MHT-CET Maths · Formula sheet

Binomial Distribution formulas

19 formulas and 51 common traps for MHT-CET Maths Binomial Distribution, grouped by subtopic.

The Binomial Setting and Probability Mass Function

The Binomial Setting — n Fixed Independent Success or Failure Trials

Binomial variable and its parameters

X \sim B(n, p),\qquad q = 1 - p,\qquad X \in \{0, 1, 2, \dots, n\}

$n$ number of trials (fixed in advance)
$p$ probability of success on a single trial
$q$ probability of failure, q = 1 − p
$X$ number of successes across the n trials

The Binomial PMF — Probability of Exactly r Successes

Binomial probability mass function

P(X = r) = {}^{n}C_r\, p^{r}\, q^{\,n-r}

ⁿCᵣnumber of ways to place the r successes among the n trials
pʳprobability of r successes
qⁿ⁻ʳprobability of the remaining n − r failures

Building the Full Probability Distribution Table

Distribution terms sum to one via the binomial expansion

\sum_{r=0}^{n} P(X = r) = \sum_{r=0}^{n} {}^{n}C_r\, p^r q^{n-r} = (q + p)^{n} = 1

Common traps

Binomial needs WITH-replacement (or constant p), not without-replacement

The binomial formula assumes every trial has the SAME success probability. Drawing objects one-by-one WITHOUT replacement changes

p

each draw — that is the hypergeometric setting, not binomial. Exam wording like 'a ball is drawn, its colour noted and REPLACED, and the process repeated' is the signal to use the binomial formula.

q = 1 − p is derived, so a binomial has only TWO parameters

A binomial distribution is fixed by exactly two numbers,

n

and

p

; the failure probability is

q = 1 - p

, never an independent third parameter. If a problem gives you

q

directly, get

p = 1 - q

before doing anything else.

'Not a swimmer is 1/5' means success p = 4/5, not p = 1/5

When the wording gives the probability of the FAILURE ('not a swimmer', 'defective', 'does not recover'), that number is

q

, not

p

. Get

p = 1 - q

first. If P(not a swimmer)

= \tfrac15

then

p = \tfrac45

, and P(4 of 5 swim)

= {}^5C_4\left(\tfrac45\right)^4\left(\tfrac15\right) = \left(\tfrac45\right)^4

'None defective' is P(X = 0) = qⁿ, and it needs WITH-replacement

For a box of 100 bulbs with 10 defective, P(a bulb is good)

= \tfrac{9}{10}

. Drawing 5 (with replacement) gives P(none defective)

= \left(\tfrac{9}{10}\right)^5

, i.e.

q^{5}

with

q =

P(good). Do NOT switch to a without-replacement (hypergeometric) count — the binomial answer is the intended one.

Don't forget the ⁿCᵣ multiplier

p^r q^{n-r}

is the probability of ONE specific arrangement of

r

successes. There are

{}^{n}C_r

arrangements, so the full probability is

{}^{n}C_r\,p^r q^{n-r}

. Writing just

p^r q^{n-r}

(omitting the coefficient) is the most common single-term slip — except at the extremes

r = 0

r = n

, where

{}^{n}C_r = 1

Match the exponents to r and n − r, in that order

{}^{n}C_r\,p^r q^{n-r}

, the success probability

p

carries the exponent

r

(the number of successes) and the failure probability

q

carries

n - r

. Swapping them — e.g. writing

{}^{6}C_5\,q^5 p^1

when 5 successes are wanted — is a standard distractor built into MHT-CET option sets.

Order the table by ascending r — P(X = 0) uses qⁿ, P(X = n) uses pⁿ

In a distribution table the first entry is

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

(all failures) and the last is

P(X = n) = p^{n}

(all successes). MHT-CET distractors reverse this order or swap the middle terms — for a die tossed twice, the correct row is

P(0) = \tfrac{25}{36},\ P(1) = \tfrac{5}{18},\ P(2) = \tfrac{1}{36}

, decreasing because

q > p

Fix which colour is 'success' before building the table

For a bag with 4 red and 3 black balls drawn with replacement, if

X

counts BLACK then

p = \tfrac37

(black) and

q = \tfrac47

(red) — so

P(X=0) = \left(\tfrac47\right)^3

uses the RED probability. Assigning

p

to the wrong colour flips the whole table and lands on the mirror-image distractor.

