MHT-CET Maths · Binomial Distribution

Parameter Estimation and the Probability Ratio

Use the ratio of two adjacent binomial probabilities to turn a condition like P(X=a) = c·P(X=b) into a simple linear equation in p and q, and read off the unknown parameter p (or n) — the engine behind almost every 'find p' MHT-CET question.

All Binomial Distribution notes MHT-CET Maths strategy

Why this matters

This subtopic is a reliable single-mark scorer: 15 PYQs sit here (4 HARD, 10 MODERATE, 1 EASY). The whole subtopic runs on one idea — the successive-term ratio P(X=k)/P(X=k−1) = ((n−k+1)/k)·(p/q) — which lets the huge factorials cancel so a condition collapses to a linear relation in p and q. The recurring shapes are always the same: a·P(X=i) = b·P(X=j) to find p, the identity ⁿCₐ = ⁿC_b ⇒ a+b = n to find n, and the most-probable value (mode). Master the cancellation once and every variant falls out.

Concept 1 of 6

The Binomial PMF, Mean and Variance (Recall)

Intuition

Everything on this page rests on one distribution: n independent Bernoulli trials, each a success with probability p and failure with probability q = 1 − p. The probability of exactly r successes is a single-term binomial expansion. Keep the mean np and variance npq at your fingertips — several questions finish by asking for the variance once p is known.

Definition

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

Probability mass function: $P(X = r) = {}^{n}C_{r}\,p^{r}q^{n-r}$ for $r = 0, 1, \dots, n$ .
Mean: $np$ ; Variance: $npq$ ; Standard deviation: $\sqrt{npq}$ .
Because $q < 1$ , the variance $npq$ is always LESS than the mean $np$ — a quick sanity check.

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}

nnumber of independent trials
pprobability of success on one trial
qprobability of failure, q = 1 − p
rnumber of successes counted

Worked example

For

X \sim B(6, \tfrac13)

, write

P(X=2)

and compute the mean and variance.

PMF: $P(X=2) = {}^{6}C_{2}\left(\tfrac13\right)^{2}\left(\tfrac23\right)^{4} = 15\cdot\tfrac{1}{9}\cdot\tfrac{16}{81} = \dfrac{240}{729} = \dfrac{80}{243}$ .
Mean $= np = 6\cdot\tfrac13 = 2$ .
Variance $= npq = 6\cdot\tfrac13\cdot\tfrac23 = \tfrac43$ ; note $\tfrac43 < 2$ , as expected.

Answer:

P(X=2) = \dfrac{80}{243}

, mean

= 2

, variance

= \dfrac{4}{3}

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

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

1.
State $P(X=r)$ for $X \sim B(n,p)$ .
2.
Mean and variance of $B(n,p)$ ?
3.
For $B(10, \tfrac13)$ , find the variance.
4.
Is the variance of a binomial ever larger than its mean?

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.

Concept 2 of 6

The Successive-Term Ratio of a Binomial Distribution

Intuition

Dividing P(X=k) by P(X=k−1) makes the powers of p and q drop by one each and the binomial coefficients divide down to a single fraction. The result is a compact ratio that is the engine for finding p, comparing probabilities, and locating the mode — it turns messy factorials into one clean expression.

Definition

For $X \sim B(n, p)$ , take the ratio of two consecutive probabilities:

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

The coefficient ratio simplifies to

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

, so:

The full ratio is $\dfrac{P(X=k)}{P(X=k-1)} = \dfrac{n-k+1}{k}\cdot\dfrac{p}{q}$ .
The powers of $p$ and $q$ always contribute a single factor $\dfrac{p}{q}$ , never a squared one — one step up in $k$ means one more $p$ and one fewer $q$ .

Ratio of consecutive binomial probabilities

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

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

Visualization · why the coefficient is C(n, k)

Each leaf is one ordered outcome of 3 trials; with success probability p and failure q, a path with 2 successes and 1 failure has probability p²q regardless of the order. Exactly 3 of the 8 paths have 2 successes — that count is C(3, 2) = 3, so P(X = 2) = C(3, 2)·p²q. In general the number of length-n paths with k successes is C(n, k).

