MHT-CET Maths · Binomial Distribution

Mean, Variance and Standard Deviation of a Binomial Variable

For X ~ B(n, p) you never build the distribution table — the mean is np, the variance is npq, and the standard deviation is √(npq); these three shortcuts answer almost every MHT-CET question on the topic.

All Binomial Distribution notes MHT-CET Maths strategy

Why this matters

This subtopic is pure formula-recall turned into arithmetic: 15 PYQs sit here (8 EASY, 5 MODERATE, 2 HARD). The EASY band is direct np or npq once you read n and p off a with-replacement or coin-toss setup; the MODERATE and HARD bands reverse the process — given the mean and the variance you recover n and p, then compute a tail probability like P(X ≥ 1). The single most reliable check across every question is that the variance npq is always LESS than the mean np (because q < 1) — an answer with variance ≥ mean is wrong on sight.

Concept 1 of 4

The Mean of a Binomial Variable is np

Intuition

If a single trial succeeds with probability p, then over n independent trials you expect a fraction p of them to succeed — so the average number of successes is simply n times p. No distribution table is needed: the mean of X ~ B(n, p) is np, full stop.

Definition

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

Mean (expected value): $\mu = E(X) = np$ .
This is the number of trials times the single-trial success probability — it needs no summation, no $P(X=r)$ table.
The mean is the balance point of the distribution: for $B(n, \tfrac12)$ it sits at the centre $\tfrac{n}{2}$ .

Mean of a binomial variable

\mu = E(X) = np

nnumber of independent trials
pprobability of success on a single trial
qprobability of failure, q = 1 − p

Worked example

A biased coin with

P(\text{head}) = \tfrac{1}{3}

is tossed 9 times. Find the expected number of heads.

This is $X \sim B(n, p)$ with $n = 9$ , $p = \tfrac13$ .
Mean $= np = 9 \times \tfrac13$ .
So $\mu = 3$ .

Answer:

\mu = 3

heads

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Give the mean of $X \sim B(20, 0.1)$ .
2.
A die is rolled 12 times; X = number of sixes. Find the mean.
3.
For $B(n, \tfrac12)$ , where does the mean sit?
4.
State the mean of a binomial variable in words.

The mean is np, never p or p^n

For

X \sim B(n, p)

the mean is

np

— the trial count multiplied by the success probability. Writing the mean as

p

alone, or as

p^n

, or as

\tfrac{p}{n}

, are the standard distractors. With

n = 2

p = \tfrac1{13}

the mean is

\tfrac{2}{13}

, NOT

\tfrac{1}{169} = p^2

'With replacement' is what makes the trials binomial

Drawing cards WITH replacement keeps

p

constant across draws, so

X

is binomial and the mean is

np

. WITHOUT replacement the trials are dependent (hypergeometric) and

np

no longer applies. Read the words 'with replacement' before using the shortcut.

Concept 2 of 4

Variance is npq and Standard Deviation is the Square Root of npq

Intuition

Spread accumulates across trials the same way the mean does: each of the n trials contributes pq to the variance, so the total variance is npq. The standard deviation is just its square root, √(npq). Because q is a probability less than 1, the variance npq is always smaller than the mean np.

Definition

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

Variance: $\sigma^2 = \operatorname{Var}(X) = npq$ .
Standard deviation: $\sigma = \sqrt{npq}$ .
Key inequality: since $0 < q < 1$ , we always have $npq < np$ , i.e. variance is always less than the mean for a binomial variable.

You can also get the variance the long way via $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$ , but for a genuine binomial $npq$ is far faster.

Variance and standard deviation of a binomial variable

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

nnumber of independent trials
psuccess probability, q = 1 − p
npqthe variance — always less than the mean np

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 die is rolled 18 times. If X is the number of times a multiple of 3 appears, find the variance and standard deviation of X.

A multiple of 3 on a die is $\{3, 6\}$ , so $p = \tfrac26 = \tfrac13$ , $q = \tfrac23$ , $n = 18$ .
Variance $= npq = 18 \times \tfrac13 \times \tfrac23 = 4$ .
Standard deviation $= \sqrt{npq} = \sqrt{4} = 2$ .
Check: mean $= np = 6$ , and variance $4 < 6$ , as required.

