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The problem is asking to fill in the missing probabilities in a table to create a legitimate probability distribution for the discrete random variable \( X \). 

We know that for a valid probability distribution:
1. All probabilities must be non-negative.
2. The sum of the probabilities must equal 1.

### Given table:
\[
\begin{array}{|c|c|}
\hline
\text{Value } x \text{ of } X & P(X = x) \\
\hline
-3 & 0.25 \\
0  & ? \\
4  & ? \\
5  & 0.24 \\
6  & 0.18 \\
\hline
\end{array}
\]

### Step 1: Add the known probabilities.
\[
P(X = -3) = 0.25, \quad P(X = 5) = 0.24, \quad P(X = 6) = 0.18
\]
Summing these known probabilities:
\[
0.25 + 0.24 + 0.18 = 0.67
\]

### Step 2: Use the fact that the total probability must equal 1.
\[
P(X = -3) + P(X = 0) + P(X = 4) + P(X = 5) + P(X = 6) = 1
\]
Substituting the known probabilities:
\[
0.67 + P(X = 0) + P(X = 4) = 1
\]
\[
P(X = 0) + P(X = 4) = 1 - 0.67 = 0.33
\]

### Step 3: Assign values to \( P(X = 0) \) and \( P(X = 4) \).
Since \( P(X = 0) + P(X = 4) = 0.33 \), there are various ways to distribute this. If no further constraints are given, one possible distribution is:
\[
P(X = 0) = 0.15, \quad P(X = 4) = 0.18
\]

Thus, the completed table could be:

\[
\begin{array}{|c|c|}
\hline
\text{Value } x \text{ of } X & P(X = x) \\
\hline
-3 & 0.25 \\
0  & 0.15 \\
4  & 0.18 \\
5  & 0.24 \\
6  & 0.18 \\
\hline
\end{array}
\]

Would you like to explore more or adjust these probabilities?

Here are 5 questions to expand on this:
1. What are the characteristics of a discrete probability distribution?
2. How do we handle cases with multiple missing probabilities?
3. What are some other ways to distribute the remaining 0.33 between \( P(X = 0) \) and \( P(X = 4) \)?
4. How would you verify if this is a valid probability distribution?
5. Can we use these values to compute expected value or variance?

**Tip**: When assigning missing probabilities, always ensure the sum of all probabilities equals 1 and the individual values are valid (between 0 and 1).

Fill in the P(X = x) values to give a legitimate probability distribution for the discrete random variable X, whose possible values are -3, 0, 4, 5, and 6.

Learn how to complete a discrete probability distribution by filling in missing probabilities for a random variable X with values -3, 0, 4, 5, and 6. This solution uses basic probability rules, ensuring the sum equals 1 and each probability is valid. Suitable for Grades 10-12.

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