Title: Default for 37.1 Discrete Probability Distributions
Course: HELM Workbooks 2023-2024

Discrete probability distributions

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.

Permutations and Combinations

If we define the probability that an event \(A\) occurs by using the definition: \[\text{P}(A)=\frac{\text{The number of equally likely experimental outcomes favourable to }A}{\text{The total number of equally likely outcomes forming the sample space}}=\frac{a}{n}\] then we can only find \(\text{P}(A)\) 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.

Factorials

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

\(\qquad n!=n\times (n-1)\times (n-2)\times\dots\times 3\times 2\times 1\qquad\qquad n\ge 1\)

Simple examples are:
\(\qquad 3!=3\times 2\times 1 = 6\)
\(\qquad 5!=5\times 4\times 3\times 2\times 1=120\)
\(\qquad 8!=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1=40320\)

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
\(\qquad 0!=1\)

Permutations

A permutation of a set of distinct objects places the objects in order. For example the set of three numbers \(\{ 1, 2, 3\}\) can be placed in the following orders: \[\qquad 1,2,3\quad 1,3,2\quad 2,1,3\quad 2,3,1\quad 3,2,1\quad 3,1,2\] 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 \(\{ 1,2,3\}\) has the distinct permutations \[\qquad 1,2,3\quad 1,3,2\quad 2,1,3\quad 2,3,1\quad 3,2,1\quad 3,1,2\]

Example

Write out the possible permutations of the letters \(A, B, C\) and \(D\).

Solution

The possible permutations are \[\begin{matrix}ABCD& ABDC& ADBC& ADCB &ACBD& ACDB\\ BADC& BACD& BCDA& BCAD& BDAC& BDCA\\ CABD& CADB& CDBA& CDAB& CBAD& CBDA\\ DABC& DACB& DCAB& DCBA& DBAC& DBCA \end{matrix}\] 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 \(n_1+n_2+\dots +n_r=n\) and if \(p\) is the number of permutations possible from \(n\) objects we may write
\[\qquad p\times (n_1!\times n_2!\times\dots \times n_r!)=n!\]
and so \(p\) is given by the formula
\[\displaystyle{\qquad p=\frac{n!}{n_1!\times n_2!\times\dots \times n_r!}}.\]
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

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
\(\qquad n(n-1)(n-2)\dots (n-r+1)\qquad \) which is the same as \(\displaystyle{\quad\frac{n!}{(n-r)!}}\)
This is usually denoted by \(^nP_r\) so that   \[\displaystyle{\qquad ^nP_r=\frac{n!}{(n-r)!}}.\] If we allow repetitions the number of permutations becomes \(n^r\) (can you see why?).

Example

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 \(\qquad 4\times 3=\dfrac{4\times 3\times 2\times 1}{1\times 2}=\dfrac{4!}{2!}=12\quad\) permutations

A combination of objects takes no account of order whereas a permutation does. The formula \(^nP_r=\dfrac{n!}{(n-r)!}\) 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
\(\qquad C\times r!=\dfrac{n!}{(n-r)!}\)
and so \(C\) is given by the formula
\(\qquad C=\dfrac{n!}{r!(n-r)!}\)
We normally denote the right-hand side of this expression by \(^nC_r\) so that
\(\qquad ^nC_r=\dfrac{n!}{r!(n-r)!}\)
A common alternative notation for \(^nC_r\) is \(\left(\begin{matrix}n\\r\end{matrix}\right)\).

Example

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 \(23^3=12167\).

Exercise

1. How many different signals consisting of \([value]\) symbols can be sent using the dot and dash of Morse code?
2. How many can be sent if [value] symbols or less can be sent?
Feedback: Solution 1. Clearly, the order of the symbols is important. We can choose each symbol in two ways, either a dot or a dash. The number of distinct signals is \(\qquad 2\times \cdots \times 2= 2^[value]=[value]\). 2. If [value] or less symbols may be used, the total number of signals may be calculated as follows: Using one symbol \(2^1 = 2\) ways Using two symbols:\(2\times 2= 2^2=4\) ways ... Using [value] symbols:\(2^[value]=[value]\) ways The total number of signals which may be sent is: \[\sum_{n=1}^[value] 2^n = 2^1 + 2^2 + \cdots + 2^[value] = [value].\]

