4.7 Discrete random variables and expected valueIB Maths: Applications and Interpretation SL: Revision notes

A discrete random variable $X$ takes separate values (usually whole numbers) according to chance, such as the score on a die or the number of goals in a match. A probability distribution lists each value $x$ with its probability $P(X=x)$. Every probability satisfies $0\leq P(X=x)\leq1$, and all the probabilities add up to 1: $\sum P(X=x)=1.$

A distribution can be given as a table, for example $x=1,2,3,4,5$ with $P(X=x)=0.1,0.2,0.15,0.05,0.5$, or as a formula, for example $P(X=x)=\frac{4+x}{18}$ for $x\in\{1,2,3\}$. Substitute each value to build the table: here $\frac{5}{18},\frac{6}{18},\frac{7}{18}$, which sum to 1. For a probability such as $P(X\geq2)$, add the probabilities of every value that qualifies.

The expected value (mean) of a discrete random variable is $\mathrm{E}(X)=\sum xP(X=x).$ It is the long-run average of $X$ over many trials, not necessarily a value that $X$ can take. Example: $P(X=1)=0.1$, $P(X=2)=0.3$, $P(X=3)=0.4$, $P(X=4)=0.2$ gives $\mathrm{E}(X)=0.1+0.6+1.2+0.8=2.7$. A GDC can also do this: enter the values and probabilities as two lists and calculate the one-variable statistics with the probabilities as frequencies.

If $X$ is the gain of a player (prize minus entry fee), then $\mathrm{E}(X)$ is the expected gain per game. If $\mathrm{E}(X)=0$ the game is fair. A negative value favours the organiser and a positive value favours the player. Example: a game costs 4 AED and pays 0, 6 or 10 AED with probabilities $0.5$, $0.3$, $0.2$. Expected prize $=3.8$ AED, so the expected gain is $3.8-4=-0.2$ AED: not fair. Over 500 games, the organiser expects to make $500\times0.2=100$ AED.

Many problems build a distribution from counting outcomes. For two fair four-sided dice with $X$ the larger number, there are 16 equally likely outcomes, and $P(X=3)=\frac{5}{16}$ because five outcomes have a larger number of 3. Building the full table gives $\frac{1}{16},\frac{3}{16},\frac{5}{16},\frac{7}{16}$ and $\mathrm{E}(X)=\frac{50}{16}=3.125$. To find a total over repeated plays, multiply the expected value for one play by the number of plays.