MHT-CET Maths · Binomial Distribution

The Binomial Setting and Probability Mass Function

Fix n independent trials, each a success (p) or failure (q = 1 − p); then X = number of successes follows B(n, p), and P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ — the single formula every question in this subtopic runs on.

All Binomial Distribution notes MHT-CET Maths strategy

Why this matters

This is the foundation of the whole chapter: 10 PYQs sit here (5 EASY, 4 MODERATE, 1 HARD). Every later idea (mean np, variance npq, at-least/at-most tails) is built on top of this one PMF. The recurring skills are three: reading n, p and q correctly from the wording (with-replacement draws, 'success = …'), evaluating a single P(X = r), and writing out the full P(X = 0…n) distribution table for a small experiment.

Concept 1 of 3

The Binomial Setting — n Fixed Independent Success or Failure Trials

Intuition

Picture the same simple experiment repeated a fixed number of times, each time with only two possible outcomes labelled success and failure, and each repeat unaffected by the others. Count how many successes you get — that count is a binomial random variable. Tossing a coin 10 times and counting heads, or drawing a ball with replacement 3 times and counting reds, are the everyday pictures.

Definition

Four conditions define a binomial setting (Bernoulli trials):

Fixed number of trials $n$ , decided in advance.
Two outcomes per trial — a success (probability $p$ ) and a failure (probability $q = 1 - p$ ).
**Constant $p$ ** — the success probability is the same on every trial (drawing WITH replacement keeps this true; without replacement breaks it).
Independent trials — one trial's result does not change another's.

Then $X$ , the number of successes in the $n$ trials, is a binomial variable, written $X \sim B(n, p)$ . It takes values $0, 1, 2, \dots, n$ .

Binomial variable and its parameters

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

nnumber of trials (fixed in advance)
pprobability of success on a single trial
qprobability of failure, q = 1 − p
Xnumber of successes across the n trials

Worked example

A fair coin is tossed 8 times and

X

counts the number of tails. Identify

n

p

and

q

, and state which values

X

can take.

Each toss is one trial with two outcomes; tossing 8 times fixes $n = 8$ .
Success = a tail, so $p = \tfrac12$ and $q = 1 - \tfrac12 = \tfrac12$ .
The tosses are independent and $p$ is constant, so $X \sim B\!\left(8, \tfrac12\right)$ .
A count of tails ranges from none to all: $X \in \{0, 1, 2, \dots, 8\}$ .

Answer:

n = 8,\ p = q = \tfrac12,\ X \sim B\!\left(8, \tfrac12\right),\ X \in \{0,\dots,8\}

Practice this conceptself-check · 4 quick reps

Try it yourself

A card is drawn from a well-shuffled deck, its suit noted, and it is REPLACED; this is done 4 times. If success is drawing a spade, find

n

p

and

q

Practice — Level 1 (4 reps)

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

1.
How many parameters does a binomial distribution have, and what are they?
2.
If $p = 0.3$ is the success probability, what is the failure probability $q$ ?
3.
Does drawing balls WITHOUT replacement give a binomial setting?
4.
For $X \sim B(5, p)$ , what values can $X$ take?

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.

Concept 2 of 3

The Binomial PMF — Probability of Exactly r Successes

Intuition

To get exactly

r

successes in

n

trials you need

r

successes (each contributing

p

) and

n - r

failures (each contributing

q

); that is

p^r q^{n-r}

for one particular ordering. Since the

r

successes can sit in any of

{}^{n}C_r

positions, multiply by that count. Reading

n

p

q

off the wording is half the battle.

Definition

For $X \sim B(n, p)$ , the probability mass function (the probability of exactly $r$ successes) is

P(X = r) = {}^{n}C_r\, p^{r} q^{\,n-r},\qquad r = 0, 1, 2, \dots, n.

The ${}^{n}C_r$ counts the ways to choose WHICH trials succeed; $p^{r} q^{n-r}$ is the probability of any one such pattern.
All-successes: $P(X = n) = p^{n}$ . No-successes: $P(X = 0) = q^{n}$ (both binomial coefficients are 1).
Set up $p$ , $q$ , $n$ first, then plug in $r$ . 'None defective' means $r = 0$ with success = defective, i.e. $q^{n}$ where $q$ = P(good).

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

Worked example

A fair die is rolled 5 times. Find the probability of getting a six exactly twice.

Success = a six, so $p = \tfrac16$ , $q = \tfrac56$ , $n = 5$ .
Here $r = 2$ , so $P(X = 2) = {}^{5}C_2\left(\tfrac16\right)^{2}\left(\tfrac56\right)^{3}$ .
${}^{5}C_2 = 10$ and $\left(\tfrac56\right)^3 = \tfrac{125}{216}$ , so $P = 10\cdot\tfrac{1}{36}\cdot\tfrac{125}{216} = \tfrac{1250}{7776}$ .
Simplify: $\tfrac{1250}{7776} = \tfrac{625}{3888}$ .

Answer:

P(X = 2) = \dfrac{625}{3888}

Practice this conceptself-check · 4 quick reps

Try it yourself

The probability that a seed germinates is

0.8

. If 4 seeds are sown, find the probability that all 4 germinate.

Practice — Level 1 (4 reps)

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

1.
For $X \sim B(n, p)$ , write $P(X = 0)$ in terms of $q$ .
2.
For $X \sim B(n, p)$ , write $P(X = n)$ .
3.
A coin is tossed 6 times. Probability of exactly 6 heads?
4.
$X \sim B(3, \tfrac13)$ . Compute $P(X = 1)$ .

From the bank · past-year question

Example 2Binomial DistributionEASY

The probability that a student is not a swimmer is

\frac{1}{5}

. The probability that out of 5 students selected at random 4 are swimmers is

[Q112 · 25 April Shift II · 2025]

'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.

