Given,

$P\left(A\right)\ge \frac{4}{5}$

$A$ is subset of $S$ hence

$A$ can have elements;

type $1$:$\left\{\right\}$

type $2$: $\left\{E_1\right\},\left\{E_2\right\},\dots \dots \left\{E_8\right\}$

type $3$: $\left\{E_1,E_2\right\},\left\{E_1,E_3\right\}\dots \dots \left\{E_1,E_8\right\}$

$⋮$

$⋮$

type $6$: $\left\{E_1,E_2,\dots \dots .E_5\right\},\dots \dots \left\{E_4,E_5,E_6,E_7,E_8\right\}$

type $7$: $\left\{E_1,E_2,\dots \dots .E_6\right\},\dots \dots \left\{E_3,E_4,\dots \dots \dots .E_8\right\}$

type $8$: $\left\{E_1,E_2,\dots \dots .E_7\right\}\left\{E_2,E_3,\dots \dots .E_8\right\}$

type $9$: $\left\{E_1,E_2,\dots \dots E_8\right\}$

As $P\left(A\right)\ge \frac{4}{5}$;

Note: Type $1$ to Type $4$ elements can not be in set $A$ as maximum probability of type $4$ elements.

$\left\{E_5,E_6,E_7,E_8\right\}$ 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 $P_5$

$P_5=\frac{n_1+n_2+n_3+n_4+n_5}{36}\ge \frac{4}{5}$

$\Rightarrow n_1+n_2+n_3+n_4+n_3\ge 28.8$

Hence, $2$ possible ways $\left\{E_5,E_6,E_7,E_8,E_3 \mathrm{or} E_4\right\}$

$P_6=n_1+n_1+n_3+n_4+n_3+n_6\ge 28.8$

$\Rightarrow 9$ possible ways

$P_7\Rightarrow n_1+n_1+\dots \dots \dots +n_7\ge 28.8$

$\Rightarrow 7$ possible ways

$P_8\Rightarrow n_1+n_1+\dots \dots +n_n\ge 28.8$

$\Rightarrow 1$ possible way

Total $=19$