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Week 11: Bernoulli, Binomial & Poisson Distributions

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2026-09-12

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Statistics for Data Science I — Week 11: Bernoulli, Binomial & Poisson Distributions

IITM’s Week 11 covers Bernoulli trials, independent and identically distributed random variables, binomial random variables, and Poisson random variables.

TL;DR

Week 10 gave generic formulas for any discrete random variable.

Week 11 introduces named probability models that appear repeatedly in real problems.

The main chain is:

\boxed{ \text{Bernoulli trial} \rightarrow \text{repeated IID Bernoulli trials} \rightarrow \text{Binomial count} }

and separately:

\boxed{ \text{count of events over time/space} \rightarrow \text{Poisson model} }

Core formulas:

Bernoulli

P(X=1)=p,\qquad P(X=0)=1-p

E[X]=p

\operatorname{Var}(X)=p(1-p)

Binomial

If:

X\sim\operatorname{Bin}(n,p),

then:

\boxed{ P(X=k) = \binom nkp^k(1-p)^{n-k} }

\boxed{ E[X]=np }

\boxed{ \operatorname{Var}(X)=np(1-p) }

Poisson

If:

X\sim\operatorname{Pois}(\lambda),

then:

\boxed{ P(X=k) = e^{-\lambda} \frac{\lambda^k}{k!} }

\boxed{ E[X]=\lambda }

\boxed{ \operatorname{Var}(X)=\lambda }

1. Bernoulli trial

A Bernoulli trial is a random experiment with exactly two outcomes.

Usually called:

success
failure

Examples:

heads / tails
pass / fail
defective / non-defective
click / no click
customer buys / does not buy

2. Bernoulli random variable

Define:

X= \begin{cases} 1,&\text{success}\\ 0,&\text{failure} \end{cases}

Let:

P(\text{success})=p.

Then:

P(X=1)=p

and:

P(X=0)=1-p.

We write:

\boxed{ X\sim\operatorname{Bernoulli}(p) }

3. Bernoulli PMF

The PMF is:

p_X(x)= \begin{cases} p,&x=1\\ 1-p,&x=0\\ 0,&\text{otherwise} \end{cases}

A compact version is:

\boxed{ P(X=x)=p^x(1-p)^{1-x} }

for:

x\in\{0,1\}.

4. Bernoulli expectation

From Week 10:

E[X] = 0(1-p)+1(p).

Therefore:

\boxed{ E[X]=p }

5. Bernoulli variance

Since:

X^2=X

for:

X\in\{0,1\},

we have:

E[X^2]=p.

Thus:

\operatorname{Var}(X) = E[X^2]-E[X]^2.

So:

=p-p^2

= \boxed{ p(1-p) }

6. Bernoulli standard deviation

\boxed{ SD(X) = \sqrt{p(1-p)} }

7. Repeated Bernoulli trials

Now suppose we repeat a Bernoulli experiment:

n

times.

Example:

Toss a coin 10 times.

Each toss can be represented by:

X_i= \begin{cases} 1,&\text{Head}\\ 0,&\text{Tail} \end{cases}

for:

i=1,\dots,n.

8. Independent trials

Trials are independent if the outcome of one does not change the probability distribution of another.

Formally, for appropriate events:

P(A\cap B)=P(A)P(B).

For Bernoulli trials, this means a previous success/failure does not alter the success probability of a later trial.

9. Identically distributed

Random variables:

X_1,X_2,\dots,X_n

are identically distributed if they have the same probability distribution.

For Bernoulli trials:

X_i\sim\operatorname{Bernoulli}(p)

for every:

i.

So each trial has the same success probability:

p.

10. IID

If random variables are both:

independent
identically distributed

we call them:

\boxed{\text{IID}}

meaning:

Independent and Identically Distributed.

Notation:

X_1,\dots,X_n \overset{\text{iid}}{\sim} \operatorname{Bernoulli}(p).

