01 / Binomial distribution

Toss a coin.
Count the heads.

This coin is biased: each toss has a 60% chance of heads and a 40% chance of tails. Toss it three times. Let X be the number of heads.

Heads · H 0.6Tails · T 0.4

1. Watch three tosses2. Find the matching arrangements3. Continue to Key Concepts

1One experiment = three tosses

What will happen this time?

P(H) = 0.6

Probability of this exact order

Independent tosses: multiply the probabilities along one order.

Toss 1 of 3 is next. The outcome is not known yet.

H means heads. T means tails.

What are we counting?

“Success” means heads. It is just the outcome we choose to count.

0

heads so far
0 of 3 tosses completed

H T H gives X = 2.
T H H also gives X = 2.
The order differs; the head count is the same.

Every toss is independent. Earlier results do not change the next toss: P(H) remains 0.6.

60% is a probability, not a promise. Three tosses might give 0, 1, 2 or 3 heads. The fraction of heads tends to be near 0.6 only after many independent tosses.

2Why does the combination number appear?

Find all arrangements with exactly 2 heads

All 8 outcomes

Choose a head count. The green cards are the outcomes belonging to that event. Each card records the order of all three tosses.

The 8 outcomes are not equally likely. Our coin has P(H) = 0.6, not 0.5. For example, HHH has probability 0.216, while TTT has probability 0.064.

C(3,2) = 3How many arrangements have this head count?

0.6²Multiply by 0.6 once for each head.

0.4¹Multiply by 0.4 once for each tail.

Multiply within one order; then multiply by the number of matching orders.HHT, HTH and THH each have probability 0.144. They are mutually exclusive, so 0.144 + 0.144 + 0.144 = 3 × 0.144. The factor 3 counts orders, not repeated experiments.

You have seen the experiment. Now explain why the formula works.

Binomial Distribution / Key Concepts

Understand why.
Then solve the question.

Start with a reason, connect it to a key idea, study a worked example, then try a similar exercise.

Why do we need a Binomial model? With 3 tosses we can list 8 orders. With 10 tosses there are 1,024 orders. If the question only asks how many heads, the Binomial formula groups the matching orders and adds their probabilities efficiently.

Key concept 1 / 5

1 · What are we counting, and when is it Binomial?

Why?

The model must match what X counts. A fixed n gives a known number of positions. Independence allows multiplication, and a common p makes orders with the same success count equally likely. These conditions let one Binomial formula group the orders correctly.

Key idea

Use X ∼ Bin(n, p) when there are a fixed n trials, two outcomes per trial, independent trials, and the same success probability p. X counts successes, so X = 0, 1, …, n.

Worked example

Three independent tosses, with P(H) = 0.6. Let X count heads.

Define the counted outcome: success = H, so p = 0.6.
There are n = 3 independent trials; each gives H or T, with unchanged probabilities.
Therefore X ∼ Bin(3, 0.6), with possible counts 0, 1, 2, 3.

X ∼ Bin(3, 0.6)

Watch out: Before using the formula, check independence and match p to what X counts.

Key concept 2 / 5

2 · Why multiply along one order?

Why?

HHT means all three specified results occur: first H AND second H AND third T. Independence means earlier results do not change the probabilities of later results, so the probability of this joint event is a product.

Key idea

For one specified order, multiply one factor p for each success and one factor 1 − p for each failure. With k successes in n trials, the result is pᵏ(1 − p)ⁿ⁻ᵏ.

Worked example

Find the probability of the exact order HHT.

First H: probability 0.6. Second H: probability 0.6. Third T: probability 0.4.
Multiply because all three results are required and the tosses are independent.
0.6 × 0.6 × 0.4 = 0.144. This is one order only.

P(HHT) = 0.6² × 0.4 = 0.144

Watch out: A probability for one exact order is different from a probability for a head count.

Key concept 3 / 5

3 · Why does the combination number appear?

Why?

