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Step-by-Step Solution
Step 1: Understand the problem
We have a sample space S = \{ E_1, E_2, \dots, E_8 \} with P(E_n) = \frac{n}{36} for each n = 1, 2, \dots ,8 . For any subset A \subseteq S , the probability of A is
P(A) \;=\; \sum_{E_n \in A} \frac{n}{36} \;=\; \frac{\sum_{E_n \in A} n}{36}.
We need to find the number of subsets A for which P(A) \ge \frac{4}{5} .
Step 2: Convert the probability condition into a sum condition
The condition P(A) \ge \frac{4}{5} translates to
\frac{\sum_{E_n \in A} n}{36} \;\ge\; \frac{4}{5},
which implies
\sum_{E_n \in A} n \;\ge\; \frac{4}{5} \times 36 \;=\; 28.8.
Since the sum of the elements in a subset must be an integer, we get
\sum_{E_n \in A} n \;\ge\; 29.
Step 3: Use the total sum of elements in S
The elements are labeled by n = 1, 2, \dots, 8 . Their total sum is
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 \;=\; 36.
Therefore, for a subset A , we have
\sum_{E_n \in A} n + \sum_{E_n \in A^c} n \;=\; 36,
where A^c is the complement of A in S .
Step 4: Relate subset sums to its complement
From \sum_{E_n \in A} n \ge 29 , we get
\sum_{E_n \in A^c} n = 36 - \sum_{E_n \in A} n \;\le\; 36 - 29 \;=\; 7.
Thus, counting subsets A with sum of elements at least 29 is equivalent to counting complements A^c whose sum of elements is at most 7.
Step 5: Count all subsets of S whose sum is at most 7
We look for all subsets of \{1,2,3,4,5,6,7,8\} that sum to 7 or less. They include:
The empty set: sum is 0 \le 7.
Single-element subsets: \{1\}, \{2\}, \{3\}, \{4\}, \{5\}, \{6\}, \{7\} (each has sum \le 7 ). This gives 7 subsets.
Two-element subsets with sum \le 7 : for example, \{1,2\} (sum=3), \{1,3\} (sum=4), ..., up to \{3,4\} (sum=7). Counting systematically gives 9 such pairs.
Three-element subsets with sum \le 7 : only \{1,2,3\} (sum=6) and \{1,2,4\} (sum=7) qualify. So that adds 2 subsets.
Subsets of more than three elements: any such subset exceeds sum 7 because even \{1,2,3,4\} sums to 10, which is greater than 7, so they do not qualify.
Hence, the total number of subsets of \{1,2,3,4,5,6,7,8\} having sum at most 7 is
1 \;(\text{empty set}) + 7 + 9 + 2 \;=\; 19.
Step 6: Conclude the number of subsets
Because each such subset of sum at most 7 corresponds exactly to a complement subset with sum at least 29, there are
19
subsets A \subseteq S for which P(A) \ge \frac{4}{5}.
Final Answer: 19
