Let S = \ E, E _ 2 E _ 8 \ be a sample space of raddom experiment...

Given,

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

$A$ is subset of $S$ hence

$A$ can have elements;

type $1$:\left\right)

type $2$: \left\{{E}_{1}\right\}, \left\{{E}_{2}\right\}, \ldots \ldots \left\{{E}_{8}\right\}$</p><p>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\}$</p><p>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}_{7}\right\}\left\{{E}_{2}, {E}_{3}, \ldots \ldots . {E}_{8}\right\}$</p><p>type$9$: \left\{{E}_{1}, {E}_{2}, \ldots \ldots {E}_{8}\right\}$

As $P\left(A\right) \ge \dfrac{4}{5}$;

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\}$is$\dfrac{5}{36} + \dfrac{6}{36} + \dfrac{7}{36} + \dfrac{8}{36} = \dfrac{13}{18} < \dfrac{4}{5}$</p><p>Now for Type$5$acceptable elements let's call probability as${P}_{5}$</p><p>${P}_{5} = \dfrac{{n}_{1} + {n}_{2} + {n}_{3} + {n}_{4} + {n}_{5}}{36} \ge \dfrac{4}{5}$</p><p>$⇒ {n}_{1} + {n}_{2} + {n}_{3} + {n}_{4} + {n}_{3} \ge 28 . 8$</p><p>Hence,$2$possible ways \left\{{E}_{5}, {E}_{6}, {E}_{7}, {E}_{8}, {E}_{3}or{E}_{4}\right\}$

${P}_{6} = {n}_{1} + {n}_{1} + {n}_{3} + {n}_{4} + {n}_{3} + {n}_{6} \ge 28 . 8$

$⇒ 9$ possible ways

${P}_{7} ⇒ {n}_{1} + {n}_{1} + \ldots \ldots \ldots + {n}_{7} \ge 28 . 8$

$⇒ 7$ possible ways

${P}_{8} ⇒ {n}_{1} + {n}_{1} + \ldots \ldots + {n}_{n} \ge 28 . 8$

$⇒ 1$ possible way

Total $= 19$