Let's go through the problem step-by-step.

Problem Summary:

We are rolling a die 10 times. Let $X$ be a random variable representing the number of times we get a multiple of 3 (i.e., 3 or 6). The task is to:

1. Determine the probability distribution of $X$.
2. Calculate $E[X]$ (the expected value of $X$).
3. Calculate $\text{Var}(X)$ (the variance of $X$).

Solution:

1. Define the Probability of Success:

   • Since we are interested in the outcome being a multiple of 3, the favorable outcomes are 3 and 6.
   • The probability of rolling a 3 or a 6 on a single roll of a die is $\frac{2}{6} = \frac{1}{3}$.

2. Distribution of $X$:

   • Since we are counting the number of times a multiple of 3 occurs in 10 independent rolls, $X$ follows a Binomial distribution.
   • Specifically, $X \sim \text{Binomial}(n = 10, p = \frac{1}{3})$.

3. Expected Value of $X$:

   • For a Binomial distribution, $E[X] = n \cdot p$.
   • Here, $E[X] = 10 \cdot \frac{1}{3} = \frac{10}{3} \approx 3.33$.

4. Variance of $X$:

   • For a Binomial distribution, $\text{Var}(X) = n \cdot p \cdot (1 - p)$.
   • Here, $\text{Var}(X) = 10 \cdot \frac{1}{3} \cdot \left(1 - \frac{1}{3}\right) = 10 \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{20}{9} \approx 2.22$.

Final Answer:

• The probability distribution of $X$ is $X \sim \text{Binomial}(10, \frac{1}{3})$.
• $E[X] = \frac{10}{3} \approx 3.33$.
• $\text{Var}(X) = \frac{20}{9} \approx 2.22$.

Would you like further details on any part of this solution?

Here are some related questions to deepen your understanding:

1. What is the probability that $X = 5$ in this scenario?
2. How does changing the number of rolls affect the expected value and variance?
3. What if the die was biased? How would that change the probability distribution?
4. How would you calculate the standard deviation of $X$?
5. What is the probability that $X$ is greater than its expected value?

Tip: Binomial distribution is commonly used when dealing with counts of events in independent trials with fixed probabilities.