11. Why IID matters

Many standard probability models assume repeated observations behave similarly and independently.

For example:

repeated fair coin tosses
independent product inspections with same defect rate
repeated independent conversion attempts

IID is a foundational assumption across statistics and machine learning.

12. Not independent example

Draw two cards without replacement.

If the first card is an ace, the probability second card is an ace changes.

So the draws are dependent.

13. Not identically distributed example

Suppose:

P(X_1=1)=0.2

and:

P(X_2=1)=0.8.

Even if independent, they are not identically distributed.

So:

\boxed{ \text{independent}\neq\text{IID} }

IID requires both properties.

14. Binomial random variable

Suppose:

fixed number of trials: $n$
each trial has two outcomes
trials are independent
success probability is constant: $p$

Let:

X=\text{number of successes}.

Then:

\boxed{ X\sim\operatorname{Binomial}(n,p) }

15. Binomial support

Since

X

counts successes among

n

trials:

\boxed{ X\in\{0,1,2,\dots,n\} }

16. Example

Toss a fair coin:

10

times.

Let:

X=\text{number of heads}.

Then:

X\sim\operatorname{Binomial}\left(10,\frac12\right).

17. Binomial PMF

For:

X\sim\operatorname{Bin}(n,p),

the probability of exactly

k

successes is:

\boxed{ P(X=k) = \binom nk p^k (1-p)^{n-k} }

for:

k=0,1,\dots,n.

18. Why the formula works

We need exactly:

k

successes and:

n-k

failures.

For one particular sequence:

SSFSF\cdots

the probability is:

p^k(1-p)^{n-k}.

But there are:

\binom nk

ways to choose which

k

trial positions are successes.

Thus:

\binom nkp^k(1-p)^{n-k}.

This directly combines:

combinations from Week 6
independence from Week 8

19. Example: exactly 3 heads

Toss a fair coin 5 times.

Let:

X=\text{number of heads}.

Then:

X\sim\operatorname{Bin}(5,0.5).

We want:

P(X=3).

So:

P(X=3) = \binom53 (0.5)^3 (0.5)^2.

Since:

\binom53=10,

P(X=3) = 10(0.5)^5

= \frac{10}{32}

= \boxed{0.3125}.

20. Exactly $k$

If the question says:

exactly $k$ successes

use:

\boxed{ P(X=k) }

directly.

21. At most $k$

"At most

k

" means:

X\leq k.

Thus:

\boxed{ P(X\leq k) = \sum_{j=0}^{k} P(X=j) }

22. At least $k$

"At least

k

" means:

X\geq k.

Thus:

P(X\geq k) = \sum_{j=k}^{n}P(X=j).

Often a complement is easier.

23. At least one success

P(X\geq1) = 1-P(X=0).

For binomial:

P(X=0) = (1-p)^n.

Therefore:

\boxed{ P(X\geq1) = 1-(1-p)^n }

This is an extremely common formula.

24. Example

Suppose probability of defect on each independently produced item is:

p=0.05.

Inspect:

10

items.

Probability at least one is defective:

1-(0.95)^{10}.

That is much easier than summing probabilities for:

1,2,\dots,10

defects.

25. Binomial expectation

Let:

X=X_1+\cdots+X_n,

where each:

X_i\sim\operatorname{Bernoulli}(p).

Then:

E[X] = E[X_1]+\cdots+E[X_n].

Since:

E[X_i]=p,

\boxed{ E[X]=np }

26. Intuition

If you repeat a trial:

100

times with success probability:

0.3,

expected number of successes:

100(0.3)=30.

This does not guarantee exactly 30.

It is the long-run average count.

27. Binomial variance

For independent Bernoulli variables:

\operatorname{Var}(X) = \sum_{i=1}^n\operatorname{Var}(X_i).

Each Bernoulli variance is:

p(1-p).

Therefore:

\boxed{ \operatorname{Var}(X) = np(1-p) }

28. Binomial standard deviation

\boxed{ SD(X) = \sqrt{np(1-p)} }

29. Binomial example

Suppose:

X\sim\operatorname{Bin}(20,0.4).

