NDA Maths · Binomial Distribution

Mean, Variance, and Recovering the Parameters

B(n, p) has mean np and variance npq, and most PYQs run this backwards — given the mean and variance, recover n and p.

All Binomial Distribution notes NDA Maths strategy

Why this matters

The other 15 PYQs live here, and they are remarkably formulaic: two facts, mean = np and variance = npq, generate almost every question. The signature move is the back-solve — divide variance by mean to get q, then read off p and n. A second family gives you a relation between two probabilities (like 9·P(X=4) = P(X=2)) and asks for p. Both reduce to one or two lines once you know the pattern.

Test yourself — a quick 17-question Binomial Distribution recall quiz, instant score.

Concept 1 of 7: Why the Mean Is np and the Variance npq

Think of

X

as a sum of

n

one-or-zero scores, one per trial. Each trial contributes an average of

p

and the contributions add up; because the trials are independent, their variances add too.

Definition

Write $X = I_1 + I_2 + \dots + I_n$ , where $I_j = 1$ on success and $0$ on failure.

One trial: mean $= p$ , variance $= pq$ .
Means always add: $E(X) = np$ .
Variances add for INDEPENDENT trials: $\operatorname{Var}(X) = npq$ .

That independence is exactly why the binomial conditions matter — drop it and the variance formula breaks.

Mean and variance of one trial

E(I) = p, \qquad \operatorname{Var}(I) = pq

Worked example

Use the single-trial values to write the mean and variance of

B(n, p)

Concept 2 of 7: Mean, Variance, and Standard Deviation

Given

n

and

p

, the three summary numbers drop straight out. Standard deviation is just the square root of the variance — and the variance is always smaller than the mean.

Definition

For $X \sim B(n, p)$ with $q = 1 - p$ :

Mean: $\mu = np$ .
Variance: $\sigma^2 = npq$ .
Standard deviation: $\sigma = \sqrt{npq}$ .

A built-in check: since $q < 1$ , the variance $npq$ is always less than the mean $np$ . If a computed variance exceeds the mean, something is wrong.

The three summary measures

\mu = np, \qquad \sigma^2 = npq, \qquad \sigma = \sqrt{npq}

Visualization · mean np at the centre, spread √(npq)

For B(10, 0.4): mean np = 4 (the dashed centre), variance npq = 2.4, so σ = √2.4 ≈ 1.55 (the shaded band). Notice σ² = 2.4 is less than the mean 4 — the variance npq is always below the mean np because q < 1, a quick sanity check on any answer.

Worked example

A fair coin is tossed 12 times. Find the mean, variance, and standard deviation of the number of heads.

The same idea in a real exam question:

NDA · Apr 2019 · Q114Moderate

Example 2 · Binomial Distribution · Mean, Variance, and Parameter Estimation in B(n, p)

In eight throws of a die, 5 or 6 is considered a success. The mean and standard deviation of total number of successes is respectively given by

Standard deviation is √(npq), not npq

Questions love to give the standard deviation and call it the variance, or vice versa. Square the SD to get the variance before using

\sigma^2 = npq

— e.g. SD

= \sqrt2

means variance

= 2

Concept 3 of 7: Recovering n and p from the Moments

Given the mean and the variance, dividing one by the other cancels

np

and hands you

q

directly. From

q

you get

p

, and from the mean you get

n

Definition

The standard back-solve:

\frac{\operatorname{Var}(X)}{E(X)} = \frac{npq}{np} = q.

Step 1: $q = \dfrac{\text{variance}}{\text{mean}}$ , then $p = 1 - q$ .
Step 2: $n = \dfrac{\text{mean}}{p}$ .

If you are handed the standard deviation, square it to the variance first.

Divide variance by mean to get q

q = \dfrac{\sigma^2}{\mu} = \dfrac{npq}{np}

Worked example

A binomial distribution has mean 8 and standard deviation 2. Find n and p.

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

NDA · Apr 2017 · Q120Moderate

Example 3 · Binomial Distribution · Mean, Variance, and Parameter Estimation in B(n, p)

The mean and standard deviation of a binomial distribution are 12 and 2 respectively. What is the number of trials?

