7/8/2025 2011 AMC 12B H reflection notes

 Q16 → Wrong due to silly mistake

Skipped Q20

• Time-sink solution requires heavy observation and quick identification.

• Involves quick use of inscribed angles and arcs.

• Q25 → Not enough time

Was not able to solve; did not understand substitution of 

n —> n-k in the solution + solution video.

Redo!

Q24 Notes:

• Easy algebra but looked difficult at first.

• Just required extra time to complete — able to solve afterwards.

Q15, Q17, Q18, Q19, Q22, Q23 → Correct 

• Q17： online solution took a long time

• I had the fastest solution.

• Q19： took a long time

• Found faster solution but took time to get to it.

Expected Value Question : 2024 Mathcounts State Sprint #24 level 1.5

2024 Mathcounts State Sprint #24

Three fair nickels and two fair dimes are tossed. What is the probability that at least two heads are showing and that at least one of the heads appears on a dime? Express your answer as a common fraction.

Try this question first before scrolling down to read the solution. 

Problem. Dennis rolls three fair six-sided dice, obtaining a, b, c ∈ {1,…,6}. Find \[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr]. \]

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Solution.

Step 1 — Linearity of expectation.

\[ \mathbb{E}\!\bigl[\,|a-b|+|b-c|+|c-a|\,\bigr] =\mathbb{E}[|a-b|]+\mathbb{E}[|b-c|]+\mathbb{E}[|c-a|] =3\,\mathbb{E}[|a-b|]. \]

Step 2 — Expected absolute difference of two dice.

Let \(X = |a-b|\). Its distribution is

\[ \begin{array}{c|cccccc} d & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \Pr(X=d) & \tfrac{6}{36} & \tfrac{10}{36} & \tfrac{8}{36} & \tfrac{6}{36} & \tfrac{4}{36} & \tfrac{2}{36} \end{array} \] \[ \mathbb{E}[|a-b|] =\frac{1}{36}\bigl(0\cdot6 + 1\cdot10 + 2\cdot8 + 3\cdot6 + 4\cdot4 + 5\cdot2\bigr) =\frac{70}{36} =\frac{35}{18}. \]