The probabilities must sum to 1 — use it as a check

Because the entries are the terms of

(q + p)^n

, they always total 1. After writing a distribution table, add the entries: if they do not sum to 1 you have miscomputed a coefficient or an exponent. This single check catches most table errors before you pick an option.

Middle term of B(2, p) carries a factor 2 (not 1)

For two trials,

P(X = 1) = {}^{2}C_1\,pq = 2pq

, because there are two orderings (success-then-failure and failure-then-success). Writing

P(X = 1) = pq

drops the count and is a classic two-trial slip; likewise

P(X = 1)

for three trials is

3pq^2

, not

pq^2

Computing Binomial Probabilities — Cumulative, Ranges and Shortcuts

Learn this subtopic in the notes

Adding PMF Terms to Get a Whole Answer

PMF term and the total-probability identity

P(X=r) = \binom{n}{r} p^r q^{\,n-r},\qquad \sum_{r=0}^{n} P(X=r) = (p+q)^n = 1

$n$ number of independent trials
$p$ probability of success on one trial
$q$ probability of failure, $q = 1-p$
$r$ number of successes, an integer from 0 to n

At Least and At Most — Cumulative Probabilities

Two-term tails you meet most often

P(X \le 1) = q^n + n\,p\,q^{\,n-1},\qquad P(X \ge n-1) = n\,p^{\,n-1}q + p^n

At Least One — the 1 minus qⁿ Shortcut

The at-least-one complement

P(X \ge 1) = 1 - q^n,\qquad \text{smallest } n:\; q^n < 1 - t

Ranges and Symmetric Events by Complement

Absolute-value condition and the complement of a range

|X - a| \le b \iff a-b \le X \le a+b,\qquad P(a{-}b \le X \le a{+}b) = 1 - \!\!\sum_{r \,\notin\, [a-b,\,a+b]} \!\!P(r)

Special Counting — Even Successes, Expected Frequency, and Fixed-Trial Events

Even-count identity and expected frequency

\sum_{r\text{ even}}\binom{n}{r} = 2^{\,n-1},\qquad \mathbb{E}[\text{count over } N] = N \cdot P(\text{event})

Finding p First When the Stem Hides It

p by counting, then the at-least-3 binomial

p = \dfrac{\text{favourable outcomes}}{\text{total outcomes}},\qquad P(X \ge 3) = \binom{4}{3}p^3 q + p^4 = p^3(4q + p)

Common traps

'At least k' includes k itself, not just above it

P(X \ge 3)

means

r = 3, 4, \dots, n

— the value 3 IS counted. Reading 'at least 3' as 'more than 3' (starting at 4) drops the largest term

P(3)

and gives the wrong answer. 'More than 3' is what excludes 3.

A compound event is a SUM of terms, not a single term

P(X \ge 3)

with

n=5

P(3)+P(4)+P(5)

, NOT just

P(3)

. Computing only the first term (e.g. only

\binom{5}{3}p^3q^2

) is the single most common slip in 'at least' questions — always list every r in the set.

'At most one defective' has two terms, not one

P(X \le 1) = P(0) + P(1) = q^n + n\,p\,q^{\,n-1}

. Computing only

P(0)=q^n

— or only

P(1)

— is the standard error. Both the all-clean case and the exactly-one case count.

Decide which outcome 'success' labels before counting

In 'unable to solve less than two problems', if

p

is the probability of SOLVING, then failing on 0 or 1 problem means solving all

n

or all-but-one:

P = p^n + n\,p^{\,n-1}q

. Mixing up which event is 'success' flips

p

and

q

and gives a completely different (wrong) option.

Factor the shared power to match the printed option

q^5 + 5pq^4

equals

q^4(q+5p)

; with

p=\tfrac1{10}

that is

\left(\tfrac9{10}\right)^4\cdot\tfrac{14}{10} = \tfrac75\left(\tfrac9{10}\right)^4

. MHT-CET options are usually pre-factored, so leaving the sum unfactored can hide the matching choice.

At least one = 1 − qⁿ, not p or np

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

where

q^n

is the probability of NO successes across all n trials. It is not

p

(one trial) and not

np

(the mean). For 10 bulbs at a 10% defective rate the answer is

1 - (9/10)^{10}

, never

10\times\tfrac1{10}

For 'smallest n', solve the inequality — don't just plug the mean

P(\text{at least one head}) > 0.99

becomes

(1/2)^n < 0.01

, i.e.

