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Quantitative Aptitude · Theoretical Distributions

Mean, Variance and Properties of Binomial Distribution

Updated 1 October 2026 · Fact-checked

For a binomial distribution with n trials and success probability p, the mean is np and the variance is npq, where q = 1 − p. To find n and p from a given mean and variance, divide variance by mean to get q, then p = 1 − q, then n = mean ÷ p.

Understand Mean, Variance and Properties of Binomial Distribution

A binomial distribution counts the number of successes in n independent trials. Each trial has only two outcomes, success or failure. The chance of success, p, is the same in every trial. The number of trials, n, is fixed in advance.

The mean is the average number of successes you expect. If you toss a fair coin 10 times, you expect 5 heads. That is n × p = 10 × 0.5. So the mean is np.

The variance measures how much the number of successes spreads around the mean. It is npq, where q = 1 − p. The standard deviation is √(npq). For 0 < p < 1, q is less than 1, so npq < np. So for a binomial distribution with 0 < p < 1, mean is greater than variance. If a question gives variance greater than mean, it cannot be binomial. The degenerate cases p = 0 and p = 1 are excluded: there the variance is 0 and the outcome is certain.

The mode is the most likely number of successes. Compute (n + 1)p. If it is not a whole number, the mode is its integer part. If it is a whole number, there are two modes: (n + 1)p and (n + 1)p − 1.

The additive property: if X is binomial with (n1, p) and Y is binomial with (n2, p), and X and Y are independent, then X + Y is binomial with (n1 + n2, p). The value of p must be the same for both. Fitting a binomial means finding n and p from data, then computing expected frequencies as N × P(X = r).

Key formulas to remember

Mean
μ = np
Average number of successes in n trials.
Variance
σ² = npq, where q = 1 − p
For 0 < p < 1, q < 1, so the variance is less than the mean.
Standard deviation
σ = √(npq)
Take the square root of the variance.
Probability of r successes
P(X = r) = nCr × p^r × q^(n − r), r = 0, 1, ..., n
Used for fitting and for mode checks.
Mode
Compute (n + 1)p. If not an integer, mode = integer part. If an integer, two modes: (n + 1)p and (n + 1)p − 1
Always check whether (n + 1)p is a whole number.
Additive property
X ~ B(n1, p), Y ~ B(n2, p), independent ⇒ X + Y ~ B(n1 + n2, p)
Needs the same p and independence.
Finding n and p
q = variance ÷ mean; p = 1 − q; n = mean ÷ p
Valid only when variance < mean, which holds for 0 < p < 1.
Expected frequency in fitting
Expected frequency = N × P(X = r)
N is the total frequency (number of sets of trials).

How to solve Mean, Variance and Properties of Binomial Distribution questions

Use this method for any question on mean, variance or parameters of a binomial distribution.

  1. 1Write down what is given: n, p, mean, variance or standard deviation.
  2. 2Check the conditions: fixed n, two outcomes, constant p, independent trials.
  3. 3If mean and variance are both given, check that variance < mean. If not, the distribution cannot be binomial.
  4. 4Find q = variance ÷ mean, then p = 1 − q, then n = mean ÷ p.
  5. 5Check that n comes out as a whole number. Standard deviation is the square root of variance, so square it first if needed.
  6. 6If asked for the mode, compute (n + 1)p and apply the integer rule.
  7. 7For fitting, find P(X = r) for each r and multiply by N.
  8. 8Verify by checking np and npq against the original data.

Quickest way: Ratio shortcut and option elimination

When to use it: Use in MCQs where mean and variance are given and you must find n, p or q.

  1. Compute variance ÷ mean. This is q directly.
  2. Get p = 1 − q. Often p is a simple fraction like 1/3 or 1/4.
  3. Compute n = mean ÷ p. Mentally check np equals the mean.
  4. If variance ≥ mean, eliminate any option that calls it binomial.
  5. For mode, compute (n + 1)p only. Do not list all probabilities.
  6. If the question gives standard deviation, square it first. Many students forget this.
  7. Skip any question that needs a long table of nCr values unless you have time left. A wrong answer costs 0.25.

