$5$ 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 $2$: $\left\{E_{1}\right\},\left\{E_{2}\right\}, \ldots \ldots .\left\{E_{8}\right\}$
type $3$: $\left\{ E _{1}, E _{2}\right\},\left\{ E _{1}, E _{3}\right\} \ldots \ldots .\left\{ E _{1}, E _{ 8 }\right\}$
.
.
.
type $6$: $\left\{ E _{1}, E _{2}, \ldots \ldots E _{5}\right\}, \ldots \ldots\left\{ E _{4}, E _{5}, E _{6}, E _{7}, E _{8}\right\}$
type $7$: $\left\{ E _{1}, E _{2}, \ldots \ldots . . E _{6}\right\}, \ldots \ldots .\left\{ E _{3}, E _{4}, \ldots \ldots \ldots . . E _{ 8 }\right\}$
type $8$: $\left\{ E _{1}, E _{2}, \ldots \ldots . E _{9}\right\}\left\{ E _{2}, E _{3}, \ldots \ldots \ldots . E _{8}\right\}$
type $9$: $\left\{ E _{1}, E _{2}, \ldots \ldots . . E _{ 8 }\right\}$
As $P ( A ) \geq \frac{4}{5}$
Note : Type $1$ to Type $4$ elements can not be in set
$A$ as maximum probability of type $4$ elements.
$\left\{ E _{5}, E _{6}, E _{ 7 }, E _{ s }\right\}$ 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 $P _{ 5 }$
$P _{5}=\frac{ n _{1}+ n _{2}+ n _{3}+ n _{4}+ n _{5}}{36} \leq \frac{4}{5}$
$\Rightarrow n _{1}+ n _{2}+ n _{3}+ n _{4}+ n _{5} \geq 28.8$
Hence, $2$ possible ways $\left\{ E _{9}, E _{6}, E _{\eta}, E _{\varepsilon}, E _{3}\right.$ or $\left.E _{4}\right\}$
$P _{6}= n _{1}+ n _{2}+ n _{3}+ n _{4}+ n _{5}+ n _{6} \geq 28.8$
$\Rightarrow 9$ possible ways
$P _{8} \Rightarrow n _{1}+ n _{2}+\ldots \ldots \ldots+ n _{1} \geq 288$
$\Rightarrow 7$ possible ways
$P _{ 8 } \Rightarrow n _{1}+ n _{2}+\ldots \ldots \ldots+ n _{ 8 } \geq 28.8$
$\Rightarrow 1$ possible way
Total $=19$
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