Worked example

For

X \sim B(n, p)

, simplify

\dfrac{P(X=2)}{P(X=1)}

in terms of

n, p, q

Write both probabilities: $P(X=2) = {}^{n}C_{2}\,p^{2}q^{\,n-2}$ , $P(X=1) = {}^{n}C_{1}\,p\,q^{\,n-1}$ .
Divide: $\dfrac{{}^{n}C_{2}}{{}^{n}C_{1}}\cdot\dfrac{p^{2}}{p}\cdot\dfrac{q^{\,n-2}}{q^{\,n-1}} = \dfrac{{}^{n}C_{2}}{{}^{n}C_{1}}\cdot\dfrac{p}{q}$ .
Coefficient ratio: $\dfrac{{}^{n}C_{2}}{{}^{n}C_{1}} = \dfrac{n(n-1)/2}{n} = \dfrac{n-1}{2}$ .
So $\dfrac{P(X=2)}{P(X=1)} = \dfrac{n-1}{2}\cdot\dfrac{p}{q}$ — matching $\dfrac{n-k+1}{k}\cdot\dfrac{p}{q}$ at $k=2$ .

Answer:

\dfrac{P(X=2)}{P(X=1)} = \dfrac{n-1}{2}\cdot\dfrac{p}{q}

Practice this conceptself-check · 4 quick reps

Try it yourself

For

X \sim B(n, p)

, find

\dfrac{P(X=k)}{P(X=k-1)}

as a single simplified expression.

Practice — Level 1 (4 reps)

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

1.
Simplify $\dfrac{{}^{n}C_{k}}{{}^{n}C_{k-1}}$ .
2.
State $\dfrac{P(X=k)}{P(X=k-1)}$ for $B(n,p)$ .
3.
What power of $p/q$ appears in the ratio of two ADJACENT binomial probabilities?
4.
For $B(n,p)$ , evaluate $\dfrac{P(X=1)}{P(X=0)}$ .

From the bank · past-year question

Example 2Binomial DistributionMODERATE

X\sim B(n,p)

then

\frac{P(X=k)}{P(X=k- 1)}=

[Q107 · 19 April Shift II · 2025]

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.

Drill 1 more on the successive-term ratio of a binomial distribution

Concept 3 of 6

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

Intuition

The most common shape: you are told two binomial probabilities are related, e.g. 5P(X=0) = P(X=1). Write both with the PMF, cancel the common binomial coefficient and powers, and you are left with a linear relation between p and q. Combine with p + q = 1 to solve for p — then finish with whatever is asked (often the variance npq).

Definition

Procedure to find $p$ from a condition like $a\,P(X=i) = b\,P(X=j)$ :

Substitute the PMF on both sides: $a\,{}^{n}C_{i}\,p^{i}q^{\,n-i} = b\,{}^{n}C_{j}\,p^{j}q^{\,n-j}$ .
Cancel the common powers of $p$ and $q$ and the numerical coefficients; you get a linear equation in $p$ and $q$ (e.g. $4q = 3p$ ).
Substitute $q = 1 - p$ and solve the linear equation for $p$ .
If the answer wants the variance, compute $npq$ with the $p$ just found.

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

Worked example

For

X \sim B(6, p)

4P(X=1) = P(X=2)

. Find

p

and hence the variance.

Write the PMFs: $4\,{}^{6}C_{1}\,p\,q^{5} = {}^{6}C_{2}\,p^{2}q^{4}$ .
Insert coefficients: $4\cdot 6\,p\,q^{5} = 15\,p^{2}q^{4}$ , i.e. $24 p q^{5} = 15 p^{2}q^{4}$ .
Cancel $p\,q^{4}$ : $24 q = 15 p$ , so $8q = 5p$ .
Use $q = 1-p$ : $8(1-p) = 5p \Rightarrow 8 = 13p \Rightarrow p = \tfrac{8}{13}$ , $q = \tfrac{5}{13}$ .
Variance $= npq = 6\cdot\tfrac{8}{13}\cdot\tfrac{5}{13} = \dfrac{240}{169}$ .

Answer:

p = \dfrac{8}{13}

, variance

= \dfrac{240}{169}

Practice this conceptself-check · 4 quick reps

Try it yourself

For

X \sim B(5, p)

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

. Find

p

Practice — Level 1 (4 reps)

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

1.
For $B(n,p)$ , $P(X=1) = P(X=0)\cdot n$ . What relation between $p,q$ is this?
2.
For $B(4,p)$ , $P(X=1) = P(X=3)$ . Find $p$ .
3.
After cancelling, a condition gives $3p = 2q$ . Find $p$ .
4.
Once $p = \tfrac14$ for $B(8,p)$ , what is the variance?

From the bank · past-year question

Example 3Binomial DistributionEASY

If X is a binomial variable with range

\mathbf{\{ 0,1,2,3,4\}}

and

P(X = 3) = 3P(X = 4)

then the parameter '

p

' of the binomial distribution is.

