NDA Maths · Binomial Distribution

The Binomial Setting and Computing Probabilities

A binomial experiment is n independent repeats of the same two-outcome trial, and P(X = k) counts how many of the n trials end in success.

All Binomial Distribution notes NDA Maths strategy

Why this matters

This is the workhorse half of the chapter — 15 PYQs, mostly EASY and MODERATE, and they recur almost verbatim (coins, dice, ships, disease in workers). Three question shapes cover nearly all of them: an exact count P(X = k), an at-least / at-most count handled by the complement, and a short tail you sum directly. Lock the formula and the complement reflex and these become near-free marks.

Concept 1 of 7

Bernoulli Trials: One Success, One Failure

Intuition

Before counting over many trials, fix the single trial. A Bernoulli trial has exactly two outcomes — call one 'success' (probability

p

) and the other 'failure'. Since something must happen, the failure probability is whatever is left over.

Definition

A Bernoulli trial is a single experiment with two outcomes, success and failure.

Success probability: $p$ .
Failure probability: $q = 1 - p$ .
Always $p + q = 1$ , so $0 \le p \le 1$ .

Which outcome you label 'success' is your choice — but once chosen, $p$ must be the probability of THAT event for the rest of the problem.

Failure complements success

q = 1 - p, \qquad p + q = 1

Worked example

A spinner lands on red with probability

0.3

. Taking 'lands on red' as success, what are

p

and

q

Success is 'red', so $p = 0.3$ .
Failure (not red) is the complement: $q = 1 - 0.3 = 0.7$ .

Answer:

p = 0.3,\ q = 0.7

Practice this concept2 quick reps

Practice — Level 1 (2 reps)

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

1.
If $p = \tfrac{1}{4}$ , what is $q$ ?
2.
A die is rolled; success = 'getting a 6'. Find $p$ and $q$ .

Concept 2 of 7

When Is It Binomial? The Four Conditions

Intuition

A process is binomial only when you repeat the SAME Bernoulli trial a fixed number of times, the trials do not affect each other, and

p

never changes. If any of those breaks, the simple formula does not apply.

Definition

A random variable $X$ is binomial, written $X \sim B(n, p)$ , when all four hold:

Fixed n: the number of trials is decided in advance (not 'keep going until...').
Two outcomes: each trial is success or failure.
Independence: the result of one trial does not change the others.
Constant p: the success probability is the same on every trial.

Here $X$ = the number of successes in the $n$ trials, taking values $0, 1, 2, \dots, n$ . Drawing cards WITHOUT replacement fails 'constant $p$ ' and 'independence', so it is not binomial.

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

Is 'draw 3 cards from a deck without replacement and count the kings' a binomial experiment?

Fixed number of draws (3)? Yes.
Two outcomes per draw (king / not king)? Yes.
Independent with constant $p$ ? NO — once a card is removed, the next draw's probability changes.
Because $p$ is not constant and the draws are dependent, the binomial model does not apply (this is hypergeometric).

Answer:Not binomial — sampling without replacement breaks constant

p

and independence.

From the bank · past-year question

Example 2Binomial DistributionEASY

Under which of the following conditions may binomial distribution be used? I. The number of trials is infinite and not fixed. II. The trials are independent. III. Each trial has two possible outcomes. Select the correct answer using the code given below.

[Q106 · Sep · 2025]

'Until the first success' is not binomial

If the number of trials is not fixed in advance — e.g. 'toss until a head appears' — then

n

is random and the binomial formula does not apply. Binomial needs a pre-set

n

Concept 3 of 7

Reading p and the Success Event from the Story

Intuition

Half the battle is turning words into

p

. 'Thrice as likely', 'one in five on average', and 'arrives safely' each hide a probability — pin down the success event first, then its

p

Definition

Common translations:

Odds phrasing ('heads is thrice as likely as tails'): the parts are $3 : 1$ , so $p = \tfrac{3}{4}$ , $q = \tfrac{1}{4}$ .
Rate phrasing ('one in five ships is sunk on average'): $p(\text{sunk}) = \tfrac{1}{5}$ , so $p(\text{safe}) = \tfrac{4}{5}$ .
'k% chance': convert straight to a fraction, e.g. $20\% \to \tfrac{1}{5}$ .

Decide which event the question counts (hits? safe arrivals? defectives?) and make THAT the success, then read $p$ off the story.

Odds to probability

\text{odds } a : b \ \Longrightarrow\ p = \dfrac{a}{a + b}

Worked example

A player is twice as likely to win a point as to lose it. In 3 points, what is

p

for 'wins a point', and the probability of winning exactly 2?

Odds $2 : 1$ give $p = \tfrac{2}{3}$ , $q = \tfrac{1}{3}$ .
Exactly 2 wins of 3: $\binom{3}{2}\left(\tfrac{2}{3}\right)^2\left(\tfrac{1}{3}\right) = 3 \cdot \tfrac{4}{9} \cdot \tfrac{1}{3} = \tfrac{12}{27} = \tfrac{4}{9}$ .

