Probability Distributions — Binomial & Poisson

What Is a Probability Distribution?

A probability distribution assigns a probability to every possible value of a random variable. It's a complete description of all possible outcomes and how likely each is.

Random Variable X = number of heads in 3 coin flips

x     P(X=x)
0      1/8 = 0.125
1      3/8 = 0.375
2      3/8 = 0.375
3      1/8 = 0.125
       ─────────
Total: 1.000    ← probabilities must sum to 1

Expected Value and Variance of a Discrete Distribution

Expected Value (Mean)

E(X) = μ = Σ [x × P(X=x)]

For the coin example:
E(X) = 0×0.125 + 1×0.375 + 2×0.375 + 3×0.125
     = 0 + 0.375 + 0.75 + 0.375
     = 1.5

→ On average, you expect 1.5 heads in 3 flips

Variance

Var(X) = σ² = Σ [(x − μ)² × P(X=x)]
            = E(X²) − [E(X)]²

SD(X) = σ = √Var(X)

The Binomial Distribution

When to Use It

The binomial distribution models the number of successes in n independent trials, each with the same probability of success p.

Four conditions (BIN):

1. Binary outcomes: each trial is success or failure
2. Independent trials
3. Number of trials (n) is fixed in advance
4. Same probability of success p on each trial

Examples:
- Number of defective items in a batch of 20 (p = defect rate)
- Number of loan defaults in a portfolio of 50 (p = default probability)
- Number of customers who click an ad from 1000 shown (p = click rate)
- Number of patients who respond to a drug from 30 (p = response rate)

Probability Mass Function (PMF)

P(X = k) = C(n,k) × pᵏ × (1−p)^(n−k)

Where:
  n = number of trials
  k = number of successes we're interested in
  p = probability of success on each trial
  C(n,k) = n! / (k!(n−k)!) = number of ways to get k successes in n trials

Binomial Parameters

If X ~ B(n, p):
E(X) = μ = np
Var(X) = σ² = np(1−p)
SD(X) = σ = √(np(1−p))

Step-by-Step Example

Problem: A customer call centre handles 10 calls per hour. Each call has a 30% chance of resulting in a sale. What is the probability of exactly 3 sales in one hour?

n = 10, p = 0.30, k = 3

P(X = 3) = C(10,3) × 0.30³ × 0.70⁷
         = 120 × 0.027 × 0.0824
         = 120 × 0.002225
         = 0.267

→ 26.7% chance of exactly 3 sales

Cumulative Probabilities

P(X ≤ 3) = P(0) + P(1) + P(2) + P(3)

P(X=0) = C(10,0) × 0.30⁰ × 0.70¹⁰ = 1 × 1 × 0.0282 = 0.0282
P(X=1) = C(10,1) × 0.30¹ × 0.70⁹ = 10 × 0.30 × 0.0404 = 0.1211
P(X=2) = C(10,2) × 0.30² × 0.70⁸ = 45 × 0.09 × 0.0576 = 0.2335
P(X=3) = 0.2668

P(X ≤ 3) = 0.0282 + 0.1211 + 0.2335 + 0.2668 = 0.6496

→ 65% chance of 3 or fewer sales
→ P(X > 3) = 1 − 0.6496 = 0.3504 (35% chance of more than 3 sales)

Binomial Distribution Shape

p = 0.5: Symmetric bell shape
p < 0.5: Right-skewed (most probability at lower values)
p > 0.5: Left-skewed (most probability at higher values)
n large, p moderate: approaches Normal distribution (Central Limit Theorem preview)

The Poisson Distribution

When to Use It

The Poisson distribution models the number of rare events in a fixed interval (time, space, area) when events occur independently at a constant average rate.

Conditions:

1. Events occur independently
2. Average rate (λ) is constant over the interval
3. Two events cannot occur at exactly the same instant
4. Appropriate for rare events (small probability per tiny interval)

Examples:
- Number of server crashes per day (λ = 2 per day)
- Number of customer complaints per week (λ = 5 per week)
- Number of accidents on a highway per month (λ = 3 per month)
- Number of α-particle emissions per second from a radioactive source
- Number of defaults in a loan portfolio of 10,000 (if rate is very low)

Probability Mass Function

P(X = k) = (e^(−λ) × λᵏ) / k!