Answer:

\sigma^2 = 4,\ \sigma = 2

Practice this conceptself-check · 4 quick reps

Try it yourself

A bag holds 8 red and 12 white balls. A ball is drawn, its colour noted, and replaced; this is done 25 times. Find the variance of the number of red balls drawn.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Find the variance of $X \sim B(3, \tfrac12)$ .
2.
Find the SD of $X \sim B(100, 0.5)$ .
3.
For $B(4, \tfrac12)$ , state the mean and variance.
4.
Can a binomial variable have mean 3 and variance 4?

From the bank · past-year question

Example 2Binomial DistributionEASY

A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one by one, with replacement, then the variance of the number of green balls drawn is

[Q150 · 16th May Shift 1 · 2023]

Variance is npq, not np or npq^2

The variance of

X \sim B(n, p)

npq

— the mean

np

multiplied by

q

. Forgetting the extra factor of

q

(using

np

) or over-counting it (using

npq^2

np^2q

) are the classic slips. For

B(10, \tfrac35)

: variance

= 10 \cdot \tfrac35 \cdot \tfrac25 = \tfrac{12}{5}

, not

6 = np

Variance is always smaller than the mean for a binomial variable

Since

q = 1 - p < 1

npq < np

— the variance can never equal or exceed the mean. Any option pairing (mean 2, variance 4) or (mean 2, variance 5) is impossible for a binomial variable and can be eliminated instantly; the only consistent partner of mean 2 with

p = \tfrac12

is variance 1.

SD is the square root of the variance, not of npq-then-forgotten

Standard deviation is

\sigma = \sqrt{npq}

. After computing the variance

npq

, take its square root: for variance 4 the SD is 2, for variance 25 the SD is 5. Reporting the variance where the SD is asked (or vice versa) loses an otherwise-correct problem.

Drill 8 more on variance is npq and standard deviation is the square root of npq

Concept 3 of 4

Recovering n and p from the Mean and Variance

Intuition

The mean np and the variance npq share the common factor np, so dividing the variance by the mean isolates q in a single step. Once you have q you have p = 1 − q, and then n = mean ÷ p. This unlocks any follow-up, because knowing n and p reconstructs the whole distribution.

Definition

Given a binomial's mean and variance, recover the parameters:

Divide variance by mean: $\dfrac{npq}{np} = q$ , so $q = \dfrac{\text{variance}}{\text{mean}}$ .
Then $p = 1 - q$ and $n = \dfrac{\text{mean}}{p}$ .
With $n$ and $p$ known you can compute any probability, e.g. $P(X = 0) = q^n$ , $P(X \ge 1) = 1 - q^n$ , or a lower tail $P(X \le r) = \sum_{k=0}^{r} {}^{n}C_{k}\,p^{k}q^{n-k}$ .

Recover q, then p and n

q = \dfrac{\sigma^2}{\mu} = \dfrac{npq}{np},\qquad p = 1 - q,\qquad n = \dfrac{\mu}{p}

Visualization · "at least 6 heads" is the shaded tail

Counts are C(8, k), each over a total of 2⁸ = 256. The shaded bars k = 6, 7, 8 give P(X ≥ 6) = (28 + 8 + 1)/256 = 37/256. Here the complement P(X ≤ 5) has six terms, so summing the three-bar tail directly is the shorter route.

Worked example

A binomial variable X has mean 6 and variance 2. Find n and p, and hence

P(X = 0)

Divide: $q = \dfrac{\sigma^2}{\mu} = \dfrac{2}{6} = \tfrac13$ .
So $p = 1 - q = \tfrac23$ .
And $n = \dfrac{\mu}{p} = \dfrac{6}{2/3} = 9$ .
Then $P(X = 0) = q^{n} = \left(\tfrac13\right)^{9} = \dfrac{1}{19683}$ .

Answer:

n = 9,\ p = \tfrac23,\ P(X=0) = \dfrac{1}{3^{9}}

Practice this conceptself-check · 4 quick reps

Try it yourself

The mean and variance of a binomial variable X are 4 and 3 respectively. Find n, p, and

P(X \ge 1)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Mean 8, variance 4: find q.
2.
Mean 8, variance 4: find n.
3.
Mean 2, variance 1: find n and p.
4.
For $B(4, \tfrac12)$ , evaluate $P(X \ge 1)$ .