Mean:

E[X] = 20(0.4) = \boxed8.

Variance:

20(0.4)(0.6) = \boxed{4.8}.

Standard deviation:

\boxed{ \sqrt{4.8} }

30. Recognizing a binomial experiment

Ask:


text
1. Fixed number n of trials?
2. Two outcomes per trial?
3. Same success probability p?
4. Independent trials?
5. X counts successes?

If all are yes:

\boxed{\text{Binomial}}

31. Non-binomial example: without replacement

Suppose 5 cards are drawn from a deck without replacement and:

X=\text{number of aces}.

Success probability changes after each draw.

The trials are dependent.

So this is not exactly a standard binomial model.

32. Non-binomial example: changing $p$

Suppose a player's success probabilities are:

0.4,\ 0.5,\ 0.6

across three attempts.

Even if independent,

p

is not constant.

Therefore the total number of successes is not standard:

\operatorname{Bin}(3,p).

33. Binomial symmetry

When:

p=\frac12,

the binomial distribution is symmetric.

For:

X\sim\operatorname{Bin}(n,0.5),

P(X=k)=P(X=n-k).

34. Binomial skew intuition

If:

p<0.5,

successes tend to be relatively rare and the distribution is generally right-skewed.

If:

p>0.5,

it tends to be left-skewed.

n

becomes large under suitable conditions, the shape can become approximately bell-like.

35. Poisson random variable

The Poisson distribution models counts of events occurring over some interval of:

time
distance
area
volume
another exposure measure

Examples:

calls arriving per minute
accidents per month
defects per meter of cable
requests per second
radioactive decays per interval

36. Poisson parameter

A Poisson random variable has parameter:

\lambda>0.

We write:

\boxed{ X\sim\operatorname{Poisson}(\lambda) }

where:

\lambda

is the average number of events in the relevant interval.

37. Poisson support

Poisson counts can be:

\boxed{ 0,1,2,3,\dots }

There is theoretically no fixed upper bound.

38. Poisson PMF

If:

X\sim\operatorname{Poisson}(\lambda),

then:

\boxed{ P(X=k) = e^{-\lambda} \frac{\lambda^k}{k!} }

for:

k=0,1,2,\dots

39. Example

Suppose calls arrive at an average rate:

\lambda=3

per minute.

Let:

X=\text{calls in one minute}.

Then:

X\sim\operatorname{Poisson}(3).

Probability of exactly 2 calls:

P(X=2) = e^{-3}\frac{3^2}{2!}.

So:

\boxed{ P(X=2)=\frac92e^{-3} }

approximately.

40. Probability of zero events

For Poisson:

P(X=0) = e^{-\lambda}\frac{\lambda^0}{0!}.

Since:

\lambda^0=1

and:

0!=1,

\boxed{ P(X=0)=e^{-\lambda} }

This is a very useful shortcut.

41. At least one Poisson event

Using complement:

P(X\geq1) = 1-P(X=0).

Therefore:

\boxed{ P(X\geq1)=1-e^{-\lambda} }

42. Poisson expectation

A remarkable property:

\boxed{ E[X]=\lambda }

The distribution parameter itself is the mean count.

43. Poisson variance

Another special property:

\boxed{ \operatorname{Var}(X)=\lambda }

So for Poisson:

\boxed{ \text{mean}=\text{variance}=\lambda }

44. Poisson standard deviation

\boxed{ SD(X)=\sqrt{\lambda} }

45. Example

If:

X\sim\operatorname{Poisson}(9),

then:

E[X]=9

\operatorname{Var}(X)=9

SD(X)=3.

46. Scaling the interval

Suppose average rate is:

4

events per hour.

For:

3

hours, expected count:

\lambda=4\cdot3=12.

Thus:

X_{3h}\sim\operatorname{Poisson}(12)

under the Poisson-process assumptions.