Where:
  λ (lambda) = average number of events in the interval (the only parameter!)
  k = number of events we're interested in
  e = Euler's number ≈ 2.71828

Poisson Parameters

If X ~ Poisson(λ):
E(X) = μ = λ
Var(X) = σ² = λ     ← unusual: mean equals variance!
SD(X) = √λ

Step-by-Step Example

Problem: A bank's ATM machine breaks down an average of 2 times per month. What is the probability of exactly 0, 1, and 3 breakdowns next month?

λ = 2, e^(−2) = 0.1353

P(X=0) = (0.1353 × 2⁰) / 0! = (0.1353 × 1) / 1 = 0.1353 (13.5%)
P(X=1) = (0.1353 × 2¹) / 1! = (0.1353 × 2) / 1 = 0.2707 (27.1%)
P(X=2) = (0.1353 × 2²) / 2! = (0.1353 × 4) / 2 = 0.2707 (27.1%)
P(X=3) = (0.1353 × 2³) / 3! = (0.1353 × 8) / 6 = 0.1804 (18.0%)
P(X=4) = (0.1353 × 2⁴) / 4! = (0.1353 × 16) / 24 = 0.0902 (9.0%)

P(X ≥ 1) = 1 − P(X=0) = 1 − 0.1353 = 0.8647
→ 86.5% chance of at least one breakdown

Scaling the Poisson Rate

λ is tied to the interval. If the rate changes, scale λ:

Average rate: 2 breakdowns per month (λ_month = 2)

For a 2-week period:   λ = 2 × (2/4) = 1
For a 6-month period:  λ = 2 × 6 = 12
For a year:            λ = 2 × 12 = 24

Comparing Binomial and Poisson

Feature	Binomial	Poisson
Trials	Fixed (n)	Not fixed
Outcomes per trial	Binary (success/fail)	Count (0,1,2,...)
Parameter(s)	n and p	λ only
Mean	np	λ
Variance	np(1−p)	λ
Mean = Variance?	No (unless special case)	Always
Used for	n trials, count successes	Rate of rare events
Shape	Depends on p	Skewed right (λ small), symmetric (λ large)

Poisson as Limit of Binomial

When n is large and p is small (rare events), Binomial(n, p) ≈ Poisson(λ = np):

n = 1000 transactions, p = 0.001 (0.1% fraud rate)
λ = np = 1000 × 0.001 = 1

P(exactly 2 fraud cases) ≈ Poisson(1):
P(X=2) = e^(−1) × 1² / 2! = 0.368 × 1 / 2 = 0.184

This saves calculating C(1000,2) × 0.001² × 0.999⁹⁹⁸

Practical Examples

Example 1: Credit Risk — Binomial

A bank approves 20 loans per month. Historical default rate = 5%.
Model: X ~ B(20, 0.05)

E(X) = 20 × 0.05 = 1 default per month on average
SD(X) = √(20 × 0.05 × 0.95) = √0.95 = 0.975

P(no defaults) = C(20,0) × 0.05⁰ × 0.95²⁰ = 0.95²⁰ = 0.358
P(2 or more defaults) = 1 − P(0) − P(1)
P(X=1) = C(20,1) × 0.05¹ × 0.95¹⁹ = 20 × 0.05 × 0.377 = 0.377
P(X≥2) = 1 − 0.358 − 0.377 = 0.265

→ 26.5% chance of 2 or more defaults in a month

Example 2: Call Centre — Poisson

Customer complaints arrive at 4 per hour on average (λ = 4).
Staff need to handle 6 or more complaints in an hour.

