$$\DeclareMathOperator{cosec}{cosec}$$

Statistics and Probability

Furthermore

Official Guidance, clarification and syllabus links:

Probability distributions will be given in the following ways:

 X P(X = x) 1 2 3 4 5 0.1 0.2 0.15 0.05 0.5

$$P(X = x) = \frac{1}{18}(4 + x)$$ for $$x \in \{1, 2, 3\}$$

$$E(X) = 0$$ indicates a fair game where $$X$$ represents the gain of a player.

Formula Booklet:
 Expected value of a discrete random variable $$X$$ $$E(X) = \sum x P(X = x)$$

A discrete random variable is a variable that can take on a countable number of distinct outcomes. The probability distribution of a discrete random variable is a list of probabilities associated with each of its possible values. It's important for students to understand how to calculate the expected value, or mean, of a discrete random variable as it gives the average outcome if an experiment is repeated a large number of times. Applications of discrete random variables and their expected values can be found in various fields such as finance, insurance, and engineering, helping to make informed decisions under uncertainty.

The key formulae for the expected value (mean) of a discrete random variable $$X$$ are given by:

$$E(X) = \sum x P(X = x)$$

Where:

$$E(X)$$ is the expected value of $$X$$,

$$x$$ are the possible values of the random variable, and

$$P(X = x)$$ is the probability that $$X$$ takes the value $$x$$.

Let's look at an example:

Consider a game where a player can score 1, 2, or 3 points with probabilities $$\frac{1}{6}$$, $$\frac{1}{6}$$, and $$\frac{4}{6}$$ respectively. The expected value of the points scored in a single play of the game is:

$$E(X) = \sum x P(X = x) = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + 3 \times \frac{4}{6} \\ = \frac{1}{6} + \frac{2}{6} + \frac{12}{6} \\ = \frac{15}{6} \\ = 2.5$$

This means that on average, a player can expect to score 2.5 points per play over the long term.

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