. Find the mean of the probability distribution of a random variable X which can take only the values 1, 2, and 3, given that P(1) = 10/33 P(2) = 1/3 P(3) = 12/33

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.

Learn how to calculate the mean (expected value) of a probability distribution for a random variable X, which can take values 1, 2, and 3 with probabilities P(1) = 10/33, P(2) = 1/3, and P(3) = 12/33. Follow a detailed step-by-step solution using the expected value formula. Suitable for advanced learners and anyone studying probability theory.

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