Answer:

p = \tfrac{2}{3}

; probability of exactly 2 wins

= \tfrac{4}{9}

From the bank · past-year question

Example 3Binomial DistributionMODERATE

A coin is biased so that heads comes up thrice as likely as tails. In four independent tosses of the coin, what is probability of getting exactly three heads?

[Q104 · Apr · 2023]

'Thrice as likely' is 3 : 1, not p = 3

Odds split the whole into parts. 'Heads thrice as likely as tails' means

3 : 1

out of 4 parts, so

p = \tfrac{3}{4}

— never read it as

p = 3

p = \tfrac{1}{3}

Drill 1 more on reading p and the success event from the story

Concept 4 of 7

The Binomial Probability Formula

Intuition

Any single sequence with

k

successes and

n - k

failures has probability

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

. There are

\binom{n}{k}

such sequences (the orderings), so multiply.

Definition

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

P(X = k) = \binom{n}{k} p^{k} q^{\,n-k}, \qquad q = 1 - p.

The three pieces:

\binom{n}{k}

counts the orderings,

p^k

is the

k

successes,

q^{\,n-k}

is the

n-k

failures. Special cases drop straight out: all failures

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

and all successes

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

Probability of exactly k successes

P(X = k) = \binom{n}{k} p^{k} q^{\,n-k}

nnumber of trials
knumber of successes counted
psuccess probability per trial
qfailure probability, 1 − p

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 5 times. What is the probability of exactly 2 heads?

Here $n = 5$ , $p = \tfrac{1}{2}$ , $k = 2$ .
$P(X = 2) = \binom{5}{2}\left(\tfrac{1}{2}\right)^2\left(\tfrac{1}{2}\right)^3 = 10 \cdot \left(\tfrac{1}{2}\right)^5$ .
$= \dfrac{10}{32} = \dfrac{5}{16}$ .

Answer:

\dfrac{5}{16}

Practice this conceptself-check · 2 quick reps

Try it yourself

The probability a seed germinates is

0.8

. Out of 5 seeds, find the probability that exactly 2 germinate.

Practice — Level 1 (2 reps)

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

1.
Probability of exactly 3 heads in 4 tosses of a fair coin.
2.
A die is rolled 3 times. Probability of getting a 6 on all three.

From the bank · past-year question

Example 4Binomial DistributionMODERATE

If a fair die is rolled 4 times, then what is the probability that there are exactly 2 sixes ?

[Q115 · Sep · 2019]

Match the exponents to the success/failure counts

The power on

p

is the number of successes

k

; the power on

q

n - k

. Swapping them — e.g.

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

— is the single most common slip, especially when

p \ne q

Drill 4 more on the binomial probability formula

Concept 5 of 7

At Least One via the Complement

Intuition

'At least one success' is the opposite of 'no successes'. Computing the single term

P(X = 0)

and subtracting from 1 is far quicker than adding

P(1) + P(2) + \dots

Definition

The complement shortcut:

P(X \ge 1) = 1 - P(X = 0) = 1 - q^{n}.

More generally, 'at most' and 'at least' counts are linked by

P(X \le k) = 1 - P(X \ge k+1)

. Always compute whichever side has FEWER terms — for 'at least one' that is the single term

P(X=0)

At least one success

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

Worked example

A die is rolled 3 times. What is the probability of getting at least one six?

Success = 'six', $p = \tfrac{1}{6}$ , $q = \tfrac{5}{6}$ , $n = 3$ .
$P(X \ge 1) = 1 - P(X = 0) = 1 - \left(\tfrac{5}{6}\right)^3$ .
$= 1 - \dfrac{125}{216} = \dfrac{91}{216}$ .

Answer:

\dfrac{91}{216}

From the bank · past-year question

Example 5Binomial DistributionEASY

The probability that a person hits a target is 0.5. What is the probability of at least one hit in 4 shots?

[Q110 · Sep · 2022]

'At most' can also flip to a complement

'At most 4 tails in 5 tosses' has five terms the long way, but its complement 'exactly 5 tails' is one term:

1 - (\tfrac12)^5 = \tfrac{31}{32}

. Read the count and complement whichever side is shorter.

Drill 1 more on at least one via the complement

Concept 6 of 7

Cumulative Probabilities: Summing the Tail

Intuition

When 'at least

k

' or 'at most

k

' lands in the middle, no single complement collapses it — you add the handful of bars on the shorter side.

Definition

A cumulative probability is a sum of exact terms:

P(X \ge k) = \sum_{j=k}^{n} \binom{n}{j} p^{j} q^{\,n-j}.

Two routes, pick the one with fewer terms:

Sum directly when the tail is short (e.g. $X \ge 6$ out of 8 is just $j = 6, 7, 8$ ).
Use the complement when the other side is shorter, e.g. $P(X \ge 2) = 1 - P(0) - P(1)$ .

