To determine whether the given distribution is a discrete probability distribution, we need to check the following two conditions:

1. Each probability $P(X = x)$ must be between 0 and 1, i.e., $0 \leq P(X = x) \leq 1$.
2. The sum of all probabilities must be 1, i.e., the sum of $P(X = -4) + P(X = 4) + P(X = 5)$ should equal 1.

Given the distribution:

\hline x & P(X = x) \\ \hline -4 & \frac{1}{4} \\ 4 & \frac{2}{3} \\ 5 & \frac{1}{3} \\ \hline \end{array}$$ ### Step 1: Check if each probability is between 0 and 1. - $$P(X = -4) = \frac{1}{4} = 0.25$$, which is between 0 and 1. - $$P(X = 4) = \frac{2}{3} \approx 0.6667$$, which is between 0 and 1. - $$P(X = 5) = \frac{1}{3} \approx 0.3333$$, which is between 0 and 1. So, each probability satisfies the condition of being between 0 and 1. ### Step 2: Check if the sum of probabilities is 1. Let's add the probabilities: $$\frac{1}{4} + \frac{2}{3} + \frac{1}{3}$$ First, find a common denominator. The least common denominator of 4 and 3 is 12. $$\frac{1}{4} = \frac{3}{12}, \quad \frac{2}{3} = \frac{8}{12}, \quad \frac{1}{3} = \frac{4}{12}$$ Now, add them: $$\frac{3}{12} + \frac{8}{12} + \frac{4}{12} = \frac{15}{12}$$ Since $$\frac{15}{12} = 1.25$$, **the sum of the probabilities is greater than 1**, which violates the second condition. ### Conclusion: The given distribution is **not** a discrete probability distribution because the sum of the probabilities is not equal to 1. Would you like more details or clarification? Here are some related questions: 1. What would happen if the sum of probabilities was exactly 1? 2. How can we adjust the probabilities to make the distribution valid? 3. How can we recognize a discrete probability distribution in other cases? 4. Can a probability distribution ever have values greater than 1? 5. How would you calculate the mean of a valid discrete probability distribution? **Tip**: Always verify that the sum of the probabilities equals 1 before concluding that a distribution is valid.