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Home » \[ 3n^2 + 2 = 425 \] - Ready Digital AB

\[ 3n^2 + 2 = 425 \] - Ready Digital AB

📅 March 8, 2026 👤 scraface

Mar 08, 2026

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📌 P(H = D) = \sum_{k=1}^{6} P(H = k) \cdot P(D = k) = \sum_{k=1}^{6} \binom{6}{k} \left(\frac{1}{2}\right)^6 \cdot \frac{1}{6}.

📌 Factor out constants:

📌 P = \frac{1}{6 \cdot 2^6} \sum_{k=1}^{6} \binom{6}{k} = \frac{1}{384} \left( \sum_{k=0}^{6} \binom{6}{k} - \binom{6}{0} \right) = \frac{1}{384} (2^6 - 1) = \frac{63}{384}.

📌 \frac{63}{384} = \frac{21}{128}.

📌 Thus, the probability that the number of heads equals the die roll is $ \boxed{\dfrac{21}{128}} $.

📌 Question: An elementary school student is learning about the solar system and draws 5 planets, coloring each one either Earth-like (E) or Jupiter-like (J), but insists that no two adjacent planets in her linear diagram are both Jupiter-like. How many valid colorings are possible?

📌 Solution: We are to count the number of binary strings of length 5 (using 'E' and 'J', or 0 and 1), where no two adjacent characters are both 'J'. This is equivalent to counting binary strings without consecutive 1s.

📌 Let $ a_n $ be the number of valid colorings of $ n $ planets with no two adjacent Jupiter-like planets.

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