Exercise

A box contains \([value]\) resistors of which \([value]\) are deemed to be ‘very high quality’ , \([value]\) ‘high quality’ and \([value]\) ‘standard’. In how many ways can a batch of \([value]\) resistors be chosen if it is to contain \([value]\) ‘very high quality’, \([value]\) ‘high quality’ and \([value]\) ‘standard’ resistors?
Feedback: Solution The order in which the resistors are chosen does not matter so that the number of ways in which the batch of \([value]\) can be chosen is: \[^{[value]}C_{[value]}\times ^{[value]}C_{[value]}\times ^{[value]}C_[value]=\dfrac{[value]!}{[value]!\times [value]!}\times \dfrac{[value]!}{[value]!\times [value]!}\times \dfrac{[value]!}{[value]!\times [value]!}= [value]\]

Random variables

A random variable \(X\) is a quantity whose value cannot be predicted with certainty. We assume that for every real number \(a\) the probability \(\text{P}(X=a)\) in a trial is well-defined. In practice, engineers are often concerned with two broad types of variables and their probability distributions: discrete random variables and their distributions, and continuous random variables and their distributions. Discrete distributions arise from experiments involving counting, for example, road deaths, car production and aircraft sales, while continuous distributions arise from experiments involving measurement, for example, voltage, corrosion and oil pressure.

Discrete random variables and probability distributions

A random variable \(X\) and its distribution are said to be discrete if the values of \(X\) can be presented as an ordered list say \(x_1,x_2,x_3,\dots\) with probability values \(p_1,p_2,p_3,\dots\). That is \(\text{P}(X=x_i)=p_i\). For example, the number of times a particular machine fails during the course of one calendar year is a discrete random variable.

More generally a discrete distribution \(f(x)\) may be defined by:
\[\qquad f(x)=\left\{\begin{matrix}p_i&\qquad\text{if}\ x=x_i\quad i=1,2,3,\dots\\ 0&\qquad\text{otherwise} \end{matrix}\right.\]

The distribution function \(F(x)\) (sometimes called the cumulative distribution function) is obtained by taking sums as defined by
\[\displaystyle{\qquad F(x)=\sum_{x_i\le x}f(x_i)=\sum_{x_i\le x}p_i.}\]

We sum the probabilities \(p_i\) for which \(x_i\) is less than or equal to \(x\). This gives a step function with jumps of size \(p_i\) at each value \(x_i\) of \(X\). The step function is defined for all values, not just the values \(x_i\) of \(X\).

Key Point

Probability Distribution of a Discrete Random Variable

Let \(X\) be a random variable associated with an experiment. Let the values of \(X\) be denoted by \(x_1,x_2,\ldots, x_n\) and let \(\text{P}(X=x_i)\) be the probability that \(x_i\) occurs. We have two necessary conditions for a valid probability distribution:
\(\bullet\)\(\text{P}(X=x_i)\ge 0\ \ \ \text{for all}\ \ x_i\)

\(\bullet\)\(\displaystyle{\sum_{i=1}^n\text{P}(X=x_i)=1}\)

Note that \(n\) may be uncountably large (infinite).

(These two statements are sufficient to guarantee that \(\text{P}(X=x_i)\le 1\) for all \(x_i\).

Example

Turbo Generators plc manufacture seven large turbines for a customer. Three of these turbines do not meet the customer’s specification. Quality control inspectors choose two turbines at random. Let the discrete random variable \(X\) be defined to be the number of turbines inspected which meet the customer’s specification.

1. Find the probabilities that \(X\) takes the values \(0, 1\) or \(2\)
2. Find and graph the cumulative distribution function.

