Probability · Lesson 6 of 9

Discrete distributions

A handful of distributions come up again and again because the same stories do: repeated trials, waiting for a success, sampling from a box, rare events. Learn the five stories and you can name the distribution from the wording, and write down its mean and variance without calculating.

← Probability moduleLesson 6 of 9 · about 40 minutes

A question to hold on to

A binomial random variable X has expectation 7 and variance 2.1. What is the probability that X = 11?

Delhi School of Economics entrance, 2017Solved at the end ↓

The building block

Bernoulli: one yes-or-no

One trial with two outcomes: the customer buys or doesn’t, the coin lands heads or doesn’t. Code success as 1 and failure as 0, with P(1) = p. That is a Bernoulli random variable, and it is the indicator from Lesson 5 under another name.

Bernoulli(p)

E (X) = p, Var (X) = p (1 - p)

Since X² = X for a 0-or-1 variable, E(X²) = p too, and the variance is p − p².

The most uncertain coin is a fair one

p(1 − p) is largest at p = ½, where it equals ¼. A coin that almost always lands heads is predictable, so its variance is small. Every binomial fact about “maximum variance” comes from this.

Interactive

Binomial: counting successes

Repeat a Bernoulli trial n times and count the successes. The count is binomial when three things hold, which Hyderabad 2019 listed as its options: a fixed number of trials, two outcomes each time, and independent trials. Implicit in “the same experiment” is that p stays the same from trial to trial.

Where does the formula come from? Any particular sequence with k successes and n − k failures, say SSF…F, has probability pᵏ(1 − p)ⁿ⁻ᵏ by independence. There are C(n, k) such sequences, one for each way of choosing which k trials succeed (Lesson 1). Add them up:

Binomial(n, p)

P (X = k) = (k n) p^{k} (1 - p)^{n - k}, E (X) = n p, Var (X) = n p (1 - p)

A binomial is a sum of n independent Bernoullis, so by Lesson 5 its mean and variance are n times a Bernoulli's.

Choose an exam set-up below or set n and p yourself. Tap a bar or use the sliders to pick the values you want; the shaded total is their probability.

Heads in 25 tosses of a fair coin. The mean, 12.5, is not a value X can take.

Trials n = 25Chance of success p = 0.5

From 13To 25

P(X ≥ 13)0.5000

P(X = k) = C(25, k) · 0.5ᵏ · 0.5^(25 − k)

Mean = np = 12.5Variance = np(1 − p) = 6.25

Tap a bar to pick a single value, or set a range with the sliders.

DSE 2008 · advanced

Each particle from a radioactive source gets through a shield with probability 0.01. Ten are emitted. What is the probability that exactly one gets through?

Name it.
10 independent trials, each a success with probability 0.01: X ~ Bin(10, 0.01).
Apply the formula with k = 1.
$(1 10) (0.01)^{1} (0.99)^{9} = 10 \times 0.01 \times 0.9 9^{9} = (0.1) (0.99)^{9}$

Answer

(0.1)(0.99)⁹ ≈ 0.0914. The “Shield” preset above shows it.

Pause and predict

A fair coin is tossed 25 times. What is the standard deviation of the number of tails? (JNU 2016)

The patterns examiners use

Binomial questions, solved

Recover n and p from the mean and variance. Divide: Var/Mean = np(1 − p)/np = 1 − p. JNU 2019 gave E(X) = 2 and Var(X) = 4/3, so 1 − p = 2/3, p = 1/3 and n = 2/(1/3) = 6. JNU 2021 gave 4 and 4/3: 1 − p = 1/3, p = 2/3, n = 6.

Compare two neighbouring probabilities. Their ratio cancels most of the formula:

Ratio of neighbours

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

Useful when two probabilities are given, or to find the most likely value.

JNU 2017 · standard

X ~ Bin(5, p), with P(X = 1) = 0.4096 and P(X = 2) = 0.2048. Find p.

Take the ratio.
P(2)/P(1) = (4/2) · p/(1 − p) = 2p/(1 − p).
Set it equal to the data.
0.2048/0.4096 = ½, so 2p/(1 − p) = ½, giving 4p = 1 − p and p = 0.2.
Check.
5 × 0.2 × 0.8⁴ = 0.4096. ✓

Answer

p = 0.2.