[Q116 · 20 April Shift I · 2025]

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.

Drill 6 more on finding p from a condition a·p(x=i) = b·p(x=j)

Concept 4 of 6

Finding p from Given Numerical Probabilities

Intuition

Sometimes the data are actual numbers — P(exactly 1 success) = 0.4096 and P(exactly 2) = 0.2048, or a single P(X=4) = a fraction. Dividing two given probabilities makes the coefficients and powers collapse to a p/q ratio (fast), or matching one given value pins p directly. Then you compute the requested probability with the recovered p.

Definition

Two flavours of 'numbers are given':

Two probabilities given: divide them. $\dfrac{P(X=1)}{P(X=2)} = \dfrac{{}^{n}C_{1}}{{}^{n}C_{2}}\cdot\dfrac{q}{p}$ ; plugging the numeric ratio gives a linear $p,q$ relation, so $q = 4p$ etc., then $p = \tfrac15$ .
One probability given: set ${}^{n}C_{r}\,p^{r}q^{\,n-r}$ equal to the given fraction and recognise $p, q$ (often $p = \tfrac14, q = \tfrac34$ ) from the powers of the fraction.
Finish by evaluating the requested $P(X = r)$ or the variance $npq$ with the recovered $p$ .

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}

Worked example

In 4 independent trials,

P(\text{exactly }1) = \tfrac{8}{27}

times

P(\text{exactly }0)

. Find

p

and

P(X=2)

Divide: $\dfrac{P(X=1)}{P(X=0)} = \dfrac{{}^{4}C_{1}}{{}^{4}C_{0}}\cdot\dfrac{p}{q} = 4\cdot\dfrac{p}{q}$ .
This equals $\tfrac{8}{27}$ ? Set $4\cdot\dfrac{p}{q} = \dfrac{8}{27}\Rightarrow \dfrac{p}{q} = \dfrac{2}{27}$ .
So $27p = 2q = 2(1-p)\Rightarrow 29p = 2 \Rightarrow p = \tfrac{2}{29}$ , $q = \tfrac{27}{29}$ .
$P(X=2) = {}^{4}C_{2}\,p^{2}q^{2} = 6\left(\tfrac{2}{29}\right)^{2}\left(\tfrac{27}{29}\right)^{2} = \dfrac{6\cdot 4\cdot 729}{29^{4}} = \dfrac{17496}{707281}$ .

Answer:

p = \dfrac{2}{29}

P(X=2) = \dfrac{17496}{707281}

Practice this conceptself-check · 4 quick reps

Try it yourself

B(5, p)

P(X=5) = \dfrac{1}{32}

. Find

p

Practice — Level 1 (4 reps)

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

1.
Two adjacent probabilities in $B(5,p)$ are in ratio $P(X=1):P(X=2) = 2:1$ . Find $q/p$ .
2.
If $q = 4p$ and $p+q=1$ , find $p$ .
3.
In $B(5,\tfrac15)$ , find $P(X=4)$ .
4.
If ${}^{6}C_{4}\,p^{4}q^{2} = \tfrac{135}{4096}$ with $15 p^4 q^2 = \tfrac{135}{4096}$ , find $p$ .

From the bank · past-year question

Example 4Binomial DistributionMODERATE

In a Binomial distribution consisting of 5 independent trials, probabilities of exactly 1 and 2 successes are 0.4096 and 0.2048 respectively, then the probability of getting exactly 4 successes is

[Q128 · 10th May Shift 1 · 2023]

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.

Drill 2 more on finding p from given numerical probabilities

Concept 5 of 6

Combination Identities: ⁿCₐ = ⁿC_b and PMF Normalisation

Intuition

Two coefficient facts finish a class of questions instantly. First, ⁿCₐ = ⁿC_b forces either a = b or a + b = n — so 'P(5 tails) = P(7 tails)' with a fair coin gives n = 12 directly. Second, all probabilities must sum to 1, so ΣⁿCᵣ(½)ⁿ = 1 means 2ⁿ = the given total, pinning n.

