Home page for accesible maths 4 Discrete random variables 4.2 Probability mass functions 4.4 Expectation

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4.3 Probability of an event

If we are interested in evaluating the probability of some event occurring for a random variable, this can easily be obtained from the pmf.

Lemma 4.10.

Let ${E\subseteq{\mathcal{S}}}$ be an event in the induced sample space. The probability of $E$ is given by

{\rm P}(R\in E)=\sum_{r\in E}p_{R}(r).

Proof.

Write ${E=\{r_{1},\ldots,r_{k}\}}\subseteq{\mathcal{S}}$ . Then

$\displaystyle{\rm P}(R\in E)$	$\displaystyle=$	$\displaystyle{\rm P}(\{R=r_{1}\}\cup\{R=r_{2}\}\cup\dots\cup\{R=r_{k}\})$
	$\displaystyle=$	$\displaystyle{\rm P}(R=r_{1})+{\rm P}(R=r_{2})+\dots+{\rm P}(R=r_{k})$
	$\displaystyle=$	$\displaystyle p_{R}(r_{1})+p_{R}(r_{2})+\dots+p_{R}(r_{k})$
	$\displaystyle=$	$\displaystyle\sum_{r\in E}p_{R}(r).$

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Example 4.11.

The length of stay in hospital after surgery is modelled as a random variable ${R}$ . The following table gives the pmf for ${R}$ .

Days stayed	r	4	5	6	7	8	9	10+	total
Probability	${p_{R}(r)}$	0.038	0.114	0.430	0.300	0.080	0.030	0.008	1

Find the probability of being in hospital for

a.

at most ${6}$ days,
b.

between ${5}$ and ${7}$ days,
c.

at least ${7}$ days.

Solution.

a.

at most ${6}$ days: ${{\rm P}(R\leq 6)={\rm P}(R=4)+{\rm P}(R=5)+{\rm P}(R=6)=0.582}$
b.

between ${5}$ and ${7}$ : ${{\rm P}(5\leq R\leq 7)={\rm P}(R=5)+{\rm P}(R=6)+{\rm P}(R=7)=0.844}$
c.

at least ${7}$ : ${{\rm P}(R\geq 7)=1-{\rm P}(R\leq 6)=1-0.582=0.418}$ .

Exercise 4.12.

Find the probability of an odd number of heads in ${3}$ tosses of a fair coin.

Solution.

${{\rm P}(\mbox{odd number Hs})={\rm P}(R\in\{1,3\})=\sum_{r=1,3}p_{R}(r)=3/8+1% /8=1/2}$ .

A specific family of events in which we are often interested, particularly for continuous random variables, is ${\{r:r\leq m\}}$ for different values of $m$ . These events are useful because for any event $E\subseteq{\mathcal{S}}$ we can calculate ${\rm P}(E)$ from the probabilities of events of the type ${\{r:r\leq m\}}$ . For example, let $E=\{4,5,6\}$ . Then

\{r\,:\,r\leq 6\}=\{r\,:\,r\leq 3\}\cup E,

a disjoint union. Therefore, by Axiom 3,

{\rm P}(\{r\,:\,r\leq 6\})={\rm P}(\{r\,:\,r\leq 3\})+{\rm P}(E).

The cumulative distribution function or cdf of a random variable ${R}$ is a function ${F_{R}:\mathbb{R}{\rightarrow}\mathbb{R}}$ given by ${F_{R}(m)={\rm P}(R\leq m).}$ For a discrete random variable ${R}$ the cumulative distribution function is given by ${F_{R}(m)={\rm P}(R\leq m)=\sum_{r=0}^{m}p_{R}(r).}$

Exercise 4.13.

What is the cumulative distribution function for a random variable ${R}$ whose pmf is specified by ${p_{R}(r)=r/10}$ for ${r=1,2,3,4}$ ?

Solution.

$\displaystyle F_{R}(0)$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle F_{R}(1)$	$\displaystyle=$	$\displaystyle p_{R}(1)=1/10$
$\displaystyle F_{R}(2)$	$\displaystyle=$	$\displaystyle p_{R}(1)+p_{R}(2)=3/10$
$\displaystyle F_{R}(3)$	$\displaystyle=$	$\displaystyle p_{R}(1)+p_{R}(2)+p_{R}(3)=6/10$
$\displaystyle F_{R}(4)$	$\displaystyle=$	$\displaystyle p_{R}(1)+p_{R}(2)+p_{R}(3)+p_{R}(4)=1$
$\displaystyle F_{R}(m)$	$\displaystyle=$	$\displaystyle 1\mbox{ for }m>4.$