This problem involves calculating the number of subsets of a given sample space that satisfy a specific probability condition. The core idea is to translate the probability condition into a condition on the sum of indices of the elementary events, and then systematically count the subsets.

1. Key Concepts and Formulas

• Probability of an Event in a Discrete Sample Space: For an event $A \subseteq S$, where $S = \{E_1, E_2, \ldots, E_N\}$ is a discrete sample space, $P(A) = \sum_{E_n \in A} P(E_n)$.
• Sum of Probabilities: The sum of probabilities of all elementary events in a sample space must equal 1: $\sum_{n=1}^N P(E_n) = 1$.
• Complement of an Event: For any event $A$, its complement $A^c = S \setminus A$. The sum of indices of elements in $A$ and $A^c$ satisfies $S_A + S_{A^c} = S_S$, where $S_S$ is the sum of indices of all elements in $S$.

2. Step-by-Step Solution

Step 1: Verify the Validity of the Probability Distribution First, we must ensure that the given probabilities for the elementary events sum to 1. The sample space is $S = \{E_1, E_2, \ldots, E_8\}$, and $P(E_n) = \frac{n}{36}$ for $n = 1, 2, \ldots, 8$. Let's sum these probabilities: $\sum_{n=1}^8 P(E_n) = P(E_1) + P(E_2) + \ldots + P(E_8)$ $= \frac{1}{36} + \frac{2}{36} + \ldots + \frac{8}{36}$ $= \frac{1+2+3+4+5+6+7+8}{36}$ The sum of the first $k$ natural numbers is given by the formula $\frac{k(k+1)}{2}$. For $k=8$, the sum is $\frac{8(8+1)}{2} = \frac{8 \times 9}{2} = 36$. $= \frac{36}{36} = 1$ Since the sum is 1, the given probability distribution is valid.

Step 2: Translate the Probability Condition into a Sum of Indices Condition We are looking for subsets $A \subseteq S$ such that $P(A) \ge \frac{4}{5}$. Let $S_A$ be the sum of the indices of the elementary events in $A$. That is, $S_A = \sum_{E_n \in A} n$. Using the definition of $P(A)$: $P(A) = \sum_{E_n \in A} P(E_n) = \sum_{E_n \in A} \frac{n}{36} = \frac{1}{36} \sum_{E_n \in A} n = \frac{S_A}{36}$ Now, we can rewrite the given condition in terms of $S_A$: $\frac{S_A}{36} \ge \frac{4}{5}$ To find the minimum integer value for $S_A$, we multiply both sides by 36: $S_A \ge \frac{4 \times 36}{5}$ $S_A \ge \frac{144}{5}$ $S_A \ge 28.8$ Since $S_A$ is a sum of integers (the indices $n$), it must be an integer. Therefore, the condition $S_A \ge 28.8$ implies $S_A \ge 29$. Our task is to find the number of subsets $A \subseteq S$ such that the sum of the indices of their elements is greater than or equal to 29.

Step 3: Utilize the Complement Event Strategy Directly counting subsets $A$ with a large sum of indices ($S_A \ge 29$) can be complex. It's often easier to count the complementary event. Let $A^c = S \setminus A$ be the complement of $A$. The sum of all indices in $S$ is $S_S = \sum_{n=1}^8 n = 36$. For any subset $A$, $S_A + S_{A^c} = S_S = 36$. We can rewrite the condition $S_A \ge 29$ in terms of $S_{A^c}$: $36 - S_{A^c} \ge 29$ $36 - 29 \ge S_{A^c}$ $S_{A^c} \le 7$ So, we need to find the number of subsets $X = A^c \subseteq S$ such that the sum of the indices of its elements is less than or equal to 7. Each such subset $X$ uniquely corresponds to a subset $A$ that satisfies the original condition.

Step 4: Systematically Enumerate Subsets $X$ with $S_X \le 7$ The elements available for forming $X$ are $\{E_1, E_2, \ldots, E_8\}$, with indices $\{1, 2, \ldots, 8\}$. We list subsets $X$ based on the sum of their indices:

• Sum = 0:
  • The empty set: $\emptyset$ (1 subset)
• Sum = 1:
  • $\{E_1\}$ (1 subset)
• Sum = 2:
  • $\{E_2\}$ (1 subset)
• Sum = 3:
  • $\{E_3\}$
  • $\{E_1, E_2\}$ ($1+2=3$) (2 subsets)
• Sum = 4:
  • $\{E_4\}$
  • $\{E_1, E_3\}$ ($1+3=4$) (2 subsets)
• Sum = 5:
  • $\{E_5\}$
  • $\{E_1, E_4\}$ ($1+4=5$)
  • $\{E_2, E_3\}$ ($2+3=5$) (3 subsets)
• Sum = 6:
  • $\{E_6\}$
  • $\{E_1, E_5\}$ ($1+5=6$)
  • $\{E_2, E_4\}$ ($2+4=6$)
  • $\{E_1, E_2, E_3\}$ ($1+2+3=6$) (4 subsets)
• Sum = 7:
  • $\{E_7\}$
  • $\{E_1, E_6\}$ ($1+6=7$)
  • $\{E_2, E_5\}$ ($2+5=7$)
  • $\{E_3, E_4\}$ ($3+4=7$)
  • $\{E_1, E_2, E_4\}$ ($1+2+4=7$) (5 subsets)

Any subset with 4 or more elements will have a sum of indices greater than 7 (e.g., the smallest sum for 4 elements is $1+2+3+4=10$). Thus, we have enumerated all possible subsets $X$ with $S_X \le 7$.

Step 5: Calculate the Total Count Adding up the number of subsets for each possible sum: Total number of subsets $X = A^c$ such that $S_X \le 7$ is: $1 + 1 + 1 + 2 + 2 + 3 + 4 + 5 = 19$.

Each of these 19 subsets $X$ corresponds to a unique subset $A = S \setminus X$ that satisfies the original condition $P(A) \ge \frac{4}{5}$. Therefore, the number of elements in the set $\left\{ {A \subseteq S:P(A) \ge {4 \over 5}} \right\}$ is 19.

3. Common Mistakes & Tips

• Forgetting to verify $\sum P(E_n) = 1$: Always perform this check to ensure the problem is well-posed.
• Rounding errors: When converting a fractional probability condition to an integer sum condition (e.g., $S_A \ge 28.8 \implies S_A \ge 29$), ensure correct rounding up for 'greater than or equal to' conditions.
• Incomplete enumeration: When listing subsets for a specific sum, use a systematic method (e.g., by number of elements, or by smallest index) to avoid missing combinations.
• Not using the complement: For conditions involving "large" sums/probabilities, counting the complement (which involves "small" sums/probabilities) is often much simpler and less prone to errors.

4. Summary

The problem required us to count subsets whose probability met a certain threshold. We first verified the probability distribution of the sample space. Then, we converted the probability condition $P(A) \ge \frac{4}{5}$ into an equivalent condition on the sum of indices of the elementary events in the subset, $S_A \ge 29$. To simplify counting, we used the complementary event approach, which meant counting subsets $A^c$ with $S_{A^c} \le 7$. A careful, systematic enumeration of these complementary subsets yielded 19.

The final answer is $\boxed{\text{19}}$.