The same ratio answers DSE 2015: P(k) = P(k + 1) exactly when (n − k)p = (k + 1)(1 − p), that is when k = (n + 1)p − 1. So (n + 1)p must be a whole number. And CUET 2024 asked you to order P(1 head in 1 toss) = 0.5, P(5 heads in 10) ≈ 0.246, P(10 in 20) ≈ 0.176 and P(50 in 100) ≈ 0.080: with more tosses, hitting exactly half becomes less likely, even though the proportion of heads gets closer to ½.

DSE 2007 · advanced

Y ~ Bin(4, 0.78). Find P(Y ≥ 3) to two decimal places.

Two terms.
P(Y = 3) + P(Y = 4) = 4(0.78)³(0.22) + (0.78)⁴.
Compute.
0.4176 + 0.3702 = 0.7878.

Answer

0.79.

Repeat until it works

Geometric: waiting for the first

Keep tossing until the first head. X, the number of tosses, is geometric: the first k − 1 tries fail and the k-th succeeds. Unlike the binomial, the number of trials is the random part. Waiting for several successes instead of one is the negative binomial.

Geometric(p), counting the trials

P (X = k) = (1 - p)^{k - 1} p, P (X \leq k) = 1 - (1 - p)^{k}, E (X) = \frac{1}{p}, Var (X) = \frac{1 - p}{p ^{2}}

Some books count failures before the first success instead; then the values start at 0 and the mean is (1 − p)/p.

The CDF by complement. X ≤ k fails only if the first k tries all fail. A fair coin: P(X ≤ 3) = 1 − (½)³ = 7/8 (DSE 2010).
The mean is the reciprocal. A fair coin takes 2 tosses on average (JAM 2025). A 1-in-6 event takes 6 tries.
No memory. Having failed five times changes nothing: P(X > m + k | X > m) = P(X > k). The dice of Lesson 3 again.
The minimum of two geometrics. JAM 2021 gave two independent variables whose CDFs, 1 − (3/4)ˣ and 1 − (2/3)ʸ, make P(X ≥ 6) = (3/4)⁵ and P(Y ≥ 6) = (2/3)⁵. The smaller is at least 6 only if both are, so P(min ≥ 6) = (3/4 × 2/3)⁵ = (1/2)⁵ = 1/32.

DSE 2005 · advanced

Each trial succeeds with probability 3/4. A gambler receives 2ⁿ if the first success is on trial n. What is the expected prize?

Expectation of a function (Lesson 5).
$E (2^{X}) = n = 1 \sum \infty 2^{n} (\frac{1}{4})^{n - 1} \frac{3}{4}$
It is a geometric series.
First term 2 × ¾ = 3/2, common ratio 2 × ¼ = ½.
Sum it.
(3/2)/(1 − ½) = 3.

Answer

3. DSE 2020 flipped the sign on odd n, prize (−2)ⁿ: the ratio becomes −½ and the sum (−3/2)/(1 + ½) = −1.

Sampling from a box

Hypergeometric: without replacement

Draw n items at once from a box of N, of which K are “successes”. The draws are not independent, because each one changes what’s left, so the count is not binomial. It is hypergeometric, and its formula is pure Lesson 1 counting: choose k of the K successes and n − k of the N − K failures, out of all C(N, n) samples.

Hypergeometric(N, K, n)

P (X = k) = \frac{( k K ) ( n - k N - K )}{( n N )}, E (X) = n \frac{K}{N}

The mean matches the binomial with p = K/N. The variance is smaller, by the factor (N − n)/(N − 1).

JNU 2020 took 5 bulbs from 100 with 10 faulty and asked for P(none faulty). Exactly, that is C(90, 5)/C(100, 5) ≈ 0.584. The answer key used (0.9)⁵ ≈ 0.590, the binomial, which treats the bulbs as put back. Lesson 1 worked through that discrepancy. With a sample this small compared with the box, the two are close; pick “Faulty bulbs” in the explorer and switch on the binomial overlay to compare.

Counting rare events

Poisson: rare events

Misprints on a page, accidents at a junction in a month, calls to a helpline in an hour. There are many chances for the event and each is unlikely, and what you know is the average count, λ. Such counts follow the Poisson distribution.

Poisson(λ)

P (X = k) = \frac{e ^{- λ} λ ^{k}}{k !}, k = 0, 1, 2, \dots E (X) = Var (X) = λ

The mean and the variance are both λ (JNU 2017). So E(X²) = Var + mean² = λ + λ² (JNU 2020).