Drill 6 more on the binomial pmf — probability of exactly r successes

Concept 3 of 3

Building the Full Probability Distribution Table

Intuition

For a small experiment you often want the whole picture: the probability of 0 successes, 1 success, up to

n

successes, laid out in a table. You just evaluate the PMF at every

r = 0, 1, \dots, n

. A neat check: all the probabilities must add to 1, exactly matching the binomial expansion of

(q + p)^n

Definition

A probability distribution of $X \sim B(n, p)$ lists each value with its probability:

Evaluate $P(X = r) = {}^{n}C_r\,p^r q^{n-r}$ for $r = 0, 1, \dots, n$ .
The successive probabilities are exactly the terms of $(q + p)^n$ , so they must sum to $(q + p)^n = 1$ — the built-in validity check.
For $n = 2$ (e.g. a die tossed twice): $P(0) = q^2$ , $P(1) = 2pq$ , $P(2) = p^2$ . For $n = 3$ : $q^3,\ 3pq^2,\ 3p^2q,\ p^3$ .

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

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 twice and

X

is the number of heads. Write the probability distribution of

X

Success = head, so $p = \tfrac12$ , $q = \tfrac12$ , $n = 2$ .
$P(X = 0) = q^2 = \tfrac14$ .
$P(X = 1) = 2pq = 2\cdot\tfrac12\cdot\tfrac12 = \tfrac12$ .
$P(X = 2) = p^2 = \tfrac14$ .
Check: $\tfrac14 + \tfrac12 + \tfrac14 = 1$ . ✓

Answer:

X

: 0, 1, 2 with probabilities

\tfrac14,\ \tfrac12,\ \tfrac14

Practice this conceptself-check · 4 quick reps

Try it yourself

A bag has 2 white and 3 black balls. A ball is drawn, its colour noted and REPLACED; this is done twice. Let

X

be the number of white balls drawn. Write the distribution of

X

Practice — Level 1 (4 reps)

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

1.
For $X \sim B(2, p)$ , write $P(X = 1)$ .
2.
For $X \sim B(3, p)$ , write the four probabilities in order $r=0,1,2,3$ .
3.
A die is tossed twice, $X$ = number of fours. Find $P(X = 0)$ .
4.
What must the probabilities in any distribution table add up to?

From the bank · past-year question

Example 3Binomial DistributionMODERATE

A bag contains 4 red and 3 black balls. One ball is drawn and then replaced in the bag and the process is repeated. Let X denote the number of times black ball is drawn in 3 draws. Assuming that at each draw each ball is equally likely to be selected, then probability distribution of X is given by

[Q144 · 9th May Shift 2 · 2023]

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

Drill 2 more on building the full probability distribution table

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 (3)

The Binomial Setting — n Fixed Independent Success or Failure Trials
Binomial variable and its parameters
$X \sim B(n, p),\qquad q = 1 - p,\qquad X \in \{0, 1, 2, \dots, n\}$
The Binomial PMF — Probability of Exactly r Successes
Binomial probability mass function
$P(X = r) = {}^{n}C_r\, p^{r}\, q^{\,n-r}$
Building the Full Probability Distribution Table
Distribution terms sum to one via the binomial expansion
$\sum_{r=0}^{n} P(X = r) = \sum_{r=0}^{n} {}^{n}C_r\, p^r q^{n-r} = (q + p)^{n} = 1$

Watch out for (10)

Binomial needs WITH-replacement (or constant p), not without-replacement
→ The Binomial Setting — n Fixed Independent Success or Failure Trials
q = 1 − p is derived, so a binomial has only TWO parameters
→ The Binomial Setting — n Fixed Independent Success or Failure Trials
'Not a swimmer is 1/5' means success p = 4/5, not p = 1/5
→ The Binomial PMF — Probability of Exactly r Successes
'None defective' is P(X = 0) = qⁿ, and it needs WITH-replacement
→ The Binomial PMF — Probability of Exactly r Successes
Don't forget the ⁿCᵣ multiplier
→ The Binomial PMF — Probability of Exactly r Successes
Match the exponents to r and n − r, in that order
→ The Binomial PMF — Probability of Exactly r Successes
Order the table by ascending r — P(X = 0) uses qⁿ, P(X = n) uses pⁿ
→ Building the Full Probability Distribution Table
Fix which colour is 'success' before building the table
→ Building the Full Probability Distribution Table
The probabilities must sum to 1 — use it as a check
→ Building the Full Probability Distribution Table
Middle term of B(2, p) carries a factor 2 (not 1)
→ Building the Full Probability Distribution Table

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

The probability that a person who undergoes a bypass surgery will recover is 0.6. The probability that of the six patients who undergo similar operations, half of them will recover is

[Q149 · 11th May Shift 1 · 2023]

Example 2Binomial DistributionMODERATE

A fair die is tossed twice in succession. If X denotes the number of sixes in two tosses, then the probability distribution of X is given by

[Q139 · 10th May Shift 2 · 2024]

Example 3Binomial DistributionEASY

A box contains 100 bulbs, out of which 10 are defective. A sample of 5 bulbs is drawn. The probability that none is defective is

[Q146 · May Shift 1 · 2021]

Example 4Binomial DistributionEASY

A fair die is tossed twice in succession. If X denotes the number of fours in two tosses, then the probability distribution of X is given by

[Q103 · 11th May Shift 2 · 2024]

Example 5Binomial DistributionMODERATE

In a box containing 100 apples, 10 are defective. The probability that in a sample of 6 apples, 3 are defective is

[Q145 · 22 April Shift II · 2025]

Drill every past-year question on this subtopic

10 questions from the bank — paginated, with cart and Word-export support.

Drill the 10 questions