47. Example

Average calls:

6

per hour.

For:

30\text{ minutes} = 0.5\text{ hour},

parameter:

\lambda=6(0.5)=3.

So use:

\operatorname{Poisson}(3).

A common mistake is using the hourly

\lambda

without rescaling.

48. Poisson-rate intuition

If rate is:

r

events per unit time and interval length is:

t,

then:

\boxed{ \lambda=rt }

assuming a constant event rate.

49. Typical Poisson assumptions

A basic Poisson count model roughly assumes:

events occur independently
average rate is stable
events occur individually rather than in forced clusters
counts are measured over a fixed interval

The exact stochastic-process formulation can be deeper, but this intuition is enough for introductory use.

50. Binomial vs Poisson

Both model counts, but in different setups.

Binomial

Counts successes among a fixed number of trials:

n.

Upper bound:

X\leq n.

Poisson

Counts events in a fixed interval.

Possible values:

0,1,2,\dots

with no fixed finite upper bound.

51. Quick distinction

If question says:

out of 20 products, how many are defective?

Think:

\boxed{\text{Binomial}}

If question says:

how many defects occur per kilometer?

Think:

\boxed{\text{Poisson}}

52. Binomial vs Poisson table

Feature	Binomial	Poisson
Parameter(s)	$n,p$	$\lambda$
Values	$0,\dots,n$	$0,1,2,\dots$
Context	successes in fixed trials	events in fixed interval
Mean	$np$	$\lambda$
Variance	$np(1-p)$	$\lambda$

53. Poisson approximation to binomial

When:

$n$ is large
$p$ is small
$np$ is moderate

then:

\operatorname{Binomial}(n,p)

can often be approximated by:

\operatorname{Poisson}(\lambda)

with:

\boxed{ \lambda=np }

This is the classic rare-event approximation.

54. Why this makes sense

Suppose:

n=1000

and:

p=0.002.

Expected number of successes:

np=2.

We have:

many opportunities
tiny probability each time
relatively small count

This is exactly the kind of setting where Poisson behavior emerges.

55. Example approximation

Suppose:

X\sim\operatorname{Bin}(1000,0.002).

Approximate using:

Y\sim\operatorname{Poisson}(2).

Then:

P(X=0) \approx P(Y=0) = e^{-2}.

Much easier than:

(0.998)^{1000}

when manual calculations are involved.

56. Binomial exact probability example

Suppose:

X\sim\operatorname{Bin}(8,0.25).

Find:

P(X=2).

Use:

P(X=2) = \binom82 (0.25)^2 (0.75)^6.

Since:

\binom82=28,

\boxed{ P(X=2) = 28(0.25)^2(0.75)^6 }

57. Binomial cumulative probability

Find:

P(X\leq2).

Then:

\boxed{ P(X\leq2) = P(X=0)+P(X=1)+P(X=2) }

Do not confuse:

P(X=2)

with:

P(X\leq2).

58. Binomial complement example

Find:

P(X>2).

Instead of summing:

3,4,\dots,n,

use:

\boxed{ P(X>2) = 1-P(X\leq2) }

Complement strategies remain extremely useful.

59. Poisson exact probability example

Suppose:

X\sim\operatorname{Poisson}(4).

Find exactly three events:

P(X=3) = e^{-4}\frac{4^3}{3!}.

Thus:

= e^{-4}\frac{64}{6}.

60. Poisson cumulative example

Find:

P(X\leq2).

Then:

P(X=0)+P(X=1)+P(X=2).

So:

e^{-4} \left[ 1+4+\frac{4^2}{2} \right].

Thus:

\boxed{ 13e^{-4} }

61. At least three Poisson events

P(X\geq3) = 1-P(X\leq2).

So in the previous example:

\boxed{ P(X\geq3) = 1-13e^{-4} }

Complement avoids an infinite sum.

