5.4 Discrete random variables

Learning objectives

5.4.1Discrete random variables• draw up a probability distribution table relating to a given situation involving a discrete random variable X, and calculate E(X) and Var(X)5.4.2Geometric distribution• use formulae for probabilities for the binomial and geometric distributions, and recognise practical situations where these distributions are suitable models Including the notations B(n, p) and Geo(p). Geo(p) denotes the distribution in which pr = p(1 - p)r - 1 for r = 1, 2, 3, ….5.4.3Expectation and variance• use formulae for the expectation and variance of the binomial distribution and for the expectation of the geometric distribution. Proofs of formulae are not required.

For a discrete random variable $X$ , list each possible value $x$ with $P(X=x)$ . Every probability must be between 0 and 1 and the probabilities must sum to 1; use this total first to find any unknown probability.

E(X)=\sum xP(X=x),\qquad E(X^2)=\sum x^2P(X=x),\operatorname{Var}(X)=E(X^2)-[E(X)]^2.

$x$	0	1	2
$P(X=x)$	0.2	0.5	0.3
$xP(X=x)$	0	0.5	0.6
$x^2P(X=x)$	0	0.5	1.2

Thus $E(X)=1.1$ , $E(X^2)=1.7$ and $\operatorname{Var}(X)=1.7-1.1^2=0.49$ . A probability table may come from enumeration or from combining outcomes that give the same value of $X$ .

Do not average the listed $x$ -values unless they are equally likely, and do not confuse $[E(X)]^2$ with $E(X^2)$ .

Model	Random variable	Conditions	Support
$X\sim B(n,p)$	number of successes in fixed $n$ trials	independent trials, two outcomes, constant $p$	$0,1,\ldots,n$
$X\sim Geo(p)$	trial number of the first success	independent repeated trials, constant $p$	$1,2,3,\ldots$

P(X=r)=\binom nr p^r(1-p)^{n-r}\quad\text{for }X\sim B(n,p),P(X=r)=p(1-p)^{r-1}\quad\text{for }X\sim Geo(p).

If 8 independent items are defective with probability 0.1, the probability of exactly 2 defective items is $\binom82(0.1)^2(0.9)^6$ . For ranges such as at least 2, sum the relevant values or use a shorter complement.

If each attempt succeeds with probability 0.2, then the first success on attempt 4 has probability $(0.8)^3(0.2)$ . ‘After attempt 4’ means four failures, so its probability is $(0.8)^4$ .

Geometric $r$ includes the successful trial, whereas binomial $r$ counts successes. Reject either model if independence or constant $p$ is not reasonable.

Distribution	Expectation	Variance
$X\sim B(n,p)$	$E(X)=np$	$\operatorname{Var}(X)=np(1-p)$
$Y\sim Geo(p)$	$E(Y)=1/p$	not required in this syllabus objective

These are long-run summaries, not necessarily possible single outcomes. A binomial expectation can be non-integer; a geometric expectation is the average trial number on which the first success occurs.

For $X\sim B(20,0.3)$ , $E(X)=20(0.3)=6$ and $\operatorname{Var}(X)=20(0.3)(0.7)=4.2$ . The standard deviation, if requested, is $\sqrt{4.2}$ .

For $Y\sim Geo(0.2)$ , $E(Y)=1/0.2=5$ : over many repetitions, the first success occurs on trial 5 on average.

Use the parameters of the stated model. Do not use $np$ for a geometric variable or forget the factor $1-p$ in binomial variance. Proofs of these formulae are not required.

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—5.5 The normal distributionRead

5.4 Discrete random variables

Learning objectives

A probability table gives both centre and spread

Choose binomial for a fixed count, geometric for first success

Read model moments directly from their parameters

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