Let S= E, E_ 2 . E_ 8 be a sample space of random experiment such… - Vidyadip

MCQ

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 $n =1,2 \ldots .$. Then the number of elements in the set $\left\{ A \subset S : P ( A ) \geq \frac{4}{5}\right\}$ is

A
$17$
B
$18$
✓
$19$
D
$20$

✓

Answer

Correct option: C.

$19$

c
$P \left( A ^{\prime}\right)<\frac{1}{5}=\frac{36}{180}$

$5$ 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 $1:\{\}$

type $2$: $\left\{E_{1}\right\},\left\{E_{2}\right\}, \ldots \ldots .\left\{E_{8}\right\}$

type $3$: $\left\{ E _{1}, E _{2}\right\},\left\{ E _{1}, E _{3}\right\} \ldots \ldots .\left\{ E _{1}, E _{ 8 }\right\}$

.

.

.

type $6$: $\left\{ E _{1}, E _{2}, \ldots \ldots E _{5}\right\}, \ldots \ldots\left\{ E _{4}, E _{5}, E _{6}, E _{7}, E _{8}\right\}$

type $7$: $\left\{ E _{1}, E _{2}, \ldots \ldots . . E _{6}\right\}, \ldots \ldots .\left\{ E _{3}, E _{4}, \ldots \ldots \ldots . . E _{ 8 }\right\}$

type $8$: $\left\{ E _{1}, E _{2}, \ldots \ldots . E _{9}\right\}\left\{ E _{2}, E _{3}, \ldots \ldots \ldots . E _{8}\right\}$

type $9$: $\left\{ E _{1}, E _{2}, \ldots \ldots . . E _{ 8 }\right\}$

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.

$\left\{ E _{5}, E _{6}, E _{ 7 }, E _{ s }\right\}$ 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 $\left\{ E _{9}, E _{6}, E _{\eta}, E _{\varepsilon}, E _{3}\right.$ 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

$P _{8} \Rightarrow n _{1}+ n _{2}+\ldots \ldots \ldots+ n _{1} \geq 288$

$\Rightarrow 7$ possible ways

$P _{ 8 } \Rightarrow n _{1}+ n _{2}+\ldots \ldots \ldots+ n _{ 8 } \geq 28.8$

$\Rightarrow 1$ possible way

Total $=19$

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