“Exactly two heads” allows HHT OR HTH OR THH. These orders cannot occur together in the same experiment, so their probabilities add. Each has the same number of H and T factors, so they have equal probabilities. Adding equal terms becomes multiplication by the number of orders.

Key idea

C(n,k) counts the ways to choose the k success positions among n positions. Therefore P(X = k) = C(n,k)pᵏ(1 − p)ⁿ⁻ᵏ, for k = 0, 1, …, n.

Worked example

For X ∼ Bin(3, 0.6), find P(X = 2).

The matching orders are HHT, HTH and THH: C(3,2) = 3.
Each order has probability 0.6² × 0.4 = 0.144.
Add the three mutually exclusive cases: 0.144 + 0.144 + 0.144 = 3 × 0.144 = 0.432.

C(n,k) = n! / [k!(n − k)!] · 0! = 1

Watch out: C(n,k) counts possible orders. It is not the number of repeated experiments.

Key concept 4 / 5

4 · Why translate the words before calculating?

Why?

The formula gives the probability of one count k. Wording such as “at least” selects several possible counts. We must decide which integers belong to the event before adding probabilities. A complement can replace a long sum with a shorter calculation.

Key idea

Exactly r: X = r. At most r: X ≤ r. At least r: X ≥ r. More than r: X > r, or X ≥ r + 1 for integer counts. The complement of X ≥ r is X ≤ r − 1.

Worked example

For X ∼ Bin(3, 0.6), find the probability of at least one head.

At least one means X ∈ {1, 2, 3}.
The only excluded count is X = 0: all three tosses are tails. Its probability is 0.4³ = 0.064.
Use total probability 1: P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.064 = 0.936.

P(X ≥ 1) = 1 − P(X = 0) = 0.936

Watch out: At least one and more than one are different events. List the included integers.

Key concept 5 / 5

5 · Why can the mean be a non-integer?

Why?

Each trial contributes either 1 for success or 0 for failure. Its expected contribution is 1 × p + 0 × (1 − p) = p. Adding n expected contributions gives np. Each trial has variance p(1 − p); independent trials let these variances add.

Key idea

For X ∼ Bin(n, p), E(X) = np, Var(X) = np(1 − p), and SD(X) = √[np(1 − p)]. A mean describes an average count; it need not be a possible outcome of one experiment.

Worked example

For X ∼ Bin(3, 0.6), find the mean, variance and standard deviation.

Mean: E(X) = 3 × 0.6 = 1.8 heads per three-toss experiment on average.
Variance: Var(X) = 3 × 0.6 × 0.4 = 0.72.
Standard deviation: SD(X) = √0.72 ≈ 0.8485. An individual head count is still 0, 1, 2 or 3.

E(X) = 1.8 · Var(X) = 0.72 · SD(X) ≈ 0.8485

Watch out: Mean, variance and standard deviation describe different quantities. Do not use np for all three.

Check a probability after writing your method

Use X ∼ Bin(3, 0.6). Write the event and the calculation first, then use this tool to check.

heads

Calculated probabilities for X = 0, 1, 2, 3

Explain the conditions, the probability product and the combination number before using the formula.

02 / Poisson distribution

Watch the shop.
Count customers per hour.

Start with an average rate of 4 customers per hour; you can change it below. That does not mean exactly 4 arrive every hour. Let X be the number arriving during one hour.

1Watch one complete hour

One observation window

60 minutes

The animation plays one hour in 6 seconds. The clock shows minutes:seconds.

Elapsed time

00:00

0

customers
arrived so far

0 min15304560 min

Press Watch one hour. Count each customer as they arrive.

Arrival times will appear here.

Average does not mean fixed.

One hour might have 2 customers. Another might have 5. The Poisson distribution describes the probabilities of these different counts.

Average customers per hour 4

1 customer/hour12 customers/hour

Rate: 4 customers/hour
Observation time: 1 hour
Mean count λ: 4 × 1 = 4 customers

We assume customers arrive independently at a constant average rate. They do not arrive at equally spaced times. A constant average rate is a modelling assumption, not something guaranteed by knowing an average.

