4.7 Discrete random variablesIB Maths: Analysis and Approaches HL: Revision notes

A random variable is a quantity whose value is determined by the outcome of a random experiment. It is discrete if it can take only separate (countable) values, such as the number of goals in a match or the score on a dice.

We write $X$ for the random variable and $x$ for a value it takes, so $P(X = 2)$ means 'the probability that $X$ takes the value 2'.

A probability distribution lists every possible value of $X$ with its probability. It can be given

• as a table, e.g. $P(G = 0) = 0.25$, $P(G = 1) = 0.35$, …; or
• as a function, e.g. $P(X = x) = \frac{4 + x}{18}$ for $x \in \{1, 2, 3\}$.

For any distribution, every probability is between 0 and 1 and $\sum P(X = x) = 1.$ Use this to find an unknown constant: for $P(X = x) = kx$, $x \in \{1, 2, 3, 4\}$, $10k = 1$ so $k = \frac{1}{10}$.

The expected value (mean) of $X$ is $E(X) = \sum x\,P(X = x),$ which is in the formula booklet. It is the long-run average value of $X$ over many trials. It need not be a possible value of $X$: a team can have $E(G) = 1.33$ goals per match.

Example: $P(X = x) = \frac{x}{10}$ for $x = 1, 2, 3, 4$ gives $E(X) = \frac{1 + 4 + 9 + 16}{10} = 3$.

If an event has probability $p$ on each of $n$ occasions, the expected number of occurrences is $np$. With $P(G \ge 2) = 0.4$, over 30 matches the expected number with at least 2 goals is $30\times0.4 = 12$.

For two independent observations, multiply probabilities and remember different orders: $P(Y_1 + Y_2 = 5) = 2\,P(Y = 2)P(Y = 3)$ when 2 and 3 are the only values that sum to 5.

Let $X$ be the gain of a player: the amount received minus the cost to play. The game is fair if $E(X) = 0.$ If $E(X) < 0$, the player expects to lose on average and the organiser expects to profit; over $n$ games the organiser's expected profit is $-n\,E(X)$.

To make a game fair, write $E(X)$ in terms of the unknown prize or cost, set it equal to 0 and solve.