It appears there are missing probabilities for the values $x = 3$ and $x = 5$ in your table. To ensure that the sum of all probabilities $P(X = x)$ equals 1, we can calculate these missing probabilities using the given values.

From the table:

• $P(X = -3) = 0.15$
• $P(X = 2) = 0.30$
• $P(X = 4) = 0.24$

The sum of these probabilities is: $0.15 + 0.30 + 0.24 = 0.69$

Therefore, the sum of the probabilities for $x = 3$ and $x = 5$ is: $1 - 0.69 = 0.31$

Without additional information, we cannot uniquely determine $P(X = 3)$ and $P(X = 5)$ from this equation alone. However, if there's an assumption we can make about their distribution or any other given constraints, that might help split the remaining probability of 0.31 between the two values.

Would you like to specify any conditions, or should we assume they are equally likely?

Also, here are a few further inquiries that can be explored:

1. What is the expected value of $X$?
2. What is the variance of $X$?
3. How would changes to these probabilities affect the expected value?
4. Can you confirm if $P(X = 3)$ and $P(X = 5)$ should be considered equally?
5. What practical scenarios might use a probability distribution like this one?

Tip: When probabilities are missing in such tables, and no additional constraints are specified, assuming equal distribution among the missing probabilities can sometimes serve as a reasonable default assumption, though it's always best to check if any specific guidelines or conditions are mentioned.