Question

Let $S =\left\{ E , E _{2} \ldots . E _{8}\right\}$ 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

$P \left( A ^{\prime}\right)<\frac{1}{5}=\frac{36}{180}$
$5$ times the sum of missing number should be less than $36 .$
If 1 digit is missing $=7$
If 2 digit is missing $=9$
If 3 digit is missing $=2$
If 0 digit is missing $=1$
Alternate
$A$ is subset of $S$ hence
A can have elements:
type $1:\{\}$

As $P ( A ) \geq \frac{4}{5}$
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 $P _{5}$
$P _{5}=\frac{ n _{1}+ n _{2}+ n _{3}+ n _{4}+ n _{5}}{36} \leq \frac{4}{5}$
$\Rightarrow n _{1}+ n _{2}+ n _{3}+ n _{4}+ n _{5} \geq 28.8$
Hence, 2 possible ways $\left\{ E _{5}, E _{6}, E _{7}, E _{8}, E _{3}\right.$ or $\left.E _{4}\right\}$
$P _{6}= n _{1}+ n _{2}+ n _{3}+ n _{4}+ n _{5}+ n _{6} \geq 28.8$
$\Rightarrow 9$ possible ways
$P _{7} \Rightarrow n _{1}+ n _{2}+\ldots \ldots \ldots+ n _{7} \geq 288$
$\Rightarrow 7$ possible ways
$P _{8} \Rightarrow n _{1}+ n _{2}+\ldots \ldots \ldots+ n _{8} \geq 28.8$
$\Rightarrow 1$ possible way
Total $=19$