1 Discrete probability distributions

We shall look at discrete distributions in this Workbook and continuous distributions in HELM booklet  38. In order to get a good understanding of discrete distributions it is advisable to familiarise yourself with two related topics: permutations and combinations. Essentially we shall be using this area of mathematics as a calculating device which will enable us to deal sensibly with situations where choice leads to the use of very large numbers of possibilities. We shall use combinations to express and manipulate these numbers in a compact and efficient way.

1.1 Permutations and Combinations

You may recall from HELM booklet  35.2 concerned with probability that if we define the probability that an event $A$ occurs by using the definition:

$\text{P}\left(A\right)=\frac{\text{The number of equally likely experimental outcomes favourable to }A}{\text{The total number of equally likely outcomes forming the sample space}}=\frac{a}{n}$

then we can only find P $\left(A\right)$ provided that we can find both $a$ and $n$ . In practice, these numbers can be very large and difficult if not impossible to find by a simple counting process. Permutations and combinations help us to calculate probabilities in cases where counting is simply not a realistic possibility.

Before discussing permutations, we will look briefly at the idea and notation of a factorial.

1.2 Factorials

The factorial of an integer $n$ commonly called ‘factorial $n$ ’ and written $n!$ is defined as follows:

$n!=n\times \left(n-1\right)\times \left(n-2\right)\times \⋯\times 3\times 2\times 1n\ge 1$

Simple examples are:

As you can see, factorial notation enables us to express large numbers in a very compact format. You will see that this characteristic is very useful when we discuss the topic of permutations. A further point is that the definition above falls down when $n=0$ and we define

$0!=1$

1.3 Permutations

A permutation of a set of distinct objects places the objects in order . For example the set of three numbers $\left\{1,2,3\right\}$ can be placed in the following orders:

1,2,3 1,3,2 2,1,3 2,3,1 3,2,1 3,1,2

Note that we can choose the first item in 3 ways, the second in 2 ways and the third in 1 way. This gives us $3\times 2\times 1=3!=6$ distinct orders. We say that the set $\left\{1,2,3\right\}$ has the distinct permutations

1,2,3 1,3,2 2,1,3 2,3,1 3,2,1 3,1,2

Example 1

Write out the possible permutations of the letters $A,B,C$ and $D$ .

Solution

The possible permutations are

There are $4!=24$ permutations of the four letters $A,B,C$ and $D.$

In general we can order $n$ distinct objects in $n!$ ways.

Suppose we have $r$ different types of object. It follows that if we have $n_1$ objects of one kind, $n_2$ of another kind and so on then the $n_1$ objects can be ordered in $n_1!$ ways, the $n_2$ objects in $n_2!$ ways and so on. If $n_1+n_2+\⋯+n_r=n$ and if $p$ is the number of permutations possible from $n$ objects we may write

$p\times \left(n_1!\times n_2!\times \⋯\times n_r!\right)=n!$

and so $p$ is given by the formula

$p=\frac{n!}{n_1!\times n_2!\times \⋯\times n_r!}$

Very often we will find it useful to be able to calculate the number of permutations of $n$ objects taken $r$ at a time. Assuming that we do not allow repetitions, we may choose the first object in $n$ ways, the second in $n-1$ ways, the third in $n-2$ ways and so on so that the $r^{\text{th}}$ object may be chosen in $n-r+1$ ways.

Example 2

Find the number of permutations of the four letters $A,B,C$ and $D$ taken three at a time.

Solution

We may choose the first letter in 4 ways, either $A,B,C$ or $D$ . Suppose, for the purposes of illustration we choose $A$ . We may choose the second letter in 3 ways, either $B,C$ or $D$ . Suppose, for the purposes of illustration we choose $B$ . We may choose the third letter in 2 ways, either $C$ or $D$ . Suppose, for the purposes of illustration we choose $C$ . The total number of choices made is $4\times 3\times 2=24$ .

In general the numbers of permutations of $n$ objects taken $r$ at a time is

$n\left(n-1\right)\left(n-2\right)\dots \left(n-r+1\right)\text{which is the same as}\frac{n!}{\left(n-r\right)!}$

This is usually denoted by ${}_{}^n{P}_r$ so that

${}_{}^n{P}_r=\frac{n!}{\left(n-r\right)!}$

If we allow repetitions the number of permutations becomes $n^r$ (can you see why?).

Example 3

Find the number of permutations of the four letters $A,B,C$ and $D$ taken two at a time.

Solution

We may choose the first letter in 4 ways and the second letter in 3 ways giving us

$4\times 3=\frac{4\times 3\times 2\times 1}{1\times 2}=\frac{4!}{2!}=12$ permutations

1.4 Combinations

A combination of objects takes no account of order whereas a permutation does. The formula ${}_{}^n{P}_r=\frac{n!}{\left(n-r\right)!}$ gives us the number of ordered sets of $r$ objects chosen from $n$ . Suppose the number of sets of $r$ objects (taken from $n$ objects) in which order is not taken into account is $C$ . It follows that

We normally denote the right-hand side of this expression by ${}_{}^n{C}_r$ so that

Example 4

How many car registrations are there beginning with $NP05$ followed by three letters? Note that, conventionally, $I,O$ and $Q$ may not be chosen.

Solution

We have to choose 3 letters from 23 allowing repetition. Hence the number of registrations beginning with $NP05$ must be $23^3=12167$ .

Task!

1. How many different signals consisting of five symbols can be sent using the dot and dash of Morse code?
2. How many can be sent if five symbols or less can be sent?

Answer

Task!

A box contains 50 resistors of which 20 are deemed to be ‘very high quality’ , 20 ‘high quality’ and 10 ‘standard’. In how many ways can a batch of 5 resistors be chosen if it is to contain 2 ‘very high quality’, 2 ‘high quality’ and 1 ‘standard’ resistor?

Answer