Question

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

Answer 1

Answer: 19

$P\left(A^{\prime }\right)<\frac{1}{5}=\frac{36}{180}$
$5$ 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 $P(A)\ge \frac{4}{5}$
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 $P_5$
$P_5=\frac{n_1+n_2+n_3+n_4+n_5}{36}\le \frac{4}{5}$
$\Rightarrow n_1+n_2+n_3+n_4+n_5\ge 28.8$
Hence, 2 possible ways $\{E_5,E_6,E_7,E_8,E_3$ or $E_4\}$
$P_6=n_1+n_2+n_3+n_4+n_5+n_6\ge 28.8$
$\Rightarrow 9$ possible ways
$P_7\Rightarrow n_1+n_2+\dots \dots \dots +n_7\ge 288$
$\Rightarrow 7$ possible ways
$P_8\Rightarrow n_1+n_2+\dots \dots \dots +n_8\ge 28.8$
$\Rightarrow 1$ possible way
Total $=19$