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

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.

Practice this conceptself-check · 4 quick reps

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

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.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift II · Q112Easy

Example 2 · Binomial Distribution · The Binomial Setting and Probability Mass Function

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

'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

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 2 · Q144Moderate

Example 3 · Binomial Distribution · The Binomial Setting and Probability Mass Function

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

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

Test yourself on Binomial Distribution

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