P(X ≥ 6) = 1 − P(X ≤ 5)

P(X=0) = e⁻⁴ × 4⁰/0! = 0.0183
P(X=1) = e⁻⁴ × 4¹/1! = 0.0733
P(X=2) = e⁻⁴ × 4²/2! = 0.1465
P(X=3) = e⁻⁴ × 4³/3! = 0.1954
P(X=4) = e⁻⁴ × 4⁴/4! = 0.1954
P(X=5) = e⁻⁴ × 4⁵/5! = 0.1563

P(X ≤ 5) = 0.7851
P(X ≥ 6) = 1 − 0.7851 = 0.2149

→ 21.5% chance of needing 6+ staff-hours in a given hour

Example 3: A/B Test — Binomial

Current click rate: 5% (p = 0.05)
New ad shown to 200 people (n = 200)

E(clicks) = 200 × 0.05 = 10
SD = √(200 × 0.05 × 0.95) = √9.5 = 3.08

P(more than 15 clicks) = P(X > 15)?
Using normal approximation (since n is large):
z = (15 − 10) / 3.08 = 1.62
P(X > 15) ≈ P(Z > 1.62) ≈ 0.053 (about 5.3%)

→ Only 5.3% chance of 16+ clicks by chance if the true rate is 5%

Common Mistakes

1. Applying binomial when trials are not independent

Drawing 5 cards from a deck without replacement:
Trials are dependent (removing one card changes the deck)
→ Use Hypergeometric distribution, not Binomial

2. Using wrong λ for the time interval

Rate: 3 accidents/day
Question: P(0 accidents in next 8 hours)
λ for 8 hours = 3 × (8/24) = 1
P(X=0) = e^(−1) = 0.368

Mistake: using λ=3 for 8 hours → wrong!

3. Applying Poisson when rate is not constant

Website traffic is Poisson during business hours (λ=50/minute)
But NOT Poisson across the day (rate varies with time of day)
→ Need different λ values for different time periods

4. Confusing Poisson (rate) with Binomial (trials)

"10 customers arrive in an hour" sounds Poisson
But if fixed n=10 customers each buy with p=0.3, that's Binomial
→ No fixed number of trials = Poisson
   Fixed n trials, binary outcome = Binomial

Practice Exercises

1. A production line has a 4% defect rate. A sample of 25 items is inspected. Find: a) P(exactly 2 defects) b) P(3 or more defects) c) Expected number of defects and SD

2. Customer service emails arrive at an average rate of 6 per hour. Find: a) P(exactly 4 emails in one hour) b) P(10 or more emails in one hour) c) P(at most 2 emails in 30 minutes)

3. A sales team has 15 prospects. Each has a 20% probability of converting. What is the probability that at least 4 will convert? What is the expected number of conversions?

4. Traffic accidents occur at a rate of 2 per week on a stretch of highway. What is the probability of zero accidents in any given week? What is the probability of 5 or more accidents in a month (4 weeks)?

5. In a portfolio of 500 bonds, each bond has a 0.5% probability of defaulting in the next year. Use the Poisson approximation to find P(3 or more defaults).

Summary

In this chapter you learned:

• Random variable: a variable whose value is determined by a random outcome
• E(X) = Σ[x×P(X=x)] — expected value (mean); Var(X) = E(X²) − [E(X)]²
• Binomial distribution B(n,p): n independent binary trials, p = success probability
  • PMF: P(X=k) = C(n,k) × pᵏ × (1−p)^(n−k)
  • E(X) = np, Var(X) = np(1−p)
  • Use when: fixed n trials, binary outcomes, independent, same p
• Poisson distribution Poisson(λ): count of rare events in a fixed interval
  • PMF: P(X=k) = e^(−λ) × λᵏ / k!
  • E(X) = Var(X) = λ (unique: mean equals variance)
  • Use when: no fixed n, events occur at constant rate, independently
• Poisson as limit of Binomial: when n large and p small, λ = np
• Scale Poisson rate proportionally to match the time/space interval
• Complement rule for "at least one" or "k or more": 1 − P(X < k)

Next up: The Normal Distribution — the most important continuous distribution in statistics.