The Binomial Distribution (2024)

"Bi" means "two" (like a bicycle has two wheels) ... ... so this is about things with two results.

Let's Toss a Coin!

Toss a fair coin three times ... what is the chance of getting exactly two Heads?

Using H for heads and T for Tails we may get any of these 8 outcomes:

HHH
HHT
HTH
HTT
THH
THT
TTH
TTT

Which outcomes do we want?

"Two Heads" could be in any order: "HHT", "THH" and "HTH" all have two Heads (and one Tail).

So 3 of the outcomes produce "Two Heads".

What is the probability of each outcome?

Each outcome is equally likely, and there are 8 of them, so each outcome has a probability of 1/8

So the probability of event "Two Heads" is:

So the chance of getting Two Heads is 3/8

3 Heads, 2 Heads, 1 Head, None

The calculations are (P means "Probability of"):

• P(Three Heads) = P(HHH) = 1/8
• P(Two Heads) = P(HHT) + P(HTH) + P(THH) = 1/8 + 1/8 + 1/8 = 3/8
• P(One Head) = P(HTT) + P(THT) + P(TTH) = 1/8 + 1/8 + 1/8 = 3/8
• P(Zero Heads) = P(TTT) = 1/8

We can write this in terms of a Random Variable "X" = "The number of Heads from 3 tosses of a coin":

• P(X = 3) = 1/8
• P(X = 2) = 3/8
• P(X = 1) = 3/8
• P(X = 0) = 1/8

And this is what it looks like as a graph:

It is symmetrical!

Making a Formula

Now imagine we want the chances of 5 heads in 9 tosses: to list all 512 outcomes will take a long time!

So let's make a formula.

In our previous example, how can we get the values 1, 3, 3 and 1 ?

Well, they are actually in Pascal’s Triangle !

Can we make them using a formula?

Sure we can, and here it is:

The formula may look scary but is easy to use. We only need two numbers:

• n = total number
• k = number we want

The "!" means "factorial", for example 4! = 1×2×3×4 = 24

Note: it is often called "n choose k" and you can learn more here.

Let's try it:

Let's use it for a harder question:

Bias!

So far the chances of success or failure have been equally likely.

But what if the coins are biased (land more on one side than another) or choices are not 50/50.

Let's draw a tree diagram:

The "Two Chicken" cases are highlighted.

The probabilities for "two chickens" all work out to be 0.147, because we are multiplying two 0.7s and one 0.3 in each case. In other words

0.147 = 0.7 × 0.7 × 0.3

Or, using exponents:

= 0.7² × 0.3¹

The 0.7 is the probability of each choice we want, call it p

The 2 is the number of choices we want, call it k

And we have (so far):

= p^k × 0.3¹

The 0.3 is the probability of the opposite choice, so it is: 1−p

The 1 is the number of opposite choices, so it is: n−k

Which gives us:

= p^k(1-p)^(n-k)

Where

• p is the probability of each choice we want
• k is the the number of choices we want
• n is the total number of choices

Now we know the probability of each outcome is 0.147

But we need to include that there are three such ways it can happen: (chicken, chicken, other) or (chicken, other, chicken) or (other, chicken, chicken)

OK. That was a lot of work for something we knew already, but now we have a formula we can use for harder questions.

Putting it Together

Now we know how to calculate how many:

n!k!(n-k)!

And the probability of each:

p^k(1-p)^(n-k)

When multiplied together we get:

Important Notes:

• The trials are independent,
• There are only two possible outcomes at each trial,
• The probability of "success" at each trial is constant.

Quincunx

Have a play with the Quincunx (then read Quincunx Explained) to see the Binomial Distribution in action.

Throw the Die

A fair die is thrown four times. Calculate the probabilities of getting:

• 0 Twos
• 1 Two
• 2 Twos
• 3 Twos
• 4 Twos

In this case n=4, p = P(Two) = 1/6

X is the Random Variable ‘Number of Twos from four throws’.

Substitute x = 0 to 4 into the formula:

P(k out of n) = n!k!(n-k)! p^k(1-p)^(n-k)

Like this (to 4 decimal places):

• P(X = 0) = 4!0!4! × (1/6)⁰(5/6)⁴ = 1 × 1 × (5/6)⁴ = 0.4823
• P(X = 1) = 4!1!3! × (1/6)¹(5/6)³ = 4 × (1/6) × (5/6)³ = 0.3858
• P(X = 2) = 4!2!2! × (1/6)²(5/6)² = 6 × (1/6)² × (5/6)² = 0.1157
• P(X = 3) = 4!3!1! × (1/6)³(5/6)¹ = 4 × (1/6)³ × (5/6) = 0.0154
• P(X = 4) = 4!4!0! × (1/6)⁴(5/6)⁰ = 1 × (1/6)⁴ × 1 = 0.0008

Summary: "for the 4 throws, there is a 48% chance of no twos, 39% chance of 1 two, 12% chance of 2 twos, 1.5% chance of 3 twos, and a tiny 0.08% chance of all throws being a two (but it still could happen!)"

This time the graph is not symmetrical:

It is not symmetrical!

It is skewed because p is not 0.5

Sports Bikes

Your company makes sports bikes. 90% pass final inspection (and 10% fail and need to be fixed).

What is the expected Mean and Variance of the 4 next inspections?

First, let's calculate all probabilities.

• n = 4,
• p = P(Pass) = 0.9

X is the Random Variable "Number of passes from four inspections".

Substitute x = 0 to 4 into the formula:

P(k out of n) = n!k!(n-k)! p^k(1-p)^(n-k)

Like this:

• P(X = 0) = 4!0!4! × 0.9⁰0.1⁴ = 1 × 1 × 0.0001 = 0.0001
• P(X = 1) = 4!1!3! × 0.9¹0.1³ = 4 × 0.9 × 0.001 = 0.0036
• P(X = 2) = 4!2!2! × 0.9²0.1² = 6 × 0.81 × 0.01 = 0.0486
• P(X = 3) = 4!3!1! × 0.9³0.1¹ = 4 × 0.729 × 0.1 = 0.2916
• P(X = 4) = 4!4!0! × 0.9⁴0.1⁰ = 1 × 0.6561 × 1 = 0.6561

Summary: "for the 4 next bikes, there is a tiny 0.01% chance of no passes, 0.36% chance of 1 pass, 5% chance of 2 passes, 29% chance of 3 passes, and a whopping 66% chance they all pass the inspection."

Mean, Variance and Standard Deviation

Let's calculate the Mean, Variance and Standard Deviation for the Sports Bike inspections.

There are (relatively) simple formulas for them. They are a little hard to prove, but they do work!

The mean, or "expected value", is:

μ = np

The formula for Variance is:

Variance: σ² = np(1-p)

And Standard Deviation is the square root of variance:

σ = √(np(1-p))

Summary

• The General Binomial Probability Formula:

  P(k out of n) = n!k!(n-k)! p^k(1-p)^(n-k)

• Mean value of X: μ = np
• Variance of X: σ² = np(1-p)
• Standard Deviation of X: σ = √(np(1-p))

8815, 8816, 8820, 8821, 8828, 8829, 8609, 8610, 8612, 8613, 8614, 8615