To find the mean (or expected value) of the probability distribution of the random variable $X$, you use the formula:

$\mu = E(X) = \sum_{i} x_i \cdot P(x_i)$

Here, $X$ can take the values $1$, $2$, and $3$, with the corresponding probabilities $P(1) = \frac{10}{33}$, $P(2) = \frac{1}{3}$, and $P(3) = \frac{12}{33}$.

Step 1: Write down the values and probabilities.

• $x_1 = 1$, $P(x_1) = \frac{10}{33}$
• $x_2 = 2$, $P(x_2) = \frac{1}{3}$
• $x_3 = 3$, $P(x_3) = \frac{12}{33}$

Step 2: Calculate the mean.

$\mu = 1 \cdot \frac{10}{33} + 2 \cdot \frac{1}{3} + 3 \cdot \frac{12}{33}$

Step 3: Simplify the expression.

Let's simplify each term: $\mu = \frac{10}{33} + 2 \cdot \frac{1}{3} + 3 \cdot \frac{12}{33}$

Convert the fraction $\frac{1}{3}$ to a fraction with a denominator of 33: $2 \cdot \frac{1}{3} = 2 \cdot \frac{11}{33} = \frac{22}{33}$

Now, substitute and simplify: $\mu = \frac{10}{33} + \frac{22}{33} + \frac{36}{33} = \frac{10 + 22 + 36}{33} = \frac{68}{33}$

So, the mean of the probability distribution is: $\mu = \frac{68}{33} \approx 2.06$

Would you like more details or have any questions?

Here are five related questions:

1. What is the variance of the probability distribution?
2. How do you interpret the expected value in the context of probability distributions?
3. What is the probability that $X$ takes a value greater than the mean?
4. How would the mean change if $P(2)$ increased?
5. Can the mean be outside the range of possible values for $X$?

Tip: The mean (or expected value) provides a measure of the central tendency of a probability distribution, representing the average outcome if the experiment is repeated many times.