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