(−6); and P(at least one)=1−e^(−6). Using λ=2 for the three-minute interval would forget the units.

For X~Poisson(3), P(X≥2)=1−P(X=0)−P(X=1)=1−e^(−3)(1+3)=1−4e^(−3). A useful recurrence for λ>0 is P(X=k+1)=P(X=k)λ/(k+1). It helps compute several adjacent masses without repeating factorials.

Independent Poisson variables with parameters λ₁,λ₂ have a Poisson sum with parameter λ₁+λ₂. Independence is essential to this distributional claim. If Y=X, then X+Y=2X takes only even values and is not Poisson with parameter 2λ for λ>0, even though its mean is 2λ.

6. Poisson approximation to binomial

When n is large, p is small, and np=λ is of moderate size, a Poisson(λ) law can approximate Bin(n,p). This is a limiting approximation, not a universal equality and not a rule based solely on the mean. For a fixed count k, C(n,k)(λ/n)ᵏ(1−λ/n)ⁿ⁻ᵏ tends to e^(−λ)λᵏ/k! as n grows.

Example: 100 independent items each have defect probability .01. The exact probability of no defects is (.99)¹⁰⁰≈.366032. The Poisson approximation with λ=1 is e^(−1)≈.367879. State which answer is approximate. The binomial variance is .99 while the approximating Poisson variance is 1; close does not mean identical. The approximation may assign positive mass to counts greater than n, which are impossible in the exact binomial model.

Solving checklist

Identify whether the count comes from fixed independent trials or a specified Poisson model. Write support and parameter units. For binomial, check fixed n and common p; include the combination factor. For Poisson, convert rate to expected count for the exact interval. Translate at least/at most inequalities before summing, use complements for short tails, and keep exact expressions until the final step. Label approximation explicitly and do not infer a distribution from its mean alone.

Original practice: six single-correct MCQs

These are original questions, not previous-year questions. Choose exactly one option.

BP01. X~Bin(5,0.2). Find P(X=2).

A. 0.02048 B. 0.2048 C. 0.04 D. 0.67232

Answer: B

A: This omits C(5,2)=10 and counts one ordering only.

B: C(5,2)(.2)²(.8)³=.2048.

C: This omits the three failures and the number of success-position choices.

D: This is P(X≥1), not P(X=2).

BP02. For X~Bin(10,0.3), what are E[X] and Var(X)?

A. 3,3 B. 0.3,0.21 C. 3,2.1 D. 3,√2.1

Answer: C

A: Equal mean and variance is the Poisson identity, not the binomial identity.

B: These are the moments of one Bernoulli trial.

C: Mean=np=3 and variance=np(1−p)=2.1.

D: The second entry is standard deviation, not variance.

BP03. Requests follow a homogeneous Poisson process at two per minute. What is the probability of no requests in three minutes?

A. e^(−6) B. e^(−2) C. 1−e^(−6) D. 6e^(−6)

Answer: A

A: The interval mean is 2×3=6, and zero-count probability is e^(−6).

B: This uses the one-minute mean for a three-minute interval.

C: This is at least one request.

D: This is exactly one request in the interval.

BP04. X~Poisson(3). Find P(X≥2).

A. 1−e^(−3) B. (9/2)e^(−3) C. 1−3e^(−3) D. 1−4e^(−3)

Answer: D

A: This subtracts only zero and gives P(X≥1).

B: This is exactly two rather than at least two.

C: This subtracts exactly one but fails to subtract zero.

D: Subtract zero and one: 1−e^(−3)−3e^(−3).

BP05. Which experiment has an exact Bin(5,0.2) success count?

A. Five draws without replacement from ten items with two marked B. Stop at the first success with probability 0.2 per independent trial C. Five mutually independent trials, each with success probability 0.2 D. Five independent trials with probabilities 0.1,0.2,0.2,0.2,0.3

Answer: C

A: Without replacement, successes are dependent and the available composition changes.

B: The number of trials is not fixed at five.

C: Fixed n=5, common p=.2 and mutual independence satisfy the assumptions.

D: A common p is missing, even though the probabilities average .2.

BP06. For X~Bin(100,0.01), which pair gives exact P(X=0) followed by its Poisson approximation?

A. e^(−1),(.99)^100 B. (.99)^100,e^(−1) C. (.01)^100,e^(−1) D. (.99)^100,e^(−100)

Answer: B

A: This reverses exact and approximate answers.

B: Zero successes has exact probability (1−p)^n; λ=np=1 gives e^(−1) approximately.

C: (.01)^100 is all successes, not zero successes.

D: The Poisson parameter is np=1, not n=100.

Source and scope

Binomial and Poisson distributions are explicitly named in Section 1 Probability and Statistics of the official GATE 2026 CS syllabus. Original worked examples and MCQs. Uniform, normal and exponential distributions remain separate lessons; this batch does not complete the syllabus.

Official GATE 2026 CS syllabus (verified 2026-10-14).