Solution: This is a multinomial probability. Let $ X = (X_a, X_b, X_c) $ be the counts of each letter in a random 8-character word, with each character independently chosen from $\{a, b, c\}$ with equal probability $ rac{1}{3} $.

["### Understanding Multinomial Probability Through a 8-Letter Word Random Model", "When analyzing random sequences—especially in cryptography, linguistics, or text generation—understanding the distribution of character frequencies is crucial. One powerful statistical tool is the multinomial probability model, which helps quantify the likelihood of observing specific counts of outcomes in a fixed number of independent trials. In this article, we explore a concrete example: a multinomial distribution applied to an 8-character random word composed exclusively of the letters a, b, and c, each selected uniformly at random with equal probability $ \frac{1}{3} $.", "---", "#### What Is a Multinomial Distribution?", "The multinomial distribution generalizes the binomial distribution to multiple categories. Suppose you perform $ n $ independent trials, each resulting in one of $ k $ possible outcomes, with probabilities $ p_1, p_2, \dots, p_k $. For $ n $ trials, the probability of observing counts $ X_1, X_2, \dots, X_k $—where $ X_1 + X_2 + \dots + X_k = n $—is given by:", "$$\nP(X_1 = x_1, X_2 = x_2, \dots, X_k = x_k) = \frac{n!}{x_1! x_2! \cdots x_k!} p_1^{x_1} p_2^{x_2} \cdots p_k^{x_k}\n$$", "In this framework, $ X = (X_a, X_b, X_c) $ represents the counts of letters a, b, and c, respectively, over an 8-character word, with $ X_a + X_b + X_c = 8 $, and each letter chosen independently with $ P(a) = P(b) = P(c) = \frac{1}{3} $.", "---", "#### Modeling the 8-Letter Word Scenario", "Let $ X = (X_a, X_b, X_c) $ describe how many times each letter appears in a random 8-character string. Since each character is independently selected from ${a, b, c}$ with equal probability $ \frac{1}{3} $, $ X $ follows a multinomial distribution:", "$$\nX \sim \ ext{Multinomial}(n=8, p_a = \frac{1}{3}, p_b = \frac{1}{3}, p_c = \frac{1}{3})\n$$", "This setup allows us to compute meaningful probabilities about letter frequencies—answers essential in applications like:\n- Evaluating the likelihood of balanced vs. skewed character distributions in generated text\n- Testing randomness in pseudorandom letter generators\n- Analyzing letter bias in cryptographic or linguistic datasets", "---", "#### Calculating the Distribution", "For any combination $ (x_a, x_b, x_c) $ such that $ x_a + x_b + x_c = 8 $, the probability mass function becomes:", "$$\nP(X_a = x_a, X_b = x_b, X_c = x_c) = \frac{8!}{x_a! , x_b! , x_c!} \left( \frac{1}{3} \right)^{x_a} \left( \frac{1}{3} \right)^{x_b} \left( \frac{1}{3} \right)^{x_c} = \frac{8!}{x_a! , x_b! , x_c!} \left( \frac{1}{3} \right)^8\n$$", "Since $ \left( \frac{1}{3} \right)^8 = \frac{1}{6561} $, the probability is proportional to the multinomial coefficient:", "$$\n\frac{8!}{x_a! , x_b! , x_c! \cdot 6561}\n$$", "This reveals two key insights:\n1. Symmetry: All permutations of a given triple $ (x_a, x_b, x_c) $ (where order matters among letters) carry the same probability due to equal $ p_a = p_b = p_c $.\n2. Sum Constraint: Only combinations satisfying $ x_a + x_b + x_c = 8 $ are valid; others have zero probability.", "---", "#### Example: Probability for a Specific Distribution", "Suppose we compute the probability of the balanced case where $ X_a = X_b = X_c = \frac{8}{3} $. But since counts must be integers, $ (X_a, X_b, X_c) = (3, 3, 2) $ and its permutations are feasible.", "Take $ (x_a, x_b, x_c) = (3, 3, 2) $. Then:", "$$\nP(3, 3, 2) = \frac{8!}{3! , 3! , 2!} \cdot \frac{1}{6561} = \frac{40320}{6 \cdot 6 \cdot 2} \cdot \frac{1}{6561} = \frac{40320}{72} \cdot \frac{1}{6561} = 560 \cdot \frac{1}{6561} \approx 0.0852\n$$", "The $ 560 $ factor arises from arranging 3 a’s, 3 b’s, and 2 c’s in 8 positions. Moving to all such permutations yields the full set of probabilities summing to 1 over all valid integer triples.", "---", "#### Why This Model Matters", "This multinomial setup provides a rigorous foundation for analyzing text statistics:", "- Randomness Testing: If generated letters deviate significantly from expected multinomial frequencies, randomness may be compromised.\n- Cryptanalysis: Suspected biases in cipher machine outputs can be assessed via letter count distributions.\n- Natural Language Study: Helps characterize letter biases in corpora, aiding tasks like word prediction or language modeling.", "---", "#### Conclusion", "The multinomial probability model of an 8-character random word with equally likely letters a, b, and c elegantly captures how information distributes across categories. By leveraging the multinomial distribution, we gain precise tools to compute, predict, and validate letter frequencies—essential in domains ranging from computer science to linguistics. Understanding $ X = (X_a, X_b, X_c) $ through this lens transforms seemingly arbitrary sequences into analytically tractable data.", "Whether you’re building random text generators, testing cryptographic systems, or analyzing linguistic patterns, grasping this probabilistic model enhances both insight and control over discrete stochastic processes.", "---", "Keywords: multinomial probability, multinomial distribution, random word generation, letter frequency model, probability calculation, $ X = (X_a, X_b, X_c) $, $ P(X_a = x_a, X_b = x_b, X_c = x_c) $, character counting, text randomness, probability mass function."]