Variance over mean gives q, not p

The ratio

\sigma^2/\mu = q

(the FAILURE probability). Forgetting the final

p = 1 - q

step lands you on the complement and the wrong

n

Drill 7 more on recovering n and p from the moments

Concept 4 of 7: When You Are Given a Relation, Not the Values

Sometimes the question gives a relationship like 'the mean is three times the variance' instead of numbers. Substituting

np

and

npq

into the relation still collapses to a single equation in

q

Definition

Replace the words with $\mu = np$ and $\sigma^2 = npq$ , then cancel the common $np$ :

\mu = c\,\sigma^2 \ \Longrightarrow\ np = c\,(npq) \ \Longrightarrow\ 1 = cq \ \Longrightarrow\ q = \tfrac1c.

So 'mean

= c \times

variance' gives

q = \tfrac1c

immediately. With

p

known you can then find any probability, given

n

Mean equals c times variance

np = c\,(npq) \ \Longrightarrow\ q = \dfrac{1}{c}

Worked example

In a binomial distribution the mean is 4 times the variance. Find p.

Practice this concept2 quick reps

The same idea in a real exam question:

NDA · Apr 2018 · Q101Moderate

Example 4 · Binomial Distribution · Mean, Variance, and Parameter Estimation in B(n, p)

In Binomial distribution, mean is three times variance. What is the probability of exactly 3 successes in 5 trials?

Cancel np, do not cancel the wrong factor

From

np = c\,npq

, the surviving factor is

q

(giving

q = 1/c

). Cancelling to leave

p

instead — a common slip — inverts the answer.

Concept 5 of 7: Finding p from a Probability Equation

When two binomial probabilities are set equal (or in a fixed ratio), the messy parts cancel. Write both with the formula, divide, and a clean equation in

p

survives.

Definition

Given a relation between $P(X = a)$ and $P(X = b)$ , write each as $\binom{n}{k}p^k q^{\,n-k}$ and take the ratio so common powers cancel. Two handles make this fast:

Binomial coefficients are symmetric: $\binom{n}{a} = \binom{n}{n-a}$ (e.g. $\binom{6}{4} = \binom{6}{2}$ ), so they often cancel outright.
A ratio $\dfrac{P(X=b)}{P(X=a)}$ reduces to powers of $p$ and $q$ only.

Solve the resulting equation (take the positive root, since $0 < p < 1$ ).

Ratio of two probabilities

\dfrac{P(X=b)}{P(X=a)} = \dfrac{\binom{n}{b}}{\binom{n}{a}}\, p^{\,b-a}\, q^{\,a-b}

Worked example

For

X \sim B(5, p)

P(X = 2) = P(X = 3)

. Find p.

Practice this conceptself-check

The same idea in a real exam question:

NDA · Sep 2023 · Q115Moderate

Example 5 · Binomial Distribution · Mean, Variance, and Parameter Estimation in B(n, p)

B(n,p)

n=6

and

9P(X=4)=P(X=2)

. What is

p

equal to?

Use coefficient symmetry before brute force

Spotting

\binom{6}{4} = \binom{6}{2}

cancels the coefficients in one step. Expanding them numerically still works but invites arithmetic slips — and forgetting to take the positive root of

9p^2 = (1-p)^2

loses the intended answer.

Drill 3 more on finding p from a probability equation

Concept 6 of 7: Variance Is Unchanged by Y = n − X

Counting failures instead of successes (

Y = n - X

) is the same distribution with

p

and

q

swapped — and since variance is

npq

, swapping

p

and

q

leaves it untouched.

Definition

If $X \sim B(n, p)$ and $Y = n - X$ (so $Y \sim B(n, q)$ ):

Variance is symmetric in $p, q$ : $\operatorname{Var}(Y) = n q p = npq = \operatorname{Var}(X)$ .
The means are complementary: $E(Y) = nq = n - np$ .

So whenever $X + Y = n$ with $X$ binomial, $\operatorname{Var}(Y) = \operatorname{Var}(X)$ — no recomputation needed.

