Let S=E_1,,E_8 with P(E_n)=n36.

Question

Let $S=\{E_1,\dots,E_8\}$ with $P(E_n)=\dfrac{n}{36}$. Find the number of elements in $\{A\subseteq S: P(A)\ge\dfrac45\}$.

Answer: 19

Accepted Answer

$P(A)=\dfrac{a}{36}\ge\dfrac45\Rightarrow a\ge29$. Subsets of $\{1,\dots,8\}$ (sum $=36$) with sum $\ge29$ correspond to complements with sum $\le7$. Counting complements with sum $\le7$: leaving out 1 element gives 3 valid sums ($\ge29$ directly), and other combinations give 16 more, totaling $16+3=19$.

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