How Do You Calculate the Expected Value (Mean) of a Discrete Random Variable?

To calculate the expected value (mean) of a discrete random variable, multiply each possible value by its probability and add the products. For a discrete random variable $X$,

$$\operatorname{E}(X)=\sum x\operatorname{P}(X=x).$$

This is part of IB Mathematics: Analysis and Approaches, SL 4.7: Discrete random variables, and is core content for both SL and HL.

How the calculation works

A discrete random variable takes separate, countable values, each with an associated probability. Before calculating the mean, check that every probability satisfies $0\leq \operatorname{P}(X=x)\leq 1$ and that the probabilities sum to $1$.

The calculation is a weighted mean: outcomes with larger probabilities contribute more strongly to the result. Because positive and negative values can offset one another, include signs carefully. The expected value must lie between the smallest and largest possible values, providing a useful check on your answer.

Suppose a player's gain $X$, in dollars, has this probability distribution:

Therefore,

$$\operatorname{E}(X)=(-2)(0.20)+(1)(0.50)+(5)(0.30)=1.60.$$

The expected gain is therefore \1.60$per game. This does not mean the player will gain exactly$$1.60$ in one game. It represents the gain per game over many repetitions. Conceptually, expected value is not a prediction of the next outcome. It is the theoretical average approached when the same probability experiment is repeated many times under unchanged conditions. This distinction matters in context.

A common misconception is to calculate the ordinary mean of the possible values. The values are not necessarily equally likely, so they must be weighted by their probabilities. If $X$ represents a player's gain, $\operatorname{E}(X)=0$ indicates a fair game.

IB exam technique

This content can appear on AA Paper 1 or Paper 2. Show the substitution into $\operatorname{E}(X)=\sum x\operatorname{P}(X=x)$, not only the final answer. If a probability is unknown, first use $\sum \operatorname{P}(X=x)=1$, then calculate the expected value and interpret it in context.