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season 8 rick and morty - Professionaloutdoormedia

📅 May 15, 2026 👤 scraface

May 15, 2026

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📌 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} $.

📌 P(X_a = 3, X_b = 2, X_c = 3) = rac{8!}{3! \cdot 2! \cdot 3!} \cdot \left(rac{1}{3}

📌 ight)^3 \cdot \left(rac{1}{3}

📌 ight)^2 \cdot \left(rac{1}{3}

📌 ight)^3 = rac{8!}{3! \cdot 2! \cdot 3!} \cdot \left(rac{1}{3}

📌 rac{8!}{3! \cdot 2! \cdot 3!} = rac{40320}{6 \cdot 2 \cdot 6} = rac{40320}{72} = 560

📌 Now compute $ \left(rac{1}{3}

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Let $ a = \frac{79}{160} $. Then $ a^2 = \frac{6241}{25600} $,
$ a^3 = \frac{79 \cdot 6241}{160 \cdot 25600} = \frac{493039}{4096000}...
$ a^5 = a^3 \cdot a^2 = \frac{493039}{4096000} \cdot \frac{6241}{25600...
Rather than compute the full fraction, note the requirement is $ c_3 =...
$ \frac{79}{160} = 0.49375 $
$ (0.49375)^5 \approx 0.0935 $ (approximate, but we keep symbolic prec...
However, for exactness:
$ \left(\frac{79}{160}\right)^5 = \frac{79^5}{160^5} $, but instead, a...
c_3 = \frac{79}{160} - \frac{1}{5} \cdot \left(\frac{79}{160}\right)^5
But since the fifth power is not a perfect fifth, and given the model...

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