Solutions:

\( P(A') < \frac{1}{5} = \frac{36}{180} \)

\( 5 \) times the sum of missing number should be less than \( 36 \).
If \( 1 \) digit is missing \( = 7 \)
If \( 2 \) digit is missing \( = 9 \)
If \( 3 \) digit is missing \( = 2 \)
If \( 0 \) digit is missing \( = 1 \)

Alternate
A is subset of S hence
A can have elements:
Type \( 1 \) : \( \{\} \)
Type \( 2 \) : \( \{E1\}, \{E2\}, \ldots, \{E8\} \)
Type \( 3 \) : \( \{E1,E2\}, \{E1,E3\}, \ldots, \{E1,E8\} \)
:
:
Type \( 6 \) : \( \{E1,E2,\ldots,E5\}, \ldots, \{E4,E5,E6,E7,E8\} \)
Type \( 7 \) : \( \{E1,E2,\ldots,E6\}, \ldots, \{E3,E4,\ldots,E8\} \)
Type \( 8 \) : \( \{E1,E2,\ldots,E7\} \); \( \{E2,E3,\ldots,E8\} \)
Type \( 9 \) : \( \{E1,E2,\ldots,E8\} \)

As \( P(A) \ge \frac{4}{5} \)

Note : Type \( 1 \) to Type \( 4 \) elements can not be in set A as maximum probability of type \( 4 \) elements.
\( \{E5,E6,E7,E8\} \) is \( \frac{5}{36} + \frac{6}{36} + \frac{7}{36} + \frac{8}{36} = \frac{13}{18} < \frac{4}{5} \)

Now for Type \( 5 \) acceptable elements let’s call probability as \( P5 \)
\( P5 = \frac{n1 + n2 + n3 + n4 + n5}{36} \ge \frac{4}{5} \)

\( \Rightarrow \quad n1 + n2 + n3 + n4 + n5 \ge 28.8 \)

Hence, \( 2 \) possible ways \( \{E5,E6,E7,E8,E3 \text{ or } E4\} \)

\( P6 = n1 + n2 + n3 + n4 + n5 \ge 28.8 \)
\( \Rightarrow \) \( 9 \) possible ways
\( P7 \Rightarrow n1 + n2 + \ldots + n7 \ge 288 \)
\( \Rightarrow \) \( 7 \) possible ways
\( P8 \Rightarrow n1 + n2 + \ldots + n8 \ge 28.8 \)
\( \Rightarrow \) \( 1 \) possible way
Total = \( 19 \)

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