To make the table represent a legitimate probability distribution, the sum of all the probabilities $P(X = x)$ must equal 1. Let's denote the missing values as $P(X = 3)$ and $P(X = 6)$.

Given:

• $P(X = -3) = 0.29$
• $P(X = 4) = 0.29$
• $P(X = 5) = 0.30$

Let the missing values be $P(X = 3) = p_1$ and $P(X = 6) = p_2$.

The total probability should sum to 1:

$0.29 + p_1 + 0.29 + 0.30 + p_2 = 1$

Simplifying, we get:

$p_1 + p_2 + 0.88 = 1$

$p_1 + p_2 = 0.12$

Now, the values of $p_1$ and $p_2$ must be non-negative and add up to 0.12. A simple solution could be:

$p_1 = 0.06 \quad \text{and} \quad p_2 = 0.06$

So, the completed table is:

\hline \text{Value } x \text{ of } X & P(X = x) \\ \hline -3 & 0.29 \\ 3 & 0.06 \\ 4 & 0.29 \\ 5 & 0.30 \\ 6 & 0.06 \\ \hline \end{array}$$ Would you like more details on the calculations or have any other questions? --- Here are some related questions: 1. What is a probability distribution, and how is it different from a cumulative distribution? 2. Can you explain the properties of a valid probability distribution? 3. How do you determine the expected value of a discrete random variable? 4. What is the variance of a discrete random variable, and how is it calculated? 5. Can the probabilities in a distribution be negative? Why or why not? 6. How does one interpret the sum of probabilities in a probability distribution? 7. What are some common examples of discrete random variables? 8. How would you adjust the probabilities if the value of $$P(X = 5)$$ was instead 0.25? **Tip:** When working with probability distributions, always ensure that the sum of all probabilities equals 1. This is a fundamental property that must be satisfied.