Hours recorded

0

Observed mean X

—

Expected mean λ

4.00

One hour gives one count.

Read each card as “during this hour, this many customers arrived”. The most recent 12 completed hours are shown.

No complete hours recorded yet.

Notice the difference: the observation time is fixed at 1 hour, but the number of arrivals is random. An empty hour is possible, and an hour with more than the average is possible.

2Turn the hourly records into a distribution

The horizontal axis is customers in one hour. The gold bar at 3 is the fraction of recorded hours that had exactly 3 customers.

Customers arriving in one hour

X ∼ Pois(4)

Exact probabilityRecorded hourly frequency

Hourly count table and plotted tail

Put the story into a formula.

X ∼ Pois(λ)

λ is the expected number of arrivals in the chosen interval. Here λ = rate × 1 hour.

P(X = k) = e^−λ λ^k / k!

k is a whole-number count: 0, 1, 2, … . Unlike three coin tosses, there is no fixed maximum number of customers in the model.

customers

3Use the formula, one step at a time

Exactly 3 customers in an hour

1 · Define X. X counts customers arriving during one hour.

2 · State the model. Under the stated Poisson arrival assumptions, λ = (4 customers/hour)(1 hour) = 4, so X ∼ Pois(4).

3 · Translate. “Exactly 3” means X = 3.

4 · Substitute. P(X = 3) = e⁻⁴ × 4³ / 3!
= e⁻⁴ × 64 / 6 ≈ 0.1954.

5 · Interpret. About 19.54% of hours would have exactly 3 arrivals in the long run, under this model.

What about no customers?

P(X = 0) = e⁻⁴ × 4⁰ / 0! = e⁻⁴ ≈ 0.0183. Zero is a valid count. Remember 4⁰ = 1 and 0! = 1.

Keep the units consistent

The same rate of 4 customers/hour gives:

Observation time	Mean count λ
30 minutes = 0.5 hour	4 × 0.5 = 2
1 hour	4 × 1 = 4
2 hours	4 × 2 = 8

E(X) = λ · Var(X) = λ

For one hour at the original rate, mean = 4, variance = 4 and standard deviation = √4 = 2.

More than 3? Use P(X > 3) = 1 − P(X ≤ 3). Add the probabilities of 0, 1, 2 and 3, then subtract from 1.

Can you explain why a rate of 4/hour can produce 2 customers in one hour?

03 / Practice

A small check.
One step at a time.

Four questions about coin tosses, customers, event wording, and mean and variance. Each answer comes with an explanation. Your responses are not sent to Canvas.

Before calculating, write these three things.

What is X? Say exactly what you count. Which model? Give its parameters. Which event? Write the words as an equation or inequality.

Go back and replay whichever experiment you want to understand better.

Toss a coin.Count the heads.

1One experiment = three tosses

What will happen this time?

What are we counting?

2Why does the combination number appear?

Find all arrangements with exactly 2 heads

Understand why.Then solve the question.

1 · What are we counting, and when is it Binomial?

Why?

Key idea

Worked example

Your turn

2 · Why multiply along one order?

Why?

Key idea

Worked example

Your turn

3 · Why does the combination number appear?

Why?

Key idea

Worked example

Your turn

4 · Why translate the words before calculating?

Why?

Key idea

Worked example

Your turn

5 · Why can the mean be a non-integer?

Why?

Key idea

Worked example

Your turn

Check a probability after writing your method

Watch the shop.Count customers per hour.

1Watch one complete hour

One observation window

Average does not mean fixed.

One hour gives one count.

2Turn the hourly records into a distribution

Customers arriving in one hour

Put the story into a formula.

3Use the formula, one step at a time

Exactly 3 customers in an hour

Keep the units consistent

A small check.One step at a time.

Before calculating, write these three things.

Toss a coin.
Count the heads.

Understand why.
Then solve the question.

Watch the shop.
Count customers per hour.

A small check.
One step at a time.