Definition

Two identities that pin down $n$ :

Equal coefficients: ${}^{n}C_{a} = {}^{n}C_{b}$ (with $a \ne b$ ) implies $a + b = n$ . For a FAIR coin, $P(a) = P(b)$ reduces to exactly this because $\left(\tfrac12\right)^{n}$ cancels — so $P(5\text{ tails}) = P(7\text{ tails}) \Rightarrow n = 12$ .
Normalisation: $\displaystyle\sum_{r=0}^{n}{}^{n}C_{r}\left(\tfrac12\right)^{n} = \left(\tfrac12\right)^{n}\sum_{r=0}^{n}{}^{n}C_{r} = \left(\tfrac12\right)^{n}2^{n} = 1$ . If a PMF is ${}^{n}C_{r}k$ for constant $k$ , then $k\cdot 2^{n} = 1$ , so $2^{n}$ equals the reciprocal of $k$ — solve for $n$ .
After finding $n$ , evaluate any requested $P(X=r)$ with ${}^{n}C_{r}\left(\tfrac12\right)^{n}$ .

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}

Worked example

A fair coin is tossed

n

times and

P(4\text{ heads}) = P(6\text{ heads})

. Find

n

and then

P(2\text{ heads})

Fair coin: ${}^{n}C_{4}\left(\tfrac12\right)^{n} = {}^{n}C_{6}\left(\tfrac12\right)^{n}$ , so ${}^{n}C_{4} = {}^{n}C_{6}$ .
Equal coefficients with $4 \ne 6$ give $4 + 6 = n$ , i.e. $n = 10$ .
$P(2\text{ heads}) = {}^{10}C_{2}\left(\tfrac12\right)^{10} = \dfrac{45}{1024}$ .

Answer:

n = 10

P(2\text{ heads}) = \dfrac{45}{1024}

Practice this conceptself-check · 4 quick reps

Try it yourself

A PMF is

P(X=r) = {}^{n}C_{r}\left(\tfrac12\right)^{n}

for

r = 0,\dots,n

. Given the coefficients sum makes the total probability 1 and

2^{n} = 64

, find

n

and

P(X \le 1)

Practice — Level 1 (4 reps)

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

1.
${}^{n}C_{3} = {}^{n}C_{8}$ , $3 \ne 8$ . Find $n$ .
2.
For a fair coin, $P(2\text{ heads}) = P(8\text{ heads})$ . Find the number of tosses.
3.
$\displaystyle\sum_{r=0}^{n}{}^{n}C_{r} = ?$
4.
If $2^{n} = 32$ , find $n$ .

From the bank · past-year question

Example 5Binomial DistributionMODERATE

A fair coin is tossed a fixed number of times. If the probability of getting 5 tails is same as the probability of getting 7 tails, then the probability of getting 3 tails is

[Q139 · 26 April Shift II · 2025]

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.

Drill 1 more on combination identities: ⁿcₐ = ⁿc_b and pmf normalisation

Concept 6 of 6

The Most Probable Value (Mode) of a Binomial Distribution

Intuition

The mode is the value r where P(X=r) is largest — where the successive-term ratio crosses 1 (probabilities rise while the ratio exceeds 1, then fall). For a fair coin the coefficients ⁿCᵣ peak in the middle, so the most probable count sits at the centre of the range.

Definition

The mode of $X \sim B(n, p)$ is the $r$ maximising ${}^{n}C_{r}\,p^{r}q^{\,n-r}$ ; it is the value where the ratio $\dfrac{P(X=r)}{P(X=r-1)}$ drops below 1. For the fair coin $p = \tfrac12$ the probability is just ${}^{n}C_{r}\left(\tfrac12\right)^{n}$ , so the mode is wherever ${}^{n}C_{r}$ is largest:

**Even $n$ :** the coefficient peaks at the single middle value $r = \dfrac{n}{2}$ .
**Odd $n$ :** it peaks at the TWO central values $r = \dfrac{n-1}{2}$ and $r = \dfrac{n+1}{2}$ , which are equal.

So for $B(99, \tfrac12)$ the maximum is at $r = 49$ and $r = 50$ .

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}

Visualization · change n and p, watch the distribution reshape

trials n:

At p = 0.5 the bars are symmetric about the centre. Push p to 0.2 and the peak slides left (few successes likely); push it to 0.8 and it slides right. The dashed line always sits at the mean np — raising n stretches the distribution and moves that centre.

Worked example

A fair coin is tossed 8 times. For how many heads is the probability maximum, and what is that probability?

$X \sim B(8, \tfrac12)$ , so $P(X=r) = {}^{8}C_{r}\left(\tfrac12\right)^{8}$ , maximised where ${}^{8}C_{r}$ is largest.
$n = 8$ is even, so the single peak is at $r = \dfrac{8}{2} = 4$ .
$P(X=4) = {}^{8}C_{4}\left(\tfrac12\right)^{8} = \dfrac{70}{256} = \dfrac{35}{128}$ .

