The Binomial Distribution

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The Binomial Distribution

The Binomial distribution is a distribution that tells us the probability of a certain number of successes given a fixed success probability in repeated trials. To understand what it is and how it works, we first need to understand factorials and binomial coefficients.

Factorial

The factorial of a number, written n!n! is equal to n×(n1)×(n2)×...×3×2×1ntimes (n-1)times (n-2)times … times 3times 2times 1

n!n! is the number of ways you can arrange nbm{n} objects.

You can picture this as having nn choices for the first object, n1n-1 choices for the second object, n2n-2 choices for the third object, and so on.

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What if Some Objects are the Same?

Suppose we are arranging nn objects, and rr of them are the same. In any given arrangement, we can swap two objects that are the same and end up with the same arrangement. Indeed, if rr objects are the same, we can swap them into any order we like within the arrangement and we will not have changed the arrangement. There are r!r! such ways to swap rr objects. So the number of arrangements of nn objects, rr of which are the same, is n!r!dfrac{n!}{r!}

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Only Two Different Objects

Suppose we want to arrange nn objects, rr of which are of one type and the remaining nrn-r are of a second type. We can arrange the rr identical objects in r!r! ways and the nrn-r identical objects in (nr)!(n-r)! ways. So overall we have n!r!(nr)!dfrac{n!}{r!(n-r)!} arrangements. This is known as the binomial coefficient of nn and rr, written as

(nr)=nCr=n!r!(nr)!begin{pmatrix}nrend{pmatrix}=text{}^{n}C_{r}=dfrac{n!}{r!(n-r)!}

 

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The Binomial Distribution

The Binomial distribution tells us the probability of a xx successes from nn independent events where the probability of success for each event is pp.

Its probability function is:

f(x)=(nx)px(1p)nxf(x)=begin{pmatrix}nxend{pmatrix}p^{x}(1-p)^{n-x}

We can understand this as (nx)begin{pmatrix}nxend{pmatrix} is the number of possible ways to have xx successes out of nn events, then we multiply by the probability of xx successes and nxn-x failures, which is px(1p)nxp^{x}(1-p)^{n-x}

Notation: If XX is a binomial random variable with nn trials and a success probability of pp, we write XB(n,p)Xsim B(n,p)

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Binomial Tables

Your formula booklet may contain binomial tables. These give the cumulative distribution function value for the binomial distribution.

For example, if we want to find the probability of two or less successes out of five trials with a success probability of 0.150.15:

This shows that P(X2)=0.9734mathbb{P}(Xleq 2)=0.9734

If your formula booklet does not have binomial tables, you are expected to use the statistics functions of your calculator to answer questions like this. Make sure you are familiar with how these work, as they will be necessary for an exam.

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When to use the Binomial Distribution

A random variable XX is binomially distributed if:

  1. There is a fixed number of trials.
  2. Each trial has only two outcomes – success or failure.
  3. The trials are independent of each other.
  4. The success probability is the same in each trial.
  5. XX is the total number of successes in all trials.
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The Binomial Distribution Example Questions

Question 1: Find (63)begin{pmatrix}63end{pmatrix}

[2 marks]

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(63)=begin{pmatrix}63end{pmatrix}=

 

6!3!×3!=dfrac{6!}{3!times 3!}=

 

6×5×4×3×2×13×2×1×3×2×1=dfrac{6times 5times 4times 3times 2times 1}{3times 2times 1times 3times 2times 1}=

 

6×5×4×3×2×13×2×1×3×2×1=dfrac{6times 5times 4times cancel{3}times cancel{2}times cancel{1}}{cancel{3}times cancel{2}times cancel{1}times 3times 2times 1}=

 

6×5×43×2×1=dfrac{6times 5times 4}{3times 2times 1}=

 

6×5×46=dfrac{6times 5times 4}{6}=

 

6×5×46=dfrac{cancel{6}times 5times 4}{cancel{6}}=

 

5×4=5times 4=

 

2020

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Question 2: What is P(X=2)mathbb{P}(X=2) if XB(5,0.25)Xsim B(5,0.25)?

[2 marks]

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Use the formula:

 

P(X=x)=(nx)px(1p)nxmathbb{P}(X=x)=begin{pmatrix}nxend{pmatrix}p^{x}(1-p)^{n-x}

 

Here, n=5,p=0.25,x=2n=5,p=0.25,x=2

 

P(X=2)=(52)0.252(10.25)52=0.264begin{aligned}mathbb{P}(X=2)&=begin{pmatrix}52end{pmatrix}0.25^{2}(1-0.25)^{5-2}[1.2em]&=0.264end{aligned}

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Question 3: For XB(12,0.4)Xsim B(12,0.4), use a binomial table or a calculator to find P(X5)mathbb{P}(Xleq 5)

[1 mark]

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P(X5)=0.6652mathbb{P}(Xleq 5)=0.6652

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Question 4: Use a binomial table or a calculator to find P(3X6)mathbb{P}(3leq Xleq 6) where XB(8,0.6)Xsim B(8,0.6)

[1 mark]

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P(3X6)=mathbb{P}(3leq Xleq 6)=

P(X6)P(X2)mathbb{P}(Xleq 6)-mathbb{P}(Xleq 2)

We can use the tables to find these values.

0.89360.0498=0.84380.8936 – 0.0498 = 0.8438

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Question 5: Every time Jeff turns on his television, it goes to a random channel. It can receive 4040 channels. Jeff turns on his television 6060 times in a month on average. What is the probability that it will show the channel Jeff wants to watch upon turning on at least three times in a month.

[3 marks]

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Probability of correct channel is 140dfrac{1}{40}, so we have XB(60,140)Xsim B(60,dfrac{1}{40}).

Want to find P(X3)mathbb{P}(Xgeq 3)

P(X3)=1P(X2)=10.8105=0.1895begin{aligned}mathbb{P}(Xgeq 3)&=1-mathbb{P}(Xleq 2)[1.2em]&=1-0.8105[1.2em]&=0.1895end{aligned}

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Specification Points Covered

N1 – Understand and use simple, discrete probability distributions (calculation of mean and variance of discrete random variables is excluded), including the binomial distribution, as a model; calculate probabilities using the binomial distribution