From the bank · past-year question

Example 3Binomial DistributionMODERATE

If the mean and the variance of Binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than or equal to one is

[Q138 · 9th May Shift 1 · 2023]

Divide variance by mean to get q — not p

\dfrac{\text{variance}}{\text{mean}} = \dfrac{npq}{np} = q

, the FAILURE probability. Then

p = 1 - q

. Reading the quotient as

p

directly swaps the roles and (unless

p = q = \tfrac12

) gives the wrong

n

. For mean 8, variance 4:

q = \tfrac12

p = \tfrac12

n = 16

P(X = 0) is q^n, and P(X ≥ 1) = 1 − q^n

The zero-success probability uses the FAILURE probability raised to the trial count:

P(X=0) = q^{n}

, so

P(X \ge 1) = 1 - q^{n}

. For

B(4, \tfrac12)

P(X \ge 1) = 1 - \left(\tfrac12\right)^4 = \tfrac{15}{16}

. Using

p^{n}

by mistake computes the all-success probability instead.

For a lower tail sum the terms up to r, then divide by 2^n only if p = 1/2

P(X \le 2) = {}^{n}C_0 + {}^{n}C_1 + {}^{n}C_2

each times

p^{k}q^{n-k}

. When

p = q = \tfrac12

every term shares

\left(\tfrac12\right)^{n}

, so

P(X \le 2) = \dfrac{{}^{n}C_0 + {}^{n}C_1 + {}^{n}C_2}{2^{n}}

. For

n = 16

\dfrac{1 + 16 + 120}{2^{16}} = \dfrac{137}{2^{16}}

Drill 3 more on recovering n and p from the mean and variance

Concept 4 of 4

Solving When the Mean and Variance are Combined into One Equation

Intuition

Some questions do not hand you the mean and variance separately — instead they give a combination such as 'mean + variance = value' for a stated number of trials. Substitute np and npq into that equation and it becomes a single equation in p (with q = 1 − p), often a neat quadratic you solve and then reject the impossible root.

Definition

When the data is a combination of mean and variance for a known n:

Write **mean $= np$ and variance $= npq = np(1-p)$ **, with the given $n$ substituted in.
Sum condition: $np + npq = np(1 + q)$ ; replace $q = 1 - p$ and solve.
The result is an equation in $p$ alone (frequently a quadratic). **Reject any root with $p > 1$ or $p < 0$ ** — a probability must lie in $[0, 1]$ .
Once $p$ is fixed, back-substitute to report whichever quantity is asked (the variance, $p$ , etc.).

Sum of mean and variance

np + npq = np(1 + q) = np(2 - p)

Worked example

For a binomial distribution with 6 trials, the sum of the mean and the variance is

\dfrac{9}{2}

. Find p and the variance.

Mean $= 6p$ , variance $= 6p(1-p)$ , so sum $= 6p(2 - p) = \tfrac92$ .
Divide by 6: $p(2 - p) = \tfrac{3}{4}$ , i.e. $2p - p^2 = \tfrac34$ .
Rearrange: $4p^2 - 8p + 3 = 0 \Rightarrow (2p - 1)(2p - 3) = 0$ , so $p = \tfrac12$ (reject $p = \tfrac32 > 1$ ).
Variance $= 6 \times \tfrac12 \times \tfrac12 = \tfrac32$ .

Answer:

p = \tfrac12,\ \text{variance} = \tfrac32

Practice this conceptself-check · 4 quick reps

Try it yourself

The sum of the mean and the variance of a binomial distribution for 4 trials is

\dfrac{7}{4}

. Find p.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
For 10 trials, mean + variance $= \tfrac{15}{2}$ . Find q.
2.
For 10 trials with $p = \tfrac12$ , find the variance.
3.
Solve $25p^2 - 50p + 9 = 0$ for a probability.
4.
Why reject a root p > 1 in these problems?

