The Binomial Distribution (Edexcel GCSE Statistics)

Revision Note

Written by: Roger B

Reviewed by: Dan Finlay

Characteristics of the Binomial Distribution

What is a probability distribution?

A probability distribution is a list of all the possible outcomes of an experiment, along with the probabilities for each outcome
For example, if the experiment is flipping a fair coin then the distribution can be represented in a table
- x stands for a possible outcome
- P(x) is the probability of that outcome occurring

x	heads	tails
P(x)	$\frac{1}{2}$	$\frac{1}{2}$

Or if the experiment is rolling a fair dice then the distribution can also be given in a table as

x	1	2	3	4	5	6
P(x)	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$

What is a binomial distribution?

A binomial distribution can describe an experiment where
- Something is repeated a number of times
  - For example, flipping a fair coin 10 times
  - Each repeat is called a trial
- The number of successes is counted
  - For example, counting the number of heads you get in the 10 tosses
  - A success doesn't have to be a good thing
    - It's just the name used for one of the outcomes
To use a binomial distribution, the following conditions must be met
- The number of trials is fixed
- Each trial has two possible outcomes ('success' and 'failure')
- The trials are independent
- The probability of success in each trial is constant
The notation B(n, p) is used to denote a binomial distribution
- n is the number of trials
- p is the probability of success
Because there are only two possible outcomes
- the probability of failure is 1-p
- The probability of failure is often denoted by q
  - where q = 1-p
The mean of the binomial distribution B(n, p) is np
- This is the expected frequency for the number of successes in n trials

Examiner Tips and Tricks

Be sure to learn the conditions for when a binomial distribution is appropriate
- Exam questions often ask about these specifically

Worked Example

Hannah is a snowboarder who is trying to perform the Poptart trick.

Hannah would like to use a binomial distribution to find the probabilities for how many times she will successfully complete the Poptart trick, out of her next 12 attempts.

(a) Give a reason why the binomial distribution might be suitable in this case.

Consider the list of necessary conditions for using a binomial distribution
Write down one that is definitely met here
Relate it to the context

There are a fixed number of trials (her 12 attempts)

'There are only two possible outcomes, success (doing the trick) and failure (not doing the trick)' would also get the mark

(b) Suggest a reason why the binomial distribution may not be suitable in this case.

Consider the list of necessary conditions for using a binomial distribution
Write down one that might not be met here

The trials might not be independent, because she might get better each time from practising the trick

'The probability of success might not be constant' (for the same reason) would also get the mark

Hannah successfully performs the Poptart trick 20% of the time.

(c) Assuming that using a binomial distribution is suitable, write down the distribution that Hannah could use in the form B(n, p).

The number of trials, n, is 12
The probability of success, p, is 0.2 (20% as a decimal)

$B (12, 0.2)$

$B (12, \frac{1}{5})$ would also get the mark

Calculating Binomial Probabilities

What are the probabilities for a binomial distribution?

The probabilities for a binomial distribution B(n, p) can be found by expanding the bracket ${(p + q)}^{n}$
- n is the number of trials
- p is the probability of success
- q = 1-p is the probability of failure
The terms in the expansion give the probabilities for the different numbers of successes
For n=1, ${(p + q)}^{1} = p + q$
- So if x is the number of successes then the distribution for B(1, p) is

x	1	0
P(x)	$p$	$q$

For n=2, ${(p + q)}^{2} = p^{2} + 2 p q + q^{2}$
- So if x is the number of successes then the distribution for B(2, p) is

x	2	1	0
P(x)	$p^{2}$	$2 p q$	$q^{2}$

For n=3, ${(p + q)}^{3} = p^{3} + 3 p^{2} q + 3 p q^{2} + q^{3}$
- So if x is the number of successes then the distribution for B(3, p) is

x	3	2	1	0
P(x)	$p^{3}$	$3 p^{2} q$	$3 p q^{2}$	$q^{3}$

For n=4, ${(p + q)}^{4} = p^{4} + 4 p^{3} q + 6 p^{2} q^{2} + 4 p q^{3} + q^{4}$
- So if x is the number of successes then the distribution for B(4, p) is

x	4	3	2	1	0
P(x)	$p^{4}$	$4 p^{3} q$	$6 p^{2} q^{2}$	$4 p q^{3}$	$q^{4}$

For n=5, ${(p + q)}^{5} = p^{5} + 5 p^{4} q + 10 p^{3} q^{2} + 10 p^{2} q^{3} + 5 p q^{4} + q^{5}$
- So if x is the number of successes then the distribution for B(4, p) is

x	5	4	3	2	1	0
P(x)	$p^{5}$	$5 p^{4} q$	$10 p^{3} q^{2}$	$10 p^{2} q^{3}$	$5 p q^{4}$	$q^{5}$

You don't actually need to expand the brackets algebraically to find these however
- You can use Pascal's Triangle or the nCr button on your calculator (see below)

How can I calculate binomial probabilities using Pascal's triangle?