2^n > 100

. Since

2^6 = 64 < 100

but

2^7 = 128 > 100

, the minimum is

n = 7

. Stopping at

n=6

(the last value that FAILS) is the classic off-by-one.

Cap the interval at 0 and n before counting

|X-4| \le 2

reads

2 \le X \le 6

, but if

n=6

then

X

can never exceed 6 anyway — the upper end

X\le 6

is free. So the condition really excludes only

X=0,1

. Counting phantom values above n (or below 0) inflates the sum.

Use the complement when the range is most of 0…n

For

2 \le X \le 6

B(6,\tfrac12)

, adding five terms

P(2)+\dots+P(6)

is slow; subtracting the two excluded terms

1 - P(0) - P(1) = 1 - \tfrac{7}{64} = \tfrac{57}{64}

is instant. Always compare the count of included vs excluded terms and take the shorter route.

Even number of heads on a fair coin is exactly 1/2

For a FAIR coin,

P(\text{even number of heads}) = \tfrac12

regardless of how many tosses — because

\sum_{r\text{ even}}\binom{n}{r} = 2^{\,n-1}

, half of

2^n

. Do not try to add 51 separate terms for 100 tosses; the identity gives

\tfrac12

instantly. (This clean split needs

p=q=\tfrac12

Expected frequency is N × P, not N × p

When a whole experiment is repeated N times, the expected number of times an EVENT happens is

N \times P(\text{event})

, where

P(\text{event})

is worked out for one experiment first. For '4 dice thrown 27 times, at least two show a 3 or 5', find

P(X\ge 2)=\tfrac{11}{27}

for one throw, then

27\times\tfrac{11}{27}=11

'Second success at the third trial' fixes the last trial

This is NOT

\binom{3}{2}p^2q

(that would let the second win fall anywhere in three trials). The third trial MUST be a win, and exactly one of the first two is a win:

\binom21 p\,q \cdot p

. With

p=\tfrac12

this is

2\cdot\tfrac14\cdot\tfrac12 = \tfrac14

Count the favourable numbers carefully — this is where marks are lost

For 'digit product = 24' among 00–99 the only numbers are

\{38, 46, 64, 83\}

(since

3\times8

and

4\times6

are the single-digit factorisations). Listing wrong pairs — or forgetting the reversed order like 46 AND 64 — changes p and wrecks the answer.

Read the sample-space range: 00–99 is 100, 10–99 is 90

p = \dfrac{\text{favourable}}{\text{total}}

, so the denominator depends on the stated range. '00 to 99' gives 100 numbers (

p=\tfrac4{100}

); '10 to 99' gives 90 (

p=\tfrac4{90}

). Using the wrong total gives a plausible-but-wrong option — the MHT-CET distractors exploit exactly this.

After finding p, still add all the terms for 'at least 3'

'Occurs at least 3 times' in 4 trials is

\binom43 p^3 q + p^4

, not just

\binom43 p^3 q

. The

p^4

(all four) term is small but the options are built to differ by exactly it — e.g.

\tfrac{97}{25^4}

\tfrac{96}{25^4}

Mean, Variance and Standard Deviation of a Binomial Variable

Learn this subtopic in the notes

The Mean of a Binomial Variable is np

Mean of a binomial variable

\mu = E(X) = np

$n$ number of independent trials
$p$ probability of success on a single trial
$q$ probability of failure, q = 1 − p

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

$n$ number of independent trials
$p$ success probability, q = 1 − p
npqthe variance — always less than the mean 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)

Common traps

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.

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.

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}}

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.

Parameter Estimation and the Probability Ratio

Learn this subtopic in the notes

The Binomial PMF, Mean and Variance (Recall)

PMF, mean and variance of B(n, p)

P(X=r) = {}^{n}C_{r}\,p^{r}q^{\,n-r},\qquad \text{mean} = np,\qquad \text{var} = npq,\qquad \text{SD} = \sqrt{npq}

$n$ number of independent trials
$p$ probability of success on one trial
$q$ probability of failure, q = 1 − p
$r$ number of successes counted

