# Binomial Distribution Mean and Variance

> Binomial distribution mean and variance for JEE Mathematics: np, npq, mode, E(X squared), addition property and recognition traps explained.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/binomial-distribution-mean-variance
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Binomial Distribution Mean and Variance", PrepElephant, https://prepelephant.com/topics/jee/mathematics/binomial-distribution-mean-variance

## Direct answer

For X ~ B(n, p) — n independent Bernoulli trials, each succeeding with probability p — the mean is np, the variance is npq with q = 1 − p, and the standard deviation is √(npq). Because q < 1, a binomial variable always has variance strictly less than its mean, which instantly disqualifies any candidate data with Var ≥ mean. The probabilities come from P(X = r) = C(n, r)·p^r·q^(n−r); the mode is the integer part of (n + 1)p, with two adjacent modes when (n + 1)p is itself an integer. Two derived results do heavy exam duty: E(X²) = npq + n²p², and independent binomials with the same p add to B(n₁ + n₂, p).

## What you must remember

- **Mean and variance:** E(X) = np, Var(X) = npq, q = 1 − p. Variance < mean always — the standard elimination line for multiple-correct questions.
- **E(X²) shortcut:** E(X²) = Var + (mean)² = npq + n²p², so questions on E(X²) never need the summation Σr²·P(X = r).
- **Mode:** r = [(n+1)p]; if (n+1)p is an integer m, then both m − 1 and m are modes with equal probability — a favourite single-correct distinction.
- **Distribution formula:** P(X = r) = C(n, r)·p^r·q^(n−r) for r = 0 to n; "at least one" is 1 − q^n, by far the most used cumulative.
- **Addition property:** B(n₁, p) + B(n₂, p) = B(n₁ + n₂, p) requires the same p; with different success probabilities the sum is not binomial at all.
- **Symmetry case:** p = 1/2 makes the distribution symmetric about n/2, so mean = median = mode = n/2.
- **Recurrence for probabilities:** P(r+1)/P(r) = ((n − r)/(r + 1))·(p/q) — build the full table from P(0) = q^n without recomputing combinations.

## From mean to everything

A classic formulation: X ~ B(n, p) has mean 4 and variance 3 — recover everything. Divide: npq/np = q = 3/4, so p = 1/4, and then n = mean/p = 16. Now the whole question unlocks: P(X ≥ 1) = 1 − q^16 = 1 − (3/4)^16; the mode is [(17)(1/4)] = [4.25] = 4; E(X²) = 3 + 16 = 19. Three different question types collapsed in under a minute because the two given numbers were converted to structure first.

The same logic inverted appears in JEE Main numerical slots: given n and p, they ask variance (one multiplication), or they ask the ratio of mean to SD, which is √(np/q). Notice what never changes: the check that variance sits strictly below the mean. If a question hands you a "binomial" with mean 5 and variance 6, the correct response is not computation but rejection — no such p, q in (0, 1) exists. Recognition questions of this kind, where the answer is "no such distribution", reward students who know the inequality rather than only the formula.

## Recognition is the real test

JEE Main examines the formulas directly; JEE Advanced examines whether the experiment deserves the formulas. Drawing balls with replacement — binomial. Drawing without replacement — hypergeometric, not binomial, because p drifts after each draw. A tennis player serving at 60% for a whole match — binomial only while p and independence hold; if p rises with confidence, the model dies. The trap that recurs: "10 questions, each guessed from 4 options" is B(10, 1/4) with mean 2.5 and variance 1.875, but "guess until 3 correct" is not binomial — it is negative binomial with mean 3 × 4 = 12. Also watch the mode edge case: for B(9, 1/5), (n+1)p = 2 exactly, so both r = 1 and r = 2 are modes, and a question asking "the mode" singular expects you to know there are two.

## Frequently asked questions

### If a binomial variate has mean 4 and variance 3, what are n and p?

Dividing gives q = 3/4, so p = 1/4 and n = 4 × 4 = 16.

### What is the mode of a binomial distribution?

The greatest integer in (n + 1)p; when (n + 1)p is an integer, that value and the one below it are joint modes with equal probability.

### Can binomial variance ever exceed the binomial mean?

Never: npq < np because 0 < q < 1, so any "binomial" with variance ≥ mean is impossible.

### Is drawing balls without replacement a binomial experiment?

No — the success probability changes after every draw (hypergeometric); binomial requires independent trials with constant p.

### How do you compute P(X ≥ 1) without summing terms?

Use the complement: P(X ≥ 1) = 1 − P(X = 0) = 1 − q^n, a one-term calculation.