DSE 2013 · advanced

On average, two of a waiter’s customers leave no tip on a Saturday. What is the probability that next Saturday exactly three leave no tip?

Name it.
A count of fairly rare events with a known average: X ~ Poisson(2).
Apply the formula.
$P (X = 3) = \frac{e ^{- 2} 2 ^{3}}{3 !} = \frac{8}{6} e^{- 2} = \frac{4}{3} e^{- 2}$

Answer

(4/3)e⁻² ≈ 0.180.

“At least one” works as it did in Lesson 1: P(X ≥ 1) = 1 − P(X = 0) = 1 − e^(−λ). With 3 misprints a page on average and e⁻³ = 0.049787 given, P(some mistake) = 1 − 0.049787 ≈ 0.9502 (JNU 2012). CUET 2024 printed a single Poisson probability, e⁻²2³/3!, and asked for the mean: read λ = 2 off the e⁻².

Pause and predict

For which distribution is the mean greater than the variance? (JNU 2021)

Interactive

When one becomes another

The distributions are relatives, and exam questions test the family tree.

Binomial → Poisson when n is large and p is small, with λ = np held moderate. In the explorer, set n = 60 and p = 0.03 and switch on the Poisson dots: they sit on the bars. Push p to 0.5 and they drift off.
Binomial → normal when np and n(1 − p) are both large, with the same mean np and variance np(1 − p). This is why the 25-toss histogram already looks bell-shaped. Lesson 7 is about the normal itself.
Hypergeometric → binomial when the sample is a small part of the population: drawing without replacement then hardly differs from drawing with it.

CUET 2024 asked what a binomial with “a very large N” can be approximated by, with Poisson and normal both among the options. Both are correct large-n approximations; which one applies depends on p, which the question doesn’t give. The provisional answer key chose Poisson. If you meet this question, look for any hint that p is small (“rare”), which points to Poisson.

The skill that matters most

Which distribution?

Distribution	The story	P(X = k)	Mean	Variance
Bernoulli(p)	One trial: success (1) or failure (0)	P(1) = p, P(0) = 1 − p	p	p(1 − p)
Binomial(n, p)	Successes in n independent trials	C(n, k) pᵏ (1 − p)ⁿ⁻ᵏ, k = 0 … n	np	np(1 − p)
Geometric(p)	Trials up to the first success	(1 − p)ᵏ⁻¹ p, k = 1, 2, …	1/p	(1 − p)/p²
Hypergeometric(N, K, n)	Successes in n draws without replacement	C(K, k) C(N − K, n − k) / C(N, n)	nK/N	n(K/N)(1 − K/N)(N − n)/(N − 1)
Poisson(λ)	Count of rare events at rate λ	e^(−λ) λᵏ / k!, k = 0, 1, 2, …	λ	λ

Ask two questions. Is the number of trials fixed (binomial, hypergeometric) or are you waiting (geometric)? Are the trials independent, with replacement (binomial), or drawn from a finite box without it (hypergeometric)? A count with no fixed number of trials, only an average rate, is Poisson.

Name the distribution8 set-ups

A card is drawn from a well-shuffled pack n times, replaced and reshuffled each time. X is the number of aces drawn. (Hyderabad 2013)
Five bulbs are taken together from a box of 100, of which 10 are faulty. X is the number of faulty bulbs among the five. (JNU 2020)
A fair coin is tossed until a head appears. X is the number of tosses needed. (DSE 2010)
The number of printing mistakes on a page of a book. (JNU 2012)
Each customer who enters a shop buys something with probability 0.3. X is the number of buyers among 10 customers. (CUET 2023)
X is 1 if the next customer buys something and 0 if not.
The number of phone calls a call centre receives between 10 and 11 in the morning, when it averages 40 an hour.
A basketball player makes each free throw with probability 0.8. X is the number of throws until her first miss.

Solving the opening question

The opening question

The opening question · DSE 2017

A binomial random variable X has expectation 7 and variance 2.1. What is P(X = 11)?

Variance over mean gives 1 − p.
2.1/7 = 0.3, so p = 0.7.
Then n.
np = 7 gives n = 10.
Read the support.
A Bin(10, 0.7) count runs from 0 to 10. It can never be 11.

Answer

0. The formula would have you compute C(10, 11), which is 0 for the same reason: you can’t choose 11 trials from 10.

DSE 2015 · advanced

Five points are chosen independently and uniformly in the square with corners (0, 0) and (2, 2). For each, roll two dice if x + 2y ≤ 2 and one die otherwise. N is the total number of dice rolled. Find P(N = n) for 5 ≤ n ≤ 10.