Solution

1. The possible values of \(X\) are clearly \(0, 1\) or \(2\) and may occur as follows: \(\qquad\begin{matrix} \text{Sample Space}\qquad\qquad&\text{Value of}\ X\\ \text{Turbine faulty, Turbine faulty}\hfill&0\\ \text{Turbine faulty, Turbine good}\hfill&1\\ \text{Turbine good, Turbine faulty}\hfill&1\\ \text{Turbine good, Turbine good}\hfill&2 \end{matrix}\) \(\phantom{(a)\ }\)
We can easily calculate the probability that \(X\) takes the values \(0,1\) or \(2\) as follows:
\(\text{P}(X=0)=\dfrac{3}{7}\times\dfrac{2}{6}=\dfrac{1}{7}\)
\(\text{P}(X=1)=\dfrac{4}{7}\times\dfrac{3}{6}+\dfrac{3}{7}\times\dfrac{4}{6}=\dfrac{4}{7}\)
\(\text{P}(X=2)=\dfrac{4}{7}\times\dfrac{3}{6}=\dfrac{2}{7}\)
The values of \(\displaystyle{F(x)=\sum_{x_i\le x}\text{P}(X=x_i)}\) are clearly \(F(0)=\dfrac{1}{7}\qquad F(1)=\frac{5}{7}\qquad\text{and}\qquad F(2)=\dfrac{7}{7}=1\)
2. The graph of the step function \(F(x)\) is shown: [Image]

Mean and variance of a discrete probability distribution

If an experiment is performed \(N\) times in which the \(n\) possible outcomes \(X=x_1,x_2,x_3,\dots ,x_n\) are observed with frequencies \(f_1,f_2,f_3,\dots ,f_n\) respectively, we know that the mean of the distribution of outcomes is given by

\[\displaystyle{\bar x =\dfrac{f_1x_1+f_2x_2+\ldots +f_nx_n}{f_1+f_2+\ldots +f_n} =\dfrac{\sum_{i=1}^n f_ix_i}{\sum_{i=1}^n f_i} =\dfrac{1}{N}\sum_{i=1}^n f_ix_i =\sum_{i=1}^n \left(\dfrac{f_i}{N}\right)x_i.}\]
(Note that \(\displaystyle{\sum_{i=1}^nf_i=f_1+f_2+\dots +f_n=N}\))

The quantity \(\dfrac{f_i}{N}\) is called the relative frequency of the observation \(x_i\). Relative frequencies may be thought of as akin to probabilities; informally we would say that the chance of observing the outcome \(x_i\) is \(\dfrac{f_i}{N}\). Formally, we consider what happens as the number of experiments becomes very large. In order to give meaning to the quantity \(\dfrac{f_i}{N}\) we consider the limit (if it exists) of the quantity \(\dfrac{f_i}{N}\) as \(N\rightarrow\infty\) . Essentially, we define the probability \(p_i\) as
\(\qquad p_i= \lim_{N\rightarrow\infty}\dfrac{f_i}{N}\)

Replacing \(\dfrac{f_i}{N}\) with the probability \(p_i\) leads to the following definition of the mean or expectation of the discrete random variable \(X\).

Key Point

Let \(X\) be a random variable with values \(x_1,x_2,\ldots, x_n\). Let the probability that \(X\) takes the value \(x_i\) (i.e. \(\text{P}(X=x_i)\)) be denoted by \(p_i\). The mean or expected value or expectation of \(X\), which is written \(\text{E}(X)\) is defined as:

\[\displaystyle{\text{E}(X)=\sum_{i=1}^n x_i\ \text{P}(X=x_i)= p_1x_1+p_2x_2+\dots +p_nx_n}\]

The symbol \(\mu\) is sometimes used to denote \(\text{E}(X)\).

The expectation \(\text{E}(X)\) of \(X\) is the value of \(X\) which we expect on average. In a similar way we can write down the expected value of the function \(g(X)\) as \(\text{E}[g(X)]\), the value of \(g(X)\) we expect on average. We have
\[\displaystyle{\qquad \text{E}[g(X)]=\sum_i^ng(x_i)f(x_i).}\]

In particular if \(g(X)=X^2\), we obtain
\[\displaystyle{\qquad \text{E}[X^2]=\sum_i^nx_i^2f(x_i).}\]

The variance is usually written as \(\sigma^2\). For a frequency distribution it is:
\[\displaystyle{\sigma^2=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\mu)^2\qquad \text{where } \mu \text{ is the mean value}.}\]
and can be expanded and ‘simplified’ to appear as:
\[\displaystyle{\sigma^2=\frac{1}{N}\sum_{i=1}^n f_ix_i^2-\mu^2.}\]
This is often quoted in words:

\(\qquad\)The variance is equal to the mean of the squares minus the square of the mean.

