Discrete random variables

Probability distributions

A discrete random variable $X$ takes separate values, such as $0,1,2,3$, each with a probability. Its probability distribution lists every possible value with its probability, usually in a table. The table below is the distribution of the number of heads, $X$, when two fair coins are tossed. Capital $X$ is the random variable; small $x$ is a particular value it can take.

Two facts check any distribution: every probability is between 0 and 1, and the probabilities add up to 1, because $X$ must take exactly one of its values. That second fact is how you find an unknown constant: if $\mathrm{P}(X=x)=kx$ for $x=1,2,3,4$, then $k(1+2+3+4)=1$, so $k=\frac{1}{10}$.

Drawing up a distribution from a situation. First decide every value $X$ can take. Then find each probability by the methods of the probability topic: counting outcomes, combinations, or a tree diagram. Show the calculation for each value, give exact fractions where you can, and finish by checking the total is 1.

Expectation and variance

The expectation (mean) of $X$ is the long-run average value of $X$ over many repetitions:
$$\mathrm{E}(X)=\sum xp$$
Each value is weighted by how often it happens. The variance measures spread about the mean, and MF19 gives
$$\mathrm{Var}(X)=\sum x^{2}p-\{\mathrm{E}(X){\}}^2\mathrm{.}$$
This is the same idea as for data: the mean of the squares minus the square of the mean. The standard deviation is $\sqrt{\mathrm{Var}(X)}$.

For the distribution in Worked example 1: $\mathrm{E}(X)=0\times \frac{1}{10}+1\times \frac{6}{10}+2\times \frac{3}{10}=\frac{12}{10}=1.2$. Then $\sum x^{2}p=0+\frac{6}{10}+4\times \frac{3}{10}=1.8$, so $\mathrm{Var}(X)=1.8-{1.2}^2=0.36$.

$\mathrm{E}(X)$ need not be a value that $X$ can take: the expected number of red discs is 1.2. A variance can never be negative; if you get a negative one, you have probably squared the probabilities instead of the values.

The binomial distribution

Suppose an experiment is repeated a fixed number $n$ of times, each trial has just two outcomes (success or failure), the probability of success $p$ is the same for every trial, and the trials are independent. Then the number of successes $X$ has the binomial distribution, written $X\sim \mathrm{B}(n,p)$, and
$$\mathrm{P}(X=r)=\left(\genfrac{}{}{0ex}{}{n}{r}\right)p^r(1-p{)}^{n-r},r=0,1,\dots ,n\mathrm{.}$$
Why this formula? Any one particular order of $r$ successes and $n-r$ failures, such as SSF...F, has probability $p^r(1-p{)}^{n-r}$, by multiplying independent probabilities. The successes can be in any $r$ of the $n$ positions, and there are $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ ways to choose those positions. Each order is exclusive of the others, so add: $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ copies of the same probability.

Several values. "At most 2" means $\mathrm{P}(X⩽2)=\mathrm{P}(0)+\mathrm{P}(1)+\mathrm{P}(2)$. "At least 2" is quickest as $1-\mathrm{P}(0)-\mathrm{P}(1)$. "More than 2" excludes 2. Read the inequality carefully: it decides which terms you add.

Mean and variance (given in MF19; proofs are not required):
$$\mathrm{E}(X)=np,\mathrm{Var}(X)=np(1-p)\mathrm{.}$$
If you know the mean and variance, divide: $\frac{\mathrm{Var}(X)}{\mathrm{E}(X)}=1-p$, which gives $p$ and then $n$.

The geometric distribution

Now repeat independent trials, each with the same probability of success $p$, but with no fixed number: keep going until the first success. The number of trials needed, $X$, including the successful one, has the geometric distribution, written $X\sim \mathrm{Geo}(p)$, with
$$\mathrm{P}(X=r)=p(1-p{)}^{r-1},r=1,2,3,\dots$$
The reasoning is a single path on a tree: $r-1$ failures, each with probability $1-p$, followed by one success. Note that $X$ starts at 1, not 0, and has no upper limit.

Two results follow and save a lot of work:

• $\mathrm{P}(X>r)=(1-p{)}^r$, because "more than $r$ trials are needed" means "the first $r$ trials all fail";
• $\mathrm{P}(X⩽r)=1-(1-p{)}^r$.

So $\mathrm{P}(3⩽X⩽6)=\mathrm{P}(X>2)-\mathrm{P}(X>6)=(1-p{)}^2-(1-p{)}^6$.

The expectation is $\mathrm{E}(X)=\frac{1}{p}$, given in MF19 (proof not required). If a fair die is thrown until a 6 appears, the mean number of throws is 6, which matches intuition: a 6 comes up once in every six throws on average. The variance of the geometric distribution is not in the syllabus.

Choosing a model

Questions often ask you to say why a distribution is a suitable model, or to state an assumption. Tie each condition to the context.

Binomial $\mathrm{B}(n,p)$ needs:

• a fixed number of trials $n$ (12 seeds in a packet);
• two outcomes per trial (a seed germinates or does not);
• the same probability of success each time (each seed has probability 0.85 of germinating);
• independent trials (whether one seed germinates does not affect another).

Geometric $\mathrm{Geo}(p)$ needs the same last three conditions, but counts the trials up to and including the first success, so there is no fixed $n$.

Typical reasons a model fails: drawing without replacement from a small population (the probability changes from trial to trial, so not binomial); trials that affect each other (a player's confidence after a miss, so not independent); or a question about the number of successes in a fixed number of trials being modelled as geometric. The distinction between the two models is the commonest test: fixed number of trials, count the successes is binomial; count the trials until the first success is geometric.