To find the expected value $E(X)$ of a fair 4-sided die, we use the formula for the expected value of a discrete random variable:

$E(X) = \sum_{x} x \cdot P(x)$

Where:

• $x$ is a possible outcome, which in this case is the number on the bottom face of the die (1, 2, 3, or 4).
• $P(x)$ is the probability of each outcome. Since the die is fair, each outcome has an equal probability of $\frac{1}{4}$.

So, we compute the expected value as follows:

$E(X) = 1 \cdot \frac{1}{4} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{4} + 4 \cdot \frac{1}{4}$

Simplifying the terms:

$E(X) = \frac{1}{4}(1 + 2 + 3 + 4) = \frac{1}{4} \cdot 10 = 2.5$

Thus, the expected value $E(X)$ is:

$E(X) = 2.50$

Let me know if you want any further details or explanations!

Here are 5 related questions:

1. What is the variance of $X$ for this 4-sided die?
2. How would the expected value change if the die were biased?
3. What is the expected value for a 6-sided die?
4. What is the probability that $X$ is greater than 2?
5. How does the expected value change with the number of sides on the die?

Tip: The expected value for any fair $n$-sided die is always $\frac{n+1}{2}$.