Let S = {1, 2, 3, ....... 50} and ${\mathrm{A}}_{\mathrm{k}}$ be the set of multiple of k in S for k $\in$N . If ${\mathrm{x}}_{\mathrm{k}}$ is a number chosen from ${\mathrm{A}}_{\mathrm{k}}$, then match the items of List-
I with the items of List-II.

List - I	List - II
A) $\mathrm{P}({\mathrm{x}}_3 <\hspace{0.17em}30)$	I) $\frac{1}{2}$
B) $\mathrm{P}(15 <\hspace{0.17em}{\mathrm{x}}_4 \le 36)$	II) $\frac{2}{3}$
C) $\mathrm{P}({\mathrm{x}}_7 >\hspace{0.17em}35)$	III) $\frac{2}{7}$
D) $\mathrm{P}({\mathrm{x}}_{11} >\hspace{0.17em}11)$	IV) $\frac{1}{4}$
	V) $\frac{3}{4}$
	VI) $\frac{9}{16}$

Given S={1, 2, 3, ............ 50}

for option A )

let ${\mathrm{A}}_3=\{3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 36, 42, 45, 48\}$

Now $\mathrm{P}({\mathrm{x}}_3<30)=\frac{9}{16}$

$\therefore \mathrm{A}\to \mathrm{VI}$

for option B)

let ${\mathrm{A}}_4=\{4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48\}$

Now $\mathrm{P}(15<{\mathrm{x}}_4\le 36)=\frac{6}{12}=\frac{1}{2}$

$\therefore \mathrm{B}\to \mathrm{I}$

for option C)

let ${\mathrm{A}}_7=\{7, 14, 21, 28, 35, 42, 49\}$

Now $\mathrm{P}({\mathrm{x}}_7>35)=\frac{2}{7}$

$\therefore \mathrm{C}\to \mathrm{III}$

for option D)

let ${\mathrm{A}}_{11}=\{11, 22, 33, 44\}$

Now $\mathrm{P}({\mathrm{x}}_{11}>11)=\frac{3}{4}$

$\therefore \mathrm{D}\to \mathrm{V}$

Hence $\mathrm{A}\to \mathrm{VI}, \mathrm{B}\to \mathrm{I}, \mathrm{C}\to \mathrm{III}, \mathrm{D}\to \mathrm{V}$