Common mistakes in Mean, Variance and Properties of Binomial Distribution

  • Using variance = np instead of npq.

    Students mix up the mean and variance formulas.

    Fix: Remember that variance has the extra q. For 0 < p < 1, q < 1, so variance is smaller than mean.

  • Treating the given standard deviation as the variance.

    The question says SD, but students plug it directly into npq.

    Fix: Square the SD first. Then use npq = SD².

  • Accepting a distribution with variance greater than mean as binomial.

    Students do not check the condition before computing.

    Fix: Check variance < mean first. If it fails, no such binomial exists.

  • Giving one mode when (n + 1)p is an integer.

    Students take the integer part without checking.

    Fix: If (n + 1)p is a whole number, there are two modes: (n + 1)p and (n + 1)p − 1.

  • Applying the additive property when p differs.

    Students see two binomials and add n values automatically.

    Fix: Check p is the same in both. Otherwise the sum is not binomial.

  • Using n = variance ÷ pq without finding p first, or getting a fractional n.

    Steps get mixed up under time pressure.

    Fix: Follow the order q, then p, then n. n must be a whole number.

Worked examples

Example 1

The mean and variance of a binomial distribution are 6 and 4 respectively. What is the value of n? Options: (a) 12 (b) 18 (c) 9 (d) 24

Show the solution
  1. q = variance ÷ mean = 4 ÷ 6 = 2/3.
  2. p = 1 − 2/3 = 1/3.
  3. n = mean ÷ p = 6 ÷ (1/3) = 18.
  4. Check: np = 18 × 1/3 = 6 and npq = 6 × 2/3 = 4. Both match.

Answer: (b) 18

Example 2

If X and Y are independent binomial variables with parameters (n = 5, p = 1/3) and (n = 7, p = 1/3), what is the variance of X + Y? Options: (a) 4 (b) 8/3 (c) 16/3 (d) 12

Show the solution
  1. The p is the same and the variables are independent, so X + Y is binomial with n = 5 + 7 = 12 and p = 1/3.
  2. q = 2/3.
  3. Variance = npq = 12 × 1/3 × 2/3 = 8/3.

Answer: (b) 8/3

Example 3

For a binomial distribution with n = 10 and p = 0.5, what is the mode? Options: (a) 4 (b) 5 (c) 6 (d) 4 and 5

Show the solution
  1. Compute (n + 1)p = 11 × 0.5 = 5.5.
  2. 5.5 is not a whole number, so the mode is its integer part.
  3. Mode = 5.
  4. Check with the mean: np = 5, and for p = 0.5 the distribution is symmetric about 5.

Answer: (b) 5

Exam tips

  • Always test variance < mean first. Questions often include a trap where the given values are impossible for a binomial.
  • When SD is given, square it before using npq.
  • For mode, compute (n + 1)p and check whether it is a whole number. This is a favourite trap.
  • Look for friendly fractions after finding q. If n is not a whole number, recheck your arithmetic.
  • Fitting questions are long. Attempt them only after finishing the quicker mean and variance questions.

Practice questions from Theoretical Distributions

Mean, Variance and Properties of Binomial Distribution: frequently asked questions

Why is the mean of a binomial distribution greater than the variance?

Mean is np and variance is npq. When 0 < p < 1, q = 1 − p is less than 1, so npq is less than np. So the variance is smaller than the mean. In the degenerate cases p = 0 or p = 1 the variance is 0 (the outcome is certain), so these cases are excluded.

How do I find n and p from mean and variance?

Divide variance by mean to get q. Then p = 1 − q. Finally n = mean ÷ p. Check that n is a whole number.

What is fitting a binomial distribution?

You use observed data to find n and p, usually from the mean. Then you compute P(X = r) for each r and multiply by the total frequency N. This gives the expected frequencies.

When does a binomial distribution have two modes?

When (n + 1)p is a whole number. The two modes are (n + 1)p and (n + 1)p − 1.