The Successive-Term Ratio of a Binomial Distribution

Ratio of consecutive binomial probabilities

\frac{P(X=k)}{P(X=k-1)} = \frac{n-k+1}{k}\cdot\frac{p}{q}

$k$ the higher of the two success counts
$n−k+1$ the coefficient ratio numerator ⁿC_k / ⁿC_(k−1)
$p/q$ one extra success over one fewer failure

Finding p from a Condition a·P(X=i) = b·P(X=j)

Cancelling a condition to a linear relation

a\,{}^{n}C_{i}\,p^{i}q^{\,n-i} = b\,{}^{n}C_{j}\,p^{j}q^{\,n-j}\;\Longrightarrow\;\text{linear in }p,q,\quad q = 1-p

Finding p from Given Numerical Probabilities

Divide two given probabilities to expose p/q

\frac{P(X=i)}{P(X=j)} = \frac{{}^{n}C_{i}}{{}^{n}C_{j}}\left(\frac{p}{q}\right)^{\,i-j}

Combination Identities: ⁿCₐ = ⁿC_b and PMF Normalisation

The two n-pinning identities

{}^{n}C_{a} = {}^{n}C_{b}\;(a\ne b)\;\Rightarrow\; a+b = n,\qquad \sum_{r=0}^{n}{}^{n}C_{r} = 2^{n}

The Most Probable Value (Mode) of a Binomial Distribution

Most probable value for a fair coin B(n, ½)

n\text{ even: } r = \tfrac{n}{2};\qquad n\text{ odd: } r = \tfrac{n-1}{2}\ \text{and}\ \tfrac{n+1}{2}

Common traps

Variance is npq, not np or np·q with q = p

The mean of

B(n,p)

np

and the variance is

npq

with

q = 1-p

. Using

np

for the variance, or writing

q = p

, is the classic slip. Since

q<1

, variance

npq

is always strictly smaller than the mean

np

The exponent of q is n − r, not r

P(X=r) = {}^{n}C_{r}\,p^{r}q^{\,n-r}

, success power

p^{r}

counts the

r

successes and failure power

q^{\,n-r}

counts the remaining

n-r

trials. Swapping the exponents (

p^{n-r}q^{r}

) flips success and failure and is a frequent distractor.

The coefficient ratio is (n−k+1)/k, not (n−k)/k or (n−k+1)/(k+1)

\dfrac{{}^{n}C_{k}}{{}^{n}C_{k-1}} = \dfrac{n-k+1}{k}

. The +1 comes from

(n-(k-1)) = n-k+1

, and the denominator is exactly

k

. Distractors like

\dfrac{n-k}{k-1}

\dfrac{n-k+1}{k+1}

are the standard wrong options — verify by testing

k=1

(the ratio must be

n\cdot\tfrac{p}{q}

Do not invert the ratio: it is p/q, not q/p

Moving from

k-1

up to

k

adds one success, so the ratio carries

\dfrac{p}{q}

(one more

p

, one fewer

q

). Writing

\dfrac{q}{p}

inverts the direction and is a frequent trap — the option

\dfrac{n+1}{k}\cdot\dfrac{q}{p}

is designed to catch exactly this.

Cancel powers of BOTH p and q before solving

From

a\,{}^{n}C_{i}\,p^{i}q^{\,n-i} = b\,{}^{n}C_{j}\,p^{j}q^{\,n-j}

, cancel the smaller power of

p

and the smaller power of

q

from each side so a clean linear relation like

4q = 3p

survives. Forgetting to cancel the

q

-powers, or keeping a stray

p^{2}

, leaves a quadratic that cannot match the intended linear answer.

Always substitute q = 1 − p at the end, not p = 1 − q inconsistently

After cancelling you have a relation between

p

and

q

(say

3p = q

). Replace

q

with

1-p

to get a single-variable equation:

3p = 1-p \Rightarrow p = \tfrac14

. Solving without eliminating

q

leaves two unknowns; mixing up which is

1 -

the other flips the answer to

q

instead of

p

Read what the question finally asks — p, or the variance/probability that follows

Many of these questions do NOT stop at

p

: after finding

p

they ask for

npq

, or another probability like

P(X=4)

. Compute

p

first, then plug into whatever is requested. Reporting

p

when the option list is variances is a careless-miss trap.