Variance survives the swap

\operatorname{Var}(n - X) = npq = \operatorname{Var}(X)

Worked example

X \sim B(50, \tfrac35)

and

Y = 50 - X

. Find

\operatorname{Var}(Y)

The same idea in a real exam question:

NDA · Apr 2023 · Q105Easy

Example 6 · Binomial Distribution · Mean, Variance, and Parameter Estimation in B(n, p)

Let

X

and

Y

be two random variables such that

X+Y=100

. If

X

follows Binomial distribution with parameters

n=100

and

p=\frac{4}{5}

, what is the variance of

Y

Variance does not flip; the mean does

Y = n - X

leaves the variance equal to

\operatorname{Var}(X)

, but its mean becomes

n - np

. Do not 'adjust' the variance for the swap — only the mean changes.

Concept 7 of 7: The Symmetric Case: Mean = n/2 when p = ½

When success and failure are equally likely, the distribution is symmetric and balances exactly in the middle, at

n/2

. A distribution whose frequencies are the binomial coefficients is precisely this case.

Definition

For $p = \tfrac12$ : $\mu = np = \tfrac{n}{2}$ and $\sigma^2 = npq = \tfrac{n}{4}$ . A variable taking values $0, 1, \dots, n$ with frequencies $\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}$ has total frequency $2^n$ and mean

\frac{\sum_{k} k\binom{n}{k}}{\sum_{k}\binom{n}{k}} = \frac{n\,2^{\,n-1}}{2^{n}} = \frac{n}{2},

the same

n/2

— because those frequencies are exactly

B(n, \tfrac12)

up to the factor

2^n

Symmetric binomial mean

p = \tfrac12 \ \Longrightarrow\ \mu = \dfrac{n}{2}

Worked example

A fair coin is tossed 8 times. Without summing, state the mean number of heads.

The same idea in a real exam question:

NDA · Sep 2017 · Q115Moderate

Example 7 · Binomial Distribution · Mean, Variance, and Parameter Estimation in B(n, p)

If a variable takes values

0, 1, 2, 3, \ldots, n

with frequencies

1, C(n,1), C(n,2), C(n,3), \ldots, C(n,n)

respectively, then the arithmetic mean is

Symmetry needs p = ½, not just 'two outcomes'

The mean is

n/2

ONLY when

p = q = \tfrac12

. With a biased trial the peak shifts to

np

— do not default to

n/2

unless success and failure are equally likely.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (7)

Why the Mean Is np and the Variance npq
Mean and variance of one trial
$E(I) = p, \qquad \operatorname{Var}(I) = pq$
Mean, Variance, and Standard Deviation
The three summary measures
$\mu = np, \qquad \sigma^2 = npq, \qquad \sigma = \sqrt{npq}$
Recovering n and p from the Moments
Divide variance by mean to get q
$q = \dfrac{\sigma^2}{\mu} = \dfrac{npq}{np}$
When You Are Given a Relation, Not the Values
Mean equals c times variance
$np = c\,(npq) \ \Longrightarrow\ q = \dfrac{1}{c}$
Finding p from a Probability Equation
Ratio of two probabilities
$\dfrac{P(X=b)}{P(X=a)} = \dfrac{\binom{n}{b}}{\binom{n}{a}}\, p^{\,b-a}\, q^{\,a-b}$
Variance Is Unchanged by Y = n − X
Variance survives the swap
$\operatorname{Var}(n - X) = npq = \operatorname{Var}(X)$
The Symmetric Case: Mean = n/2 when p = ½
Symmetric binomial mean
$p = \tfrac12 \ \Longrightarrow\ \mu = \dfrac{n}{2}$

Watch out for (6)

Standard deviation is √(npq), not npq
→ Mean, Variance, and Standard Deviation
Variance over mean gives q, not p
→ Recovering n and p from the Moments
Cancel np, do not cancel the wrong factor
→ When You Are Given a Relation, Not the Values
Use coefficient symmetry before brute force
→ Finding p from a Probability Equation
Variance does not flip; the mean does
→ Variance Is Unchanged by Y = n − X
Symmetry needs p = ½, not just 'two outcomes'
→ The Symmetric Case: Mean = n/2 when p = ½

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