Pascal’s triangle is a way of finding the numbers in front of the powers of p and q when expanding ${(p + q)}^{n}$
- The first row has just the number 1
- Each row begins and ends with the number 1
- From the third row each number in between the 1s is the sum of the two numbers above it

An image showing the numbers in the first six rows of Pascal's Triangle

The first row is the n=0 row
- ${(p + q)}^{0} = 1$
  - You won't need this for the binomial distribution!
The second row is the n=1 row
- ${(p + q)}^{1} = p + q$
The third row is the n=2 row
- ${(p + q)}^{2} = p^{2} + 2 p q + q^{2}$
The fourth row is the n=3 row
- ${(p + q)}^{3} = p^{3} + 3 p^{2} q + 3 p q^{2} + q^{3}$
And so on for the other rows in Pascal's triangle
- You can also add extra rows to the bottom of the triangle
- Just follow the rules
  - Begin and end each row with a 1
  - Every other number in the row is the sum of the two numbers above it

How can I calculate binomial probabilities using the nCr button on my calculator?

The nCr button on your calculator can also calculate the numbers in front of the powers of p and q when expanding ${(p + q)}^{n}$
- This is useful for finding a particular probability instead of finding all the probabilities
For a binomial distribution B(n, p), the probability of getting r successes is
- $P (r successes) = (n C r) p^{r} q^{n - r}$
  - n is the number of trials
  - r is the number of successes
  - p is the probability of success
  - q = 1-p is the probability of failure
For example, for the B(4, 0.4) distribution find the probability of getting 3 successes in the 4 trials
- $n = 4$ , $r = 3$ , $p = 0.4$ , $q = 1 - 0.4 = 0.6$
- The calculator gives $4 C 3 = 4$
- So the probability is
  - $4 {(0.4)}^{3} {(0.6)}^{4 - 3} = 4 {(0.4)}^{3} {(0.6)}^{1} = 0.1536$

Worked Example

A game is played with 4 fair six-sided dice.

Each dice is rolled once, and the number of dice that land on a 1 is recorded.

Using an appropriate binomial distribution, calculate the probability that

(a) none of the dice land on a 1

There are 4 dice being rolled, so there are 4 trials (n=4)

Let a success be 'lands on a 1'
The probability of getting 1 on a fair dice is $\frac{1}{6}$ (so that is the value of p)

So the distribution to use is $B (4, \frac{1}{6})$
The probability of failure is $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$

'None of the dice lands on a 1' means there are 0 successes (r=0)

Use $P (r successes) = (n C r) p^{r} q^{n - r}$ to find the probability
(You could also use Pascal's triangle)

$4 C 0 = 1$ (from your calculator)

$(1) {(\frac{1}{6})}^{0} {(\frac{5}{6})}^{4 - 0} = {(\frac{5}{6})}^{4} = \frac{625}{1296}$

$\frac{625}{1296}$

decimal or percentage answers would also be accepted

(b) at least 2 of the dice land on a 1.

'At least 2 dice land on 1' means either 2 dice land on 1, or 3 dice land on 1, or all 4 dice land on 1

Use $P (r successes) = (n C r) p^{r} q^{n - r}$ to find the probabilities
(You could also use Pascal's triangle)

For 2 dice land on 1 (r=2)

$4 C 2 = 6$ (from your calculator)

$(6) {(\frac{1}{6})}^{2} {(\frac{5}{6})}^{2} = (6) {(\frac{1}{6})}^{2} {(\frac{5}{6})}^{2} = \frac{25}{216}$

For 3 dice land on 1 (r=3)

$4 C 3 = 4$ (from your calculator)

$(4) {(\frac{1}{6})}^{3} {(\frac{5}{6})}^{4 - 3} = (4) {(\frac{1}{6})}^{3} {(\frac{5}{6})}^{1} = \frac{5}{324}$

For 4 dice land on 1 (r=4)

$4 C 4 = 1$ (from your calculator)

$(1) {(\frac{1}{6})}^{4} {(\frac{5}{6})}^{4 - 4} = {(\frac{1}{6})}^{4} = \frac{1}{1296}$

Add those together to find the total probability

$\frac{25}{216} + \frac{5}{324} + \frac{1}{1296} = \frac{19}{144}$

Note that 'at least 2 dice land on 1' is the same as '0 or 1 dice didn't land on 1'
So you could also work this out by

finding the probabilities for 0 or 1 dice landing on one (you already know the answer for 0 dice from part (a)!)
adding those together
and subtracting that total from 1

$\frac{19}{144}$

decimal or percentage answers would also be accepted

Last updated: 18 April 2024

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The Binomial Distribution (Edexcel GCSE Statistics)

Revision Note

Characteristics of the Binomial Distribution

What is a probability distribution?

What is a binomial distribution?

Examiner Tips and Tricks

Worked Example

Calculating Binomial Probabilities

What are the probabilities for a binomial distribution?

How can I calculate binomial probabilities using Pascal's triangle?

How can I calculate binomial probabilities using the nCr button on my calculator?

Worked Example

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Author: Dan Finlay