At least two successes

P(X \ge 2) = 1 - P(X = 0) - P(X = 1)

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 fair coin is tossed 4 times. What is the probability of getting at least 3 heads?

$n = 4$ , $p = \tfrac12$ . The short side is the tail $j = 3, 4$ .
$P(X \ge 3) = \binom{4}{3}(\tfrac12)^4 + \binom{4}{4}(\tfrac12)^4 = \dfrac{4 + 1}{16}$ .
$= \dfrac{5}{16}$ .

Answer:

\dfrac{5}{16}

Practice this conceptself-check

Try it yourself

A man hits a target with probability

\tfrac15

and fires 7 times. Find the probability of at least 2 hits.

From the bank · past-year question

Example 6Binomial DistributionMODERATE

8 coins are tossed simultaneously. The probability of getting at least 6 heads is

[Q107 · Sep · 2018]

Count the terms before you sum

For 'at least

k

', the direct sum runs

j = k

n

and the complement runs

j = 0

k-1

. Pick the shorter list — and never forget a boundary term (the

j = n

all-success term is easy to drop).

Drill 3 more on cumulative probabilities: summing the tail

Concept 7 of 7

The Complementary Count Y = n − X

Intuition

X

counts successes, then

Y = n - X

counts failures. Counting failures is just a binomial with success and failure swapped — so

p

and

q

trade places.

Definition

If $X \sim B(n, p)$ then the complementary count

Y = n - X \sim B(n,\ 1 - p) = B(n, q).

The number of trials

n

is unchanged; only the success probability flips to

q

. This is why 'the number that fail' and 'the number that pass' are both binomial on the same

n

Swapping successes for failures

X \sim B(n, p) \ \Longrightarrow\ n - X \sim B(n,\, 1 - p)

Worked example

X \sim B(12, \tfrac14)

counts defective items. What distribution does the number of NON-defective items follow?

Non-defectives $= 12 - X$ .
Swap $p \to 1 - p$ : $12 - X \sim B(12,\ 1 - \tfrac14)$ .
$= B(12,\ \tfrac34)$ .

Answer:

B\!\left(12, \tfrac34\right)

From the bank · past-year question

Example 7Binomial DistributionMODERATE

Consider a random variable X which follows Binomial distribution with parameters

n=10

and

p=\frac{1}{5}

. Then

Y=10-X

follows Binomial distribution with parameters

n

and

p

respectively given by

[Q118 · Apr · 2020]

n stays the same — only p flips

A frequent distractor halves

n

or keeps

p

unchanged. The complementary count

n - X

keeps the SAME

n

and only swaps

p \leftrightarrow q

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)

Bernoulli Trials: One Success, One Failure
Failure complements success
$q = 1 - p, \qquad p + q = 1$
Reading p and the Success Event from the Story
Odds to probability
$\text{odds } a : b \ \Longrightarrow\ p = \dfrac{a}{a + b}$
The Binomial Probability Formula
Probability of exactly k successes
$P(X = k) = \binom{n}{k} p^{k} q^{\,n-k}$
At Least One via the Complement
At least one success
$P(X \ge 1) = 1 - q^{n}$
Cumulative Probabilities: Summing the Tail
At least two successes
$P(X \ge 2) = 1 - P(X = 0) - P(X = 1)$
The Complementary Count Y = n − X
Swapping successes for failures
$X \sim B(n, p) \ \Longrightarrow\ n - X \sim B(n,\, 1 - p)$

Watch out for (6)

'Until the first success' is not binomial
→ When Is It Binomial? The Four Conditions
'Thrice as likely' is 3 : 1, not p = 3
→ Reading p and the Success Event from the Story
Match the exponents to the success/failure counts
→ The Binomial Probability Formula
'At most' can also flip to a complement
→ At Least One via the Complement
Count the terms before you sum
→ Cumulative Probabilities: Summing the Tail
n stays the same — only p flips
→ The Complementary Count Y = n − X

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

During war one ship out of 5 was sunk on an average in making a certain voyage. What is the probability that exactly 3 out of 5 ships would arrive safely?

[Q112 · Sep · 2022]

Example 2Binomial DistributionEASY

Consider the following data for the items that follow: The incidence of suffering from a disease among workers in an industry has a chance of

33\frac{1}{3}\%

What is the probability that at least one out of 6 workers suffers from a disease?

[Q117 · Apr · 2022]

Example 3Binomial DistributionEASY

Five coins are tossed once. What is the probability of getting at most four tails?

[Q108 · Sep · 2022]

Example 4Binomial DistributionMODERATE

The probability of a man hitting a target is

\frac{1}{5}

. If the man fires 7 times, then what is the probability that he hits the target at least twice?

[Q118 · Apr · 2025]

Example 5Binomial DistributionEASY

Consider the following data for the items that follow: The incidence of suffering from a disease among workers in an industry has a chance of

33\frac{1}{3}\%

What is the probability that no one out of 6 workers suffers from a disease?

[Q116 · Apr · 2022]

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