We now extend the concept of variance to a random variable.

Key Point

Let \(X\) be a random variable with values \(x_1,x_2,\ldots , x_n\). The variance of \(X\), which is written \(\text{V}(X)\) is defined by \[\text{V}(X)=\sum_{i=1}^n p_i(x_i-\mu)^2\] where \(\mu\equiv \text{E}(X)\). We note that \(\text{V}(X)\) can be written in the alternative form \[\text{V}(X)=\text{E}(X^2)-[\text{E}(X)]^2\] The standard deviation \(\sigma\) of a random variable is \(\sqrt{\text{V}(X)}\).

Example

A traffic engineer is interested in the number of vehicles reaching a particular crossroads during periods of relatively low traffic flow. The engineer finds that the number of vehicles \(X\) reaching the crossroads per minute is governed by the probability distribution: \(\qquad \begin{matrix} x\quad &0&1&2&3&4\\ \text{P}(X=x)&0.37&0.39&0.19&0.04&0.01 \end{matrix}\)

1. Calculate the expected value, the variance and the standard deviation of the random variable \(X\).
2. Graph the probability distribution \(\text{P}(X=x)\) and the corresponding cumulative probability distribution \(\displaystyle{F(x)=\sum\limits_{x_i\le x}\text{P}(X=x_i)}\).

Solution

1. The expectation, variance and standard deviation and cumulative probability values are calculated as follows:
\(\qquad \begin{matrix} \quad x\quad&\quad x^2\quad&\quad \text{P}(X=x)\quad&\quad F(x)\quad\\ 0&0&0.37&0.37\\ 1&1&0.39&0.76\\ 2&4&0.19&0.95\\ 3&9&0.04&0.99\\ 4&16&0.01&1.00 \end{matrix}\) \[\begin{aligned} \qquad\text{E}(X)&=\sum_{x=0}^4 x\text{P}(X=x)\\ &=0\times 0.37+1\times0.39+2\times0.19+3\times0.04+4\times 0.01\qquad\qquad\qquad\qquad\\ &=0.93\\ \text{V}(X)&=\text{E}(X^2)-[\text{E}(X)]^2\\ &=\sum_{x=0}^4 x^2\text{P}(X=x)-\left[\sum_{x=0}^4 x\text{P}(X=x)\right]^2\\ &=0\times 0.37+1\times0.39+4\times0.19+9\times0.04+16\times 0.01-(0.93)^2\qquad\qquad\qquad\\ &=0.8051\end{aligned}\] The standard deviation is given by \(\ \sigma = \sqrt{\text{V}(X)}=0.8973\)
2. [Image: Cumulative and probability distribution function for the number of vehicles per minute reaching a crossroad]

Exercise

Find the expectation, variance and standard deviation of the number of heads when tossing [value] fair coins.

Give an exact answer, eg: \(\frac{\sqrt{2}}{2}\)
Syntax hint: \(\sqrt{2}\) can be entered using \(\texttt{sqrt(2)}\).

1. \(E(X) =\)
2. \(V(X) =\)
3. \(\sigma(X) =\)
Feedback: The number of heads on \([value]\) tosses of a fair coin follows a Binomial distribution with probability function \[f(x) = { [value] \choose x } \left(\dfrac{1}{2}\right)^x \left(\dfrac{1}{2}\right)^{[value]-x} = { [value] \choose x } \dfrac{1}{2^[value]}\quad \text{for} \quad x = 0, \ldots, [value].\] 1. The expectation is \(\text{E}(X) = \sum_{x=0}^[value] x f(x) = 0\times f(0) + 1\times f(1) + \cdots + [value]\times f([value]) = [value]\). 2. To calculate the variance, first we compute: \[\begin{align*}E(X^2) &= \sum_{x=0}^[value] x^2 f(x) \\&=0^2 \times f(0) + 1^2 \times f(1) + \cdots + [value]^2 \times f([value])\\& = [value].\end{align*}\] It follows that \(\text{V}(X) = E(X^2)-E(X)^2 = [value] - \left([value]\right)^2=[value]\). 3. Finally, the standard deviation is \(\sigma(X)=\sqrt{V(X)} = \sqrt{[value]} = [value]\).

