Let $$S = \{E, E_2 \ldots E_8\}$$ be a sample space of a random experiment such that $$P(E_n) = \frac{n}{36}$$ for every $$n = 1, 2 \ldots 8$$. Then the number of elements in the set $$\{A \subset S : P(A) \geq \frac{4}{5}\}$$ is ______

Given $$S = \{E_1, E_2, \ldots, E_8\}$$ with $$P(E_n) = \frac{n}{36}$$, and noting that $$\sum_{n=1}^{8} \frac{n}{36} = \frac{36}{36} = 1$$, for a subset $$A$$ we have $$P(A) = \frac{1}{36}\sum_{E_n \in A} n$$, so the condition $$P(A) \geq \frac{4}{5}$$ translates into $$\sum_{E_n \in A} n \geq \frac{4}{5}\times 36 = 28.8$$, i.e., $$\sum_{E_n \in A} n \geq 29$$.

If $$A^c$$ denotes the complement, then $$\sum_{A} n + \sum_{A^c} n = 36$$, so $$\sum_{A} n \geq 29 \iff \sum_{A^c} n \leq 7$$. We count subsets $$B \subseteq \{1,2,\ldots,8\}$$ with $$\text{sum}(B) \leq 7$$.

For $$|B| = 0$$ there is $$\emptyset$$ with sum 0 (1 subset). For $$|B| = 1$$ the subsets are $$\{1\}$$, $$\{2\}$$, $$\{3\}$$, $$\{4\}$$, $$\{5\}$$, $$\{6\}$$, $$\{7\}$$ (7 subsets).

For $$|B| = 2$$ the pairs with sum $$\leq 7$$ are $$\{1,2\}$$, $$\{1,3\}$$, $$\{1,4\}$$, $$\{1,5\}$$, $$\{1,6\}$$, $$\{2,3\}$$, $$\{2,4\}$$, $$\{2,5\}$$, $$\{3,4\}$$ (9 subsets). For $$|B| = 3$$ the triples with sum $$\leq 7$$ are $$\{1,2,3\}, \{1,2,4\}$$ (2 subsets). For $$|B| \geq 4$$ the minimum sum is $$1+2+3+4 = 10 > 7$$, so there are none.

$$1 + 7 + 9 + 2 = 19$$ so the answer is 19.