Uniform, Bernoulli & Binomial

A fair die. A single coin toss. Twenty coin tosses, and how many come up heads. Almost every discrete-variable question on the exam is one of these three setups dressed up in different words. Each one is a tiny recipe — point at the situation, read off the PMF, the mean, the variance. No calculus, no integrals; just arithmetic once you spot which recipe fits. The same Bernoulli-and-binomial counting reappears the day you model click-through rates, A/B-test conversions, or “how many of these 1000 requests will fail” in real data work.

The three recipes

Start with the simplest and build up. A discrete uniform says every outcome is equally likely — a fair die. A Bernoulli is a single yes/no trial — one coin flip. A Binomial counts how many yeses you get across many such flips.

Discrete uniform over the integers a, a+1, …, b (that is n = b − a + 1 values): each has probability 1/n. By symmetry the mean sits in the middle, (a+b)/2. A fair die: mean (1+6)/2 = 3.5.

Bernoulli(p) is one trial with two outcomes — success (X = 1) with probability p, failure (X = 0) with probability 1 − p.

• mean = 1·p + 0·(1−p) = p
• variance = p(1 − p) (largest at p = 0.5, zero at p = 0 or 1).

Binomial(n,p) counts the number of successes in n independent trials, each a Bernoulli(p). The number of ways to place exactly k successes among n slots is C(n,k) — the same combination from the counting lesson — and each such sequence has probability p^k (1−p)^(n−k). So

P(X = k) = C(n,k) · p^k · (1 − p)^(n − k),   k = 0, 1, …, n

Because a Binomial is the sum of n independent Bernoulli(p) trials, its mean and variance are just n copies of the Bernoulli ones added up:

• mean = np
• variance = np(1 − p)

This requires the three binomial conditions: a fixed number of trials n, independent trials, and a constant success probability p on every trial.

Here is the PMF of Binomial(n = 10, p = 0.5) — symmetric, peaking at the mean k = 5:

How GATE asks this

Almost always a NAT. The classic patterns: “a fair coin is tossed n times, find P(exactly k heads)”, or “components fail independently with probability p; find the mean/variance of the number that fail.” Spot the three conditions (fixed n, independent, constant p), read off n, p, k, and plug into the PMF or the np / np(1−p) formulas. The C(n,k) factor is the only place the counting lesson shows up.

Worked example

Let X ~ Binomial(n = 5, p = 0.5). Find P(X = 3), the mean, and the variance.

Apply the PMF with k = 3:

P(X = 3) = C(5,3) · (0.5)^3 · (0.5)^2
         = C(5,3) · (0.5)^5
         = 10 · (1/32)
         = 10/32
         = 0.3125

mean     = np        = 5 · 0.5        = 2.5
variance = np(1 − p) = 5 · 0.5 · 0.5  = 1.25

C(5,3) = 10, and since p = 1 − p = 0.5 the two powers combine into (0.5)^5 = 1/32. So P(X = 3) = 0.3125, mean = 2.5, variance = 1.25.

Quick check