Question

Let \( S=\left\{E_{1}, E_{2} \ldots . E_{8}\right\} \) be a sample space of random experiment such that \( P\left(E_{n}\right)=\frac{n}{36} \) for every \( n=1,2 \ldots 8 \). Then the number of elements in the set \( \left\{A \subset S: P(A) \geq \frac{4}{5}\right\} \) is

Answer 1

If one of the number from {1, 2, ......, 8} is left then total \(a\ge29\) 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