Exercise

A machine is operated by \([value]\) workers. There are \([value]\) workers available. How many possible teams of \([value]\) workers are there?
Feedback: Solution We pick \([value]\) workers from a total of \([value]\) workers. The order does not matter. The number of possible teams is thus a combination \[ \binom{[value]}{[value]} = \dfrac{[value]!}{[value]!([value]-[value])!} = [value] \]

Exercise

A factory has \([value]\) machines. \(2\) of these have been given an experimental modification. In the first week after this modification, problems are reported with \([value]\) of the machines.

How many different ways can we select \([value]\) machines from the \([value]\) machines in the factory?

Of the machines that did not have the experimental modification, how many ways could we have the observed number of problematic machines if both of the modified machines had a problem?

Using the above values or otherwise, what is the probability that both of the modified machines are among the \([value]\) with problems assuming that all machines are equally likely to give problems?

Write your answer as a rational number (fraction).
Feedback: Solution There are \(\binom{[value]}{[value]} = [value]\) possible different selections of \([value]\) machines in the factory. There are \([value]-2 = [value]\) machines in the factory that did not have the experimental modification. Also, if both of the modified machines had a problem then \([value]-2 = [value]\) of the remaining problematic machines did not have the modification. This means that there are \(\binom{[value]}{[value]} = [value]\) different ways of having this combination. Side note: if both of the machines with the experimental modification are problematic, then there is only \(\binom{2}{2} = 1\) way of having this combination. This means that the probability of that both of the modified machines are among the \([value]\) with problems is:\[ \dfrac{\binom{2}{2}\binom{[value]}{[value]}}{\binom{[value]}{[value]}} = \dfrac{[value]\times[value]}{[value]\times[value]} = [value] \] Remark: This is an example of sampling without replacement which we can model using the hypermetric probability distribution. Alternatively, let \( S \) be the event ``first modified machine is in the group of \([value]\)'' and \( C \) be the event ``second modified machine is in the group of \([value]\)''. Then the required probability is \[\text{P}(S)\times\text{P}(C \mid S) = \dfrac{[value]}{[value]} \times \dfrac{[value]-1}{[value]-1} = [value]\]

Exercise

A factory has \([value]\) machines. \([value]\) of these have been given an experimental modification. In the first week after this modification, problems are reported with \([value]\) of the machines. What is the probability that excatly \(2\) of the modified machines are among the \([value]\) with problems assuming that all machines are equally likely to give problems?

Write your answer as a rational number (fraction).
Feedback: Solution There are \(\binom{[value]}{[value]}\) possible different selections of \([value]\) machines and all are equally likely, \(\binom{[value]}{2}\) different choices of two modified machines, and \(\binom{[value]}{[value]}\) different choices of \([value]\) non-modified machines. Hence the required probability is \[ \dfrac{\binom{[value]}{2}\times\binom{[value]}{[value]}}{\binom{[value]}{[value]}} = [value] \]

Exercise

A random number generator produces sequences of independent digits, each of which is as likely to be any digit from 0 to 9 as any other. If \(X\) denotes any single digit, find \(\text{E}(X)\).

Write your answer as a rational number (fraction).
Feedback: Solution All digits are equally likely, so \(\text{E}(X)=\dfrac{1}{10}\{0+1+2+3+\ldots+9\}=\displaystyle [value]\)

Exercise

A hand-held calculator has a clock cycle time of 100 nanoseconds; these are positions numbered \(0, 1, \ldots, 99\). Assume a flag is set during a particular cycle at a random position. Thus, if \(X\) is the position number at which the flag is set. \[\displaystyle{\qquad \text{P}(X=k)=\frac{1}{100}\qquad k=0,1,2,\ldots , 99.}\]
Evaluate the expected position number \(\text{E}(X)\), the variance \(\text{V}(X)\) and the standard deviation \(\sigma(X)\).