62. Finding Poisson $\lambda$ from the mean

If a problem tells you:

the average count is 7

and assumes Poisson, immediately:

\boxed{ \lambda=7. }

Because:

E[X]=\lambda.

63. Finding $\lambda$ from variance

For a Poisson random variable:

\operatorname{Var}(X)=\lambda.

So if variance is:

5,

then:

\boxed{ \lambda=5. }

64. Diagnostic property

If observed/theoretical mean and variance are wildly different, a Poisson model may be questionable.

The theoretical Poisson property is:

\boxed{ E[X]=\operatorname{Var}(X) }

Though actual sample mean and variance will not generally be exactly equal in finite data.

65. IID Bernoulli sum representation

The binomial model can be written:

X=X_1+\cdots+X_n

where:

X_i \overset{\text{iid}}{\sim} \operatorname{Bernoulli}(p).

Then:

X

literally counts how many indicators equal

1

This is the cleanest conceptual definition of a binomial random variable.

66. Example

Suppose:

X_i= 1

if customer

i

buys.

Then total buyers among

n

customers:

X=\sum_{i=1}^nX_i.

Under IID Bernoulli assumptions:

\boxed{ X\sim\operatorname{Bin}(n,p). }

67. Why expectation becomes $np$

Because:

E[X] = E\left[\sum_iX_i\right].

Linearity:

=\sum_iE[X_i].

Each:

E[X_i]=p.

Therefore:

np.

This connects Weeks 10 and 11 directly.

68. Why variance becomes $np(1-p)$

For independent trials:

\operatorname{Var}(X) = \sum_i\operatorname{Var}(X_i).

Each:

\operatorname{Var}(X_i)=p(1-p).

Therefore:

np(1-p).

Common mistake: Binomial means any count

No.

A count variable is not automatically binomial.

You need the binomial structure:

fixed $n$
two outcomes
constant $p$
independent trials
count successes

Common mistake: Poisson means any count

Also no.

Poisson is designed for a particular event-counting structure with parameter:

\lambda.

A bounded count out of exactly

n

trials is usually more naturally binomial.

Common mistake: confuse

p

and

\lambda

Binomial:

p

is a probability, so:

0\leq p\leq1.

Poisson:

\lambda

is an expected count/rate parameter and can exceed

1

For example:

\lambda=20

is perfectly valid.

Common mistake: use

\lambda

as probability

If:

X\sim\operatorname{Pois}(3),

do not say:

P(X=1)=3.

Probability is obtained through:

e^{-3}\frac{3^1}{1!}.

\lambda

is the mean parameter, not a probability.

Common mistake: forgetting

\binom nk

Binomial probability is not just:

p^k(1-p)^{n-k}.

That is the probability of one particular sequence with

k

successes.

There are:

\binom nk

such sequences.

So multiply by it.

Common mistake:

n^k

instead of

\binom nk

We are choosing which:

k

of the:

n

positions contain successes.

Order within those success positions is irrelevant.

Therefore:

\binom nk,

not:

n^k.

Common mistake: wrong failure exponent

Exactly:

k

successes means:

n-k

failures.

So:

\boxed{ p^k(1-p)^{n-k} }

not:

p^k(1-p)^k.

Common mistake: "at least" vs "at most"

At least $k$

X\geq k.

At most $k$

X\leq k.

More than $k$

X>k.

Fewer than $k$

X<k.

Translate wording before calculating.

Common mistake: mean = most likely count

For:

X\sim\operatorname{Bin}(n,p),

E[X]=np.

This need not be the mode and need not even be an integer.

Example:

np=2.7.

The number of successes cannot equal:

2.7,

but the expectation can.

Common mistake: independence vs identical distribution

IID contains two independent requirements.

Random variables may be:

independent but not identically distributed
identically distributed but dependent
both
neither

Do not collapse them into one idea.

79. Distribution-identification workflow


text
Count variable?
│
├── Fixed number n of success/failure trials?
│      └── Binomial
│
├── Single success/failure trial?
│      └── Bernoulli
│
└── Number of events in time/space interval?
       └── Poisson

Then verify assumptions rather than relying only on keywords.