From the bank · past-year question

Example 4Binomial DistributionMODERATE

If the sum of mean and variance of a Binomial Distribution is

\dfrac{15}{2}

for 10 trials, then the variance is

[Q146 · 10th May Shift 2 · 2024]

Substitute q = 1 − p to reduce the sum to a single-variable equation

np + npq = np(1 + q)

. Do NOT treat

p

and

q

as independent — replace

q

with

1 - p

so you get one equation in

p

. For 10 trials,

10p(1+q) = \tfrac{15}{2}

becomes

(1-q)(1+q) = \tfrac34

, i.e.

1 - q^2 = \tfrac34

, giving

q = \tfrac12

Reject the root outside [0, 1]

The quadratic in

p

usually has two roots and one is impossible. From

25p^2 - 50p + 9 = 0

the roots are

p = \tfrac15

and

p = \tfrac95

; only

p = 0.2

is a valid probability. Choosing the root greater than 1 (or a negative root) is the built-in distractor.

Read whether p, q, or the variance is being asked

These questions sometimes ask for

p

, sometimes for the variance, sometimes for

q

. After solving for

p

, back-substitute: for 10 trials with

p = \tfrac12

the variance is

npq = 2.5

, while the answer to 'find p' would be

0.5

. Reporting the wrong quantity is the last-step slip.

Drill 1 more on solving when the mean and variance are combined into one equation

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 (4)

The Mean of a Binomial Variable is np
Mean of a binomial variable
$\mu = E(X) = np$
Variance is npq and Standard Deviation is the Square Root of npq
Variance and standard deviation of a binomial variable
$\sigma^2 = npq,\qquad \sigma = \sqrt{npq},\qquad npq < np$
Recovering n and p from the Mean and Variance
Recover q, then p and n
$q = \dfrac{\sigma^2}{\mu} = \dfrac{npq}{np},\qquad p = 1 - q,\qquad n = \dfrac{\mu}{p}$
Solving When the Mean and Variance are Combined into One Equation
Sum of mean and variance
$np + npq = np(1 + q) = np(2 - p)$

Watch out for (11)

The mean is np, never p or p^n
→ The Mean of a Binomial Variable is np
'With replacement' is what makes the trials binomial
→ The Mean of a Binomial Variable is np
Variance is npq, not np or npq^2
→ Variance is npq and Standard Deviation is the Square Root of npq
Variance is always smaller than the mean for a binomial variable
→ Variance is npq and Standard Deviation is the Square Root of npq
SD is the square root of the variance, not of npq-then-forgotten
→ Variance is npq and Standard Deviation is the Square Root of npq
Divide variance by mean to get q — not p
→ Recovering n and p from the Mean and Variance
P(X = 0) is q^n, and P(X ≥ 1) = 1 − q^n
→ Recovering n and p from the Mean and Variance
For a lower tail sum the terms up to r, then divide by 2^n only if p = 1/2
→ Recovering n and p from the Mean and Variance
Substitute q = 1 − p to reduce the sum to a single-variable equation
→ Solving When the Mean and Variance are Combined into One Equation
Reject the root outside [0, 1]
→ Solving When the Mean and Variance are Combined into One Equation
Read whether p, q, or the variance is being asked
→ Solving When the Mean and Variance are Combined into One Equation

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Binomial DistributionEASY

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Then mean of number of kings is

[Q150 · 3rd May Shift 2 · 2023]

Example 2Binomial DistributionHARD

Let a random variable X have a Binomial distribution with mean 8 and variance 4. If

P(X\leq 2) = \frac{K}{2^{16}}

, then K is

[Q111 · 11th May Shift 2 · 2024]

Example 3Binomial DistributionMODERATE

If the sum of the mean and the variance of a Binomial distribution for 5 trials is 1.8, then the value of

p

[Q124 · 13th May Shift 2 · 2024]

Example 4Binomial DistributionEASY

Three fair coins numbered 1 and 0 are tossed simultaneously. Then variance Var(X) of the probability distribution of random variable X, where X is the sum of numbers on the uppermost faces, is

[Q133 · 9th May Shift 1 · 2024]

Example 5Binomial DistributionMODERATE

If the mean and the variance of a Binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than one is equal to

[Q142 · 9th May Shift 2 · 2023]

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