Give your answers accurate to two decimal places.

1. \(\text{E}(X)=\)
2. \(\text{V}(X)=\)
3. \(\sigma(X)=\)

(Hint: The sum of the first \(k\) integers is \(k(k+1)/2\) and the sum of their squares is: \(k(k+1)(2k+1)/6.\))
Feedback: Solution 1. \(\text{E}(X)=\dfrac{1}{100}\ \{0+1+2+\ldots +99\}=\dfrac{1}{100}\left[\dfrac{99(99+1)}{2}\right]=\dfrac{99}{2} = 49.5\) 2. First, \[\begin{aligned} \text{E}(X^2) & = \dfrac{1}{100}\ \{0^2+1^2+2^2+\ldots +99^2\}\\ & =\dfrac{1}{100}\left[\dfrac{99(99+1)(2 \times 99+1)}{6}\right]\\ & = \dfrac{6567}{2} = 3283.5.\end{aligned}\] Therefore, \[\begin{aligned} \text{V}(X) &= \text{E}(X^2) - [\text{E}(X)]^2 \\&= \dfrac{6567}{2} - \left(\dfrac{99}{2}\right)^2\\ &= \dfrac{3333}{4} = 833.25.\end{aligned}\] 3. The standard deviation is \(\sigma(X) =\sqrt{\text{V}(X)} = \sqrt{833.25} \approx 28.87\) to 2 d.p.

Exercise

Concentric circles of radii \([value]\) cm and \([value]\) cm are drawn on a circular target of radius \([value]\) cm. A darts player receives \(10\), \(5\) or \(3\) points for hitting the target inside the smaller circle, middle annular region and outer annular region respectively. The player has only a \(50/50\) chance of hitting the target at all but if they do hit it they are just as likely to hit any one point on it as any other. If \(X=\) 'number of points scored on a single throw of a dart', calculate the expected value of \(X\).

Write your answer as a rational number (fraction).

\(\text{E}(X)=\)
Feedback: Solution \(X\) can take 4 values 0, 3, 5 or 10 \[\text{P}(X=0)=0.5\qquad \text{[only 50/50 chance of hitting target]}\] The probability that a particular points score is obtained is related to the areas of the annular regions which are, from the centre: \([value]^2\times \pi = [value],\ ([value]^2-[value]^2)\times \pi=[value],\ ([value]^2-[value]^2)\times \pi=[value]\) \[\begin{aligned} \text{P}(X=3\phantom{0})&=\text{P}[(3\ \text{is scored})\cap (\text{target is hit})]\\ \\ &=\text{P}(3\ \text{is scored}\,|\,\text{target is hit})\times\text{P}(\text{target is hit})\\ \\ &=\dfrac{[value]}{[value]}.\dfrac{1}{2}=[value]\\ \\ \text{P}(X=5\phantom{0})&=\text{P}(5\ \text{is scored}\,|\,\text{target is hit})\times\text{P}(\text{target is hit})\\ \\ &=\dfrac{[value]}{[value]i}.\dfrac{1}{2}=[value]\\ \\ \text{P}(X=10)&=\text{P}(10\ \text{is scored}\,|\,\text{target is hit})\times\text{P}(\text{target is hit})\\ \\ &=\dfrac{[value]}{[value]}.\dfrac{1}{2}=[value]\end{aligned}\] \[\therefore \qquad \text{E}(X)=[value]\times[value]+[value]\times[value]+[value]\times[value] = [value]\]

Introduction

It is often possible to model real systems by using the same or similar random experiments and their associated random variables. Numerical random variables may be classified in two broad but distinct categories called discrete random variables and continuous random variables. Often, discrete random variables are associated with counting while continuous random variables are associated with measuring. Generally speaking, discrete random variables can take values which are separate and can be listed. Strictly speaking, the real situation is a little more complex but it is sufficient for our purposes to equate the word discrete with a finite list. In contrast, continuous random variables can take values anywhere within a specified range. This Section will familiarize you with the idea of a discrete random variable and the associated probability distributions.