S1 5.1, S1 5.2 - Discrete probability distributions | Edexcel A-Level International Mathematics

Use the discrete random variables of S1 5.1 and the probability functions and cumulative distributions of S1 5.2 to turn experiments into probability models. Calculate event probabilities while respecting strict and inclusive boundaries.

Values of a discrete random variable

Two coins have four equally likely ordered outcomes, but their number of heads has only three possible values. Should those three values each receive probability one third? The outcomes must first be grouped by the value they produce.

Discrete random variable

A random variable assigns one numerical value to every outcome of an experiment. Use a capital letter such as X for the variable and a lower-case x for a possible value. A discrete random variable has a finite or countable set of possible values, called its support.

Discrete values need not be consecutive integers or even integers. A prize chosen from $£0$ , $£2.50$ and $£10$ is discrete. A measurement which can take any real value in an interval is continuous under that model. Merely rounding a continuous measurement for recording does not make its underlying physical model discrete.

Worked example 1

Two fair coins are tossed independently. Let X be the number of heads. Find the possible values of X and their probabilities.

Plan. List the equally likely ordered coin outcomes, then combine those giving the same head count.

Outcome	HH	HT	TH	TT
X	2	1	1	0

Each outcome has probability $1/4$ , so

P(X=0)=\frac14,\quad P(X=1)=\frac12,\quad P(X=2)=\frac14.

Answer. Support $\{0,1,2\}$ , with probabilities $1/4,1/2,1/4$ .

Check. The probabilities sum to $1$ , and the middle value has two supporting outcomes while each end value has only one. Equal outcome probabilities have not become equal value probabilities.

Change only the numerical quantity recorded: the same coin results now produce a net payment $Y=3X-1$ pounds.

Worked example 2

Find the possible values and probabilities of Y.

Plan. Transform each possible X-value and keep the probability attached to it.

X=0,1,2\quad\Longrightarrow\quad Y=-1,2,5.

$y$ , pounds	$-1$	$2$	$5$
$P(Y=y)$	$1/4$	$1/2$	$1/4$

Answer. Support $\{-1,2,5\}$ , with probabilities $1/4,1/2,1/4$ . A loss is a negative value of the variable, not a negative probability.

Check. HH gives $£5$ , HT and TH each give $£2$ , and TT gives a $£1$ loss. The four original outcomes still each have probability $1/4$ .

If different original values transform to the same new value, combine their probabilities. For instance, $(X-1)^2$ is $1$ for both $X=0$ and $X=2$ , so its probability at $1$ is $1/4+1/4=1/2$ .

Probability functions and normalisation

A probability function states how much probability belongs to each possible value, rather than how frequently the value happened in one particular observed sample.

Probability function

The probability function or probability mass function is $p(x)=P(X=x)$ . A valid distribution satisfies

p(x)\ge0\quad\text{for every possible value},\qquad \sum_{x\text{ in the support}}p(x)=1.

Values outside the support have probability zero. Finding a constant so that the total is one is called normalising the probability function.

To find an event probability, select the possible values satisfying its condition and add their probabilities. Do not assume a continuous interval of values is possible between separate support points.

Worked example 3

A random variable has $p(x)=kx$ for $x=1,2,3$ , and zero otherwise. Find k and $P(X\ge2)$ .

Plan. Make the total probability one, then add the masses at the values $2$ and $3$ .

k+2k+3k=1\quad\Longrightarrow\quad k=\frac16,

P(X\ge2)=\frac26+\frac36=\frac56.

Answer. $k=1/6$ , $P(X\ge2)=5/6$ .

Check. All probabilities $1/6,2/6,3/6$ are nonnegative and sum to $1$ . The complement has probability $P(X=1)=1/6$ , so the event probability is $1-1/6=5/6$ .

Change only the weighting rule from x to its square; the support stays the same.

Worked example 4

Now $p(x)=kx^2$ for $x=1,2,3$ , and zero otherwise. Find k and $P(X\ge2)$ .

Plan. Normalise the new weights before calculating the event probability.

k+4k+9k=1\quad\Longrightarrow\quad k=\frac1{14},

P(X\ge2)=\frac4{14}+\frac9{14}=\frac{13}{14}.

Answer. $k=1/14$ , $P(X\ge2)=13/14$ .

Check. The outside value $1$ now has probability $1/14$ , whose complement is $13/14$ . Using the previous $k=1/6$ would give total probability $14/6>1$ , so it cannot be retained.

Normalising is necessary, but the individual probabilities must also be nonnegative. Values $-0.1,0.4,0.7$ sum to one yet cannot be a probability distribution.

An inequality involving X may need simplifying first. For Example 3,

P(2X+1<6)=P(X<2.5)=P(X=1)+P(X=2)=\frac12.

There is no mass at $2.5$ ; only the supported values $1,2$ meet the condition. A table or a probability function supplies the same information. The probability-function conditions are expected knowledge, rather than a printed normalisation formula in the booklet.

Cumulative distribution functions

A cumulative distribution adds all probability up to and including a specified boundary. For a discrete variable it changes only when that boundary reaches a supported value.

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S1 5.1, S1 5.2 - Discrete probability distributions

Values of a discrete random variable

Probability functions and normalisation

Cumulative distribution functions

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