"Let S = {E1, E2, ........., E8} be a sample space of a random experiment such that $P({E_n}) = {n \over {36}}$ for every n = 1, 2, ........, 8. Then the number of elements in the set $\left\{ {A \subseteq S:P(A) \ge {4 \over 5}} \right\}$ is ___________."

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Here $P({E_n}) = {n \over {36}}$ for n = 1, 2, 3, ......, 8

Here $P(A) = {{Any\,possible\,sum\,of\,(1,2,3,\,...,\,8)( = a\,say)} \over {36}}$

$\because$ ${a \over {36}} \ge {4 \over 5}$

$\therefore$ $a \ge 29$

If one of the number from {1, 2, ......, 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

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Let S = {E₁, E₂, ........., E₈} be a sample space of a random experiment such that $P({E_n}) = {n \over {36}}$ for every n = 1, 2, ........, 8. Then the number of elements in the set $\left\{ {A \subseteq S:P(A) \ge {4 \over 5}} \right\}$ is ___________.

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Let S = {E1, E2, ........., E8} be a sample space of a random experiment such that $P({E_n}) = {n \over {36}}$ for every n = 1, 2, ........, 8. Then the number of elements in the set $\left\{ {A \subseteq S:P(A) \ge {4 \over 5}} \right\}$ is ___________.

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Let S = {E₁, E₂, ........., E₈} be a sample space of a random experiment such that $P({E_n}) = {n \over {36}}$ for every n = 1, 2, ........, 8. Then the number of elements in the set $\left\{ {A \subseteq S:P(A) \ge {4 \over 5}} \right\}$ is ___________.