E(X) = np, Var(X) = npq. Worked example.

Probability Distribution — Mean and Variance of Binomial

Question

A binomial distribution has $n = 10$ trials and probability of success $p = \frac{1}{3}$ . Find the mean and variance of the distribution.

This type appeared in JEE Main 2024 — straightforward formula application, but students consistently lose marks by confusing $q$ with $p$ .

Solution — Step by Step

From the problem: $n = 10$ , $p = \frac{1}{3}$ .

Since success and failure are complementary, $q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$ .

For a binomial distribution $X \sim B(n, p)$ , the mean (expected value) is:

E(X) = np

Substituting: $E(X) = 10 \times \frac{1}{3} = \boxed{\frac{10}{3}}$

Variance for a binomial distribution is:

\text{Var}(X) = npq

Substituting: $\text{Var}(X) = 10 \times \frac{1}{3} \times \frac{2}{3} = \frac{20}{9}$

Standard deviation $= \sqrt{\text{Var}(X)} = \sqrt{\frac{20}{9}} = \frac{2\sqrt{5}}{3}$ .

JEE sometimes asks for SD instead of variance — always check what the question is asking for.

Final answers: Mean $= \dfrac{10}{3}$ , Variance $= \dfrac{20}{9}$

Why This Works

Each trial in a binomial experiment is independent, with two outcomes: success (probability $p$ ) or failure (probability $q = 1-p$ ). After $n$ such trials, we expect $np$ successes on average — this is the mean. If $p = \frac{1}{3}$ and we run 10 trials, we expect roughly $3.33$ successes.

The variance formula $npq$ captures spread. Notice that variance is maximised when $p = q = \frac{1}{2}$ (maximum uncertainty), and shrinks toward zero when $p$ is near 0 or 1 (almost certain outcome either way). This is why $npq$ has both $p$ and $q$ in it — both probabilities contribute to uncertainty.

These two results are worth memorising as a pair. In board exams, even a 1-mark MCQ can test this directly. In JEE, you’ll need them inside longer problems involving moment generating functions or inequality questions.

X \sim B(n, p)

E(X) = np \qquad \text{Var}(X) = npq \qquad \text{SD} = \sqrt{npq}

where $q = 1 - p$

Alternative Method — Using First Principles

We can derive the mean from the definition $E(X) = \sum_{r=0}^{n} r \cdot P(X = r)$ .

E(X) = \sum_{r=0}^{n} r \cdot \binom{n}{r} p^r q^{n-r}

The $r = 0$ term vanishes. For $r \geq 1$ , use the identity $r\binom{n}{r} = n\binom{n-1}{r-1}$ :

E(X) = np \sum_{r=1}^{n} \binom{n-1}{r-1} p^{r-1} q^{n-r} = np(p+q)^{n-1} = np

Since $p + q = 1$ , the sum collapses to 1. Same result, but now you see why the formula is $np$ — it falls directly out of the binomial theorem.

You won’t need this derivation in JEE Main, but it appears in proofs asked in board exams (CBSE Class 12 Chapter 13). If the question says “derive the mean of a binomial distribution,” this is the method to show.

Common Mistake

Writing Var(X) = np instead of npq. Students who’ve just finished Poisson distribution (where mean = variance = $\lambda$ ) carry that instinct into binomial problems. For binomial, mean and variance are not equal unless $q = 1$ , which is impossible. Always write $\text{Var}(X) = npq$ — the $q$ is not optional.

Another version of this mistake: computing $q$ as $\frac{p}{1-p}$ instead of $1 - p$ . Keep it simple: $q = 1 - p$ , nothing more.

Quick Sanity Check: Variance must always be less than or equal to the mean for a binomial distribution, because $\text{Var}(X) = npq = np \cdot q \leq np = E(X)$ (since $q \leq 1$ ). If your variance comes out larger than your mean, something went wrong.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using First Principles

Common Mistake

Try These Next