Dividing the two given probabilities is faster than substituting numbers

Given

P(X=1) = 0.4096

and

P(X=2) = 0.2048

, do NOT solve for

p

from each equation separately. Divide them: the powers of

p, q

drop to a single

p/q

and the coefficients to

{}^{5}C_1/{}^{5}C_2

, giving

q = 4p

in one line. Then

p = \tfrac15

Recover p, then evaluate the REQUESTED probability — not the ones given

After finding

p = \tfrac15, q = \tfrac45

in a 5-trial problem, the question asks for

P(X=3)

P(X=4)

, e.g.

P(X=4) = {}^{5}C_{4}\left(\tfrac15\right)^{4}\tfrac45 = \tfrac{4}{625}

. Re-quoting a given value like 0.2048 is a misread.

Read a single given P(X=r) as a product of powers to spot p and q

15\,p^{4}q^{2} = \tfrac{135}{4096}

means

p^{4}q^{2} = \tfrac{9}{4096}

; recognise

\tfrac{9}{4096} = \left(\tfrac14\right)^{4}\left(\tfrac34\right)^{2}

, so

p = \tfrac14

. Trying to solve the sixth-degree equation blindly wastes time — read off the powers of the fraction.

ⁿCₐ = ⁿC_b gives a + b = n (or a = b), not a − b = n

The symmetry

{}^{n}C_{a} = {}^{n}C_{n-a}

means equal coefficients with

a \ne b

force

b = n - a

, i.e.

a + b = n

. So

P(5) = P(7)

for a fair coin gives

n = 12

, NOT

n = 2

. Subtracting the indices is the classic error.

The coefficients cancel only for a FAIR coin

P(a) = P(b)\Rightarrow {}^{n}C_{a} = {}^{n}C_{b}

works because

p = q = \tfrac12

makes

p^{a}q^{n-a} = p^{b}q^{n-b}

. If

p \ne \tfrac12

(e.g. the 100-coin question with unknown

p

), the powers do NOT cancel — you must keep

\left(\tfrac{p}{q}\right)

and solve for

p

, which is the 'find p from a condition' route instead.

Simplify the final probability into the option's power of 2

P(3\text{ tails}) = {}^{12}C_{3}\left(\tfrac12\right)^{12} = \dfrac{220}{2^{12}} = \dfrac{55}{2^{10}}

— reduce

\tfrac{220}{4096}

so it matches the answer written over

2^{10}

. Leaving it over

2^{12}

or not reducing

220 = 4\cdot 55

is why students miss the correct option.

For odd n there are TWO modes, both central

When

n

is odd (e.g.

B(99, \tfrac12)

), the maximum probability occurs at BOTH

r = \tfrac{n-1}{2}

and

r = \tfrac{n+1}{2}

(here 49 and 50) — they are exactly equal. If only one appears among the options, pick it; do not assume a unique mode for odd

n

The mode is the middle of the range, not the mean np unless p = ½

For a fair coin the mode coincides with the centre

\tfrac{n}{2}

because

{}^{n}C_{r}

is symmetric. For general

p

the mode is near

(n+1)p

, not the range midpoint — but MHT-CET most-probable-value questions are almost always fair-coin, so anchor on the central

{}^{n}C_{r}

peak.

Binomial Distribution formulas

The Binomial Setting and Probability Mass Function

The Binomial Setting — n Fixed Independent Success or Failure Trials

The Binomial PMF — Probability of Exactly r Successes

Building the Full Probability Distribution Table

Common traps

Computing Binomial Probabilities — Cumulative, Ranges and Shortcuts

Adding PMF Terms to Get a Whole Answer

At Least and At Most — Cumulative Probabilities

At Least One — the 1 minus qⁿ Shortcut

Ranges and Symmetric Events by Complement

Special Counting — Even Successes, Expected Frequency, and Fixed-Trial Events

Finding p First When the Stem Hides It

Common traps

Mean, Variance and Standard Deviation of a Binomial Variable

The Mean of a Binomial Variable is np

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

Recovering n and p from the Mean and Variance

Solving When the Mean and Variance are Combined into One Equation

Common traps

Parameter Estimation and the Probability Ratio

The Binomial PMF, Mean and Variance (Recall)

The Successive-Term Ratio of a Binomial Distribution

Finding p from a Condition a·P(X=i) = b·P(X=j)

Finding p from Given Numerical Probabilities

Combination Identities: ⁿCₐ = ⁿC_b and PMF Normalisation

The Most Probable Value (Mode) of a Binomial Distribution

Common traps

More MHT-CET Maths formula sheets