Answer:Maximum at

r = 4

, with probability

\dfrac{35}{128}

Practice this conceptself-check · 4 quick reps

Try it yourself

A fair coin is tossed 11 times. At which value(s) of

r

P(X=r)

maximum?

Practice — Level 1 (4 reps)

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

1.
Where is ${}^{10}C_{r}$ largest?
2.
For a fair coin tossed 7 times, the most probable number of heads is?
3.
The mode occurs where the ratio $P(X=r)/P(X=r-1)$ does what?
4.
For $B(99,\tfrac12)$ , the two most probable values of $r$ are?

From the bank · past-year question

Example 6Binomial DistributionMODERATE

A fair coin is tossed 99 times. If X is the number of times head occur, then

P\lbrack X = r\rbrack

is maximum when

r =

[Q150 · 20 April Shift II · 2025]

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.

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 (6)

The Binomial PMF, Mean and Variance (Recall)
PMF, mean and variance of B(n, p)
$P(X=r) = {}^{n}C_{r}\,p^{r}q^{\,n-r},\qquad \text{mean} = np,\qquad \text{var} = npq,\qquad \text{SD} = \sqrt{npq}$
The Successive-Term Ratio of a Binomial Distribution
Ratio of consecutive binomial probabilities
$\frac{P(X=k)}{P(X=k-1)} = \frac{n-k+1}{k}\cdot\frac{p}{q}$
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)
$n\text{ even: } r = \tfrac{n}{2};\qquad n\text{ odd: } r = \tfrac{n-1}{2}\ \text{and}\ \tfrac{n+1}{2}$

Watch out for (15)

Variance is npq, not np or np·q with q = p
→ The Binomial PMF, Mean and Variance (Recall)
The exponent of q is n − r, not r
→ The Binomial PMF, Mean and Variance (Recall)
The coefficient ratio is (n−k+1)/k, not (n−k)/k or (n−k+1)/(k+1)
→ The Successive-Term Ratio of a Binomial Distribution
Do not invert the ratio: it is p/q, not q/p
→ The Successive-Term Ratio of a Binomial Distribution
Cancel powers of BOTH p and q before solving
→ Finding p from a Condition a·P(X=i) = b·P(X=j)
Always substitute q = 1 − p at the end, not p = 1 − q inconsistently
→ Finding p from a Condition a·P(X=i) = b·P(X=j)
Read what the question finally asks — p, or the variance/probability that follows
→ Finding p from a Condition a·P(X=i) = b·P(X=j)
Dividing the two given probabilities is faster than substituting numbers
→ Finding p from Given Numerical Probabilities
Recover p, then evaluate the REQUESTED probability — not the ones given
→ Finding p from Given Numerical Probabilities
Read a single given P(X=r) as a product of powers to spot p and q
→ Finding p from Given Numerical Probabilities
ⁿCₐ = ⁿC_b gives a + b = n (or a = b), not a − b = n
→ Combination Identities: ⁿCₐ = ⁿC_b and PMF Normalisation
The coefficients cancel only for a FAIR coin
→ Combination Identities: ⁿCₐ = ⁿC_b and PMF Normalisation
Simplify the final probability into the option's power of 2
→ Combination Identities: ⁿCₐ = ⁿC_b and PMF Normalisation
For odd n there are TWO modes, both central
→ The Most Probable Value (Mode) of a Binomial Distribution
The mode is the middle of the range, not the mean np unless p = ½
→ The Most Probable Value (Mode) of a Binomial Distribution

Mastery check — 5 interleaved questions

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

Example 1Binomial DistributionMODERATE

X \sim B(n, p)

then

\frac{P(X=k)}{P(X=k-1)} =

[Shift || · 2025]

Example 2Binomial DistributionMODERATE

One hundred identical coins, each with probability p, of showing up heads are tossed once. If

0<p<1

and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins, then the value of p is

[Q150 · 11th May Shift 2 · 2023]

Example 3Binomial DistributionHARD

Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to

[Q103 · 2nd May Shift 2 · 2023]

Example 4Binomial DistributionHARD

The probability mass function of random variable X is given by

P[X=r]=\binom{n}{r}\left(\frac{3}{2}\right)^n

n,r\in\mathbb{N}_0

, otherwise 0, then

P[X\leq2]=

[Q129 · 9th May Shift 1 · 2024]

Example 5Binomial DistributionMODERATE

A binomial random variable X satisfies

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

when

n=6

. Then p is equal to

[Q131 · 11th May Shift 1 · 2024]

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Related notes

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