Let $S=\{E,E_2\dots .E_8\}$ be a sample space of random experiment such that $P\left(E_n\right)=\frac{n}{36}$ for every $\mathrm{n}=1,2\dots ..8$. Then the number of elements in the set $\left\{\mathrm{A}\subset \mathrm{S}:\mathrm{P}\left(\mathrm{A}\right)\ge \frac{4}{5}\right\}$ is

 Here $P\left(E_n\right)=\frac{n}{36}$ for $n=1,2,3,\dots \dots \dots ,8$
Here $P(A)$
$=\frac{\text{ Any possible sum of }(1,2,3,\dots ,8)(=a\text{ say })}{36}$
$\because \frac{a}{36}\ge \frac{4}{5}$
$\therefore a\ge 29$
If one of the number from $\{1,2,\dots ,8\}$ is left then total $a\ge 29$ by 3 ways.

Similarly by leaving terms more 2 or 3 we get 16 more combinations.
$\therefore$ Total number of different set A possible is $16+3$ $=19$