80. Formula sheet — Bernoulli

If:

X\sim\operatorname{Bernoulli}(p),

then:

\boxed{ P(X=1)=p }

\boxed{ P(X=0)=1-p }

\boxed{ E[X]=p }

\boxed{ \operatorname{Var}(X)=p(1-p) }

81. Formula sheet — Binomial

If:

X\sim\operatorname{Bin}(n,p),

then:

\boxed{ P(X=k) = \binom nk p^k(1-p)^{n-k} }

\boxed{ E[X]=np }

\boxed{ \operatorname{Var}(X) = np(1-p) }

\boxed{ SD(X) = \sqrt{np(1-p)} }

82. Formula sheet — Poisson

If:

X\sim\operatorname{Pois}(\lambda),

then:

\boxed{ P(X=k) = e^{-\lambda} \frac{\lambda^k}{k!} }

\boxed{ E[X]=\lambda }

\boxed{ \operatorname{Var}(X)=\lambda }

\boxed{ SD(X)=\sqrt{\lambda} }

83. Binomial ↔ Poisson summary

For rare events:

n\text{ large}, \qquad p\text{ small},

use approximately:

\boxed{ \operatorname{Bin}(n,p) \approx \operatorname{Pois}(np) }

when the approximation conditions are suitable.

84. Week 11 conceptual map


text
Bernoulli(p)
    │
    │ repeat n IID times
    ▼
Binomial(n,p)
    │
    │ large n, small p
    │ λ = np
    ▼
Poisson(λ)   [approximation setting]

Separately:
Poisson(λ)
= event count over interval

85. Weeks 9–11 progression


text
Week 9
Random variables + PMF/CDF
        ↓
Week 10
Expectation + variance
        ↓
Week 11
Named discrete distributions
├── Bernoulli
├── Binomial
└── Poisson

The reason named distributions are useful is that you no longer have to derive everything from scratch.

Once you identify:

X\sim\operatorname{Bin}(n,p),

you immediately know:

PMF
mean
variance
support

86. What you should actually master

By the end of Stats 1 Week 11, you should be able to:

Define a Bernoulli trial.
Recognize a Bernoulli random variable.
Compute Bernoulli mean and variance.
Explain independent vs identically distributed.
Explain what IID means.
Recognize the assumptions of a binomial experiment.
Use: $P(X=k)=\binom nkp^k(1-p)^{n-k}.$
Solve exactly / at least / at most binomial questions.
Use complement probabilities efficiently.
Know: $E[X]=np.$
Know: $\operatorname{Var}(X)=np(1-p).$
Recognize a Poisson counting problem.
Rescale $\lambda$ when the observation interval changes.
Use: $P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}.$
Know: $E[X]=\operatorname{Var}(X)=\lambda.$
Distinguish Binomial from Poisson.
Understand the rare-event Poisson approximation to Binomial.

If those are comfortable, Stats 1 Week 11 is under control.

Quick Recall & Flashcards

Test your active recall before proceeding to practice problems. Format:

Question :: Answer.

Week 11 Diagnostic: If X follows a Binomial distribution with n = 10 and p = 0.3, what is E[X]? :: Answer: 3 — For X ~ Bin(n, p), E[X] = n * p = 10 * 0.3 = 3.
Week 11 Diagnostic: For a Poisson random variable Y with parameter lambda = 4, what is the variance Var(Y)? :: Answer: 4 — For Y ~ Pois(lambda), Var(Y) = lambda = 4.

Practice & Self-Check

Attempt these diagnostic questions to verify your operational understanding.

Question 1

If X follows a Binomial distribution with n = 10 and p = 0.3, what is E[X]?

NAT

3 Unit Assessment

Question 2

For a Poisson random variable Y with parameter lambda = 4, what is the variance Var(Y)?

NAT

3 Unit Assessment