The chance one point earns two dice.
x + 2y ≤ 2 cuts off a triangle with corners (0, 0), (2, 0), (0, 1): area 1, out of 4. So p = ¼.
Count the points that do.
K, the number of such points among 5, is Bin(5, ¼), and N = 5 + K.
Translate.
$P (N = n) = P (K = n - 5) = (n - 5 5) (\frac{1}{4})^{n - 5} (\frac{3}{4})^{10 - n}$

Answer

The binomial hides behind a geometry question. Spot the independent yes-or-no trials, and the rest is the formula.

Watch out for

The mistakes examiners are fishing for

Binomial without replacement. Drawing from a finite box without putting items back is hypergeometric, not binomial.
Forgetting the support. A Bin(n, p) count cannot exceed n; a geometric count cannot be 0 if it counts trials.
Mixing up the geometric’s two versions. Trials until the first success has mean 1/p; failures before it has mean (1 − p)/p.
The Poisson parameter is the mean, not a probability. λ can be 40.

Revise

The lesson on one screen

Bernoulli: One trial. Mean p, variance p(1 − p), at most ¼, at p = ½.
Binomial: Successes in n independent trials: C(n, k)pᵏ(1 − p)ⁿ⁻ᵏ. Mean np, variance np(1 − p) ≤ n/4.
Find n and p: Var/Mean = 1 − p, then n = Mean/p.
Geometric: Trials to the first success: (1 − p)ᵏ⁻¹p. P(X ≤ k) = 1 − (1 − p)ᵏ. Mean 1/p. No memory.
Hypergeometric: Successes in n draws without replacement: C(K, k)C(N − K, n − k)/C(N, n). Mean nK/N.
Poisson: Rare events at average rate λ: e^(−λ)λᵏ/k!. Mean = variance = λ.
Approximations: Binomial ≈ Poisson(np) for large n, small p; ≈ normal for large np and n(1 − p).

Practice

Check yourself

0 of 10 correct so far

Compute

Binomial mean

Each of 20 admitted students completes the degree with probability 0.7 (CUET 2023). Expected number who complete?

E(X)

Binomial probability

A fair coin is tossed 6 times. What is the probability of exactly 2 heads?

P(X = 2)

Find n

A binomial variable has mean 6 and variance 4.2. Find n.

Geometric

A die is rolled until the first six. What is the probability that it takes more than 3 rolls?

P(X > 3)

Poisson

Calls arrive at an average of 2 per minute, following a Poisson distribution. What is the probability of no calls in a given minute?

P(X = 0)

At least one

Three fair dice are rolled (JAM 2021). What is the probability of at least one six?

Quick questions

1For a Poisson distribution

2In a binomial distribution with p = q, P(X = x) equals (CUET 2020)

3Cards are drawn n times from a shuffled pack, with replacement. The number of aces follows (Hyderabad 2013)

4The variance of a binomial distribution with n trials cannot exceed

Real questions

Past papers, by level

Foundation

Definitions and one-step problems. Most CUET and UGC NET questions sit here.

14 questionsPractise →

Standard

Two or three steps, up to the hardest JAM, JNU and Hyderabad questions.

13 questionsPractise →

Delhi School level

This topic’s hardest questions, from the DSE entrance, are worked through in Lesson 9: Distributions.

42 questionsLesson 9 →

All 27 past-paper questions for this lesson, easiest first: practise the full set →

These are questions from the TestStop bank. A free account answers 10 a day, each with its worked solution; All-Access opens every set with no daily limit.

← Lesson 5Random variables, expectation and variance Lesson 7 →Continuous distributions

Discrete distributions

Bernoulli: one yes-or-no

Binomial: counting successes

Binomial questions, solved

Geometric: waiting for the first

Hypergeometric: without replacement

Poisson: rare events

When one becomes another

Which distribution?

The opening question

The mistakes examiners are fishing for

The lesson on one screen

Check yourself

Compute

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next

Discrete distributions

Bernoulli: one yes-or-no

Binomial: counting successes

Binomial questions, solved

Geometric: waiting for the first

Hypergeometric: without replacement

Poisson: rare events

When one becomes another

Which distribution?

The opening question

The mistakes examiners are fishing for

The lesson on one screen

Check yourself

Compute

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next