Binomial Distribution Mean and Variance
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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.