CTET 2 Math and Science 7 Jan 2022 Paper

${\mathrm{t}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}$
Here,

$\mathit{a}=$ first term
$\mathrm{n}=\mathrm{n}$. of terms
$\mathrm{d}=$ common difference
$\mathrm{P}\left(\mathrm{A}\right)=\mathrm{n}\left(\mathrm{A}\right)∕\mathrm{n}\left(\mathrm{S}\right)$
$\mathrm{P}\left(\mathrm{A}\right)$ is the probability of an event "A" $\mathrm{n}\left(\mathrm{A}\right)$ is the number of favorable outcomes. $\mathrm{n}\left(\mathrm{S}\right)$ is the total number of events in the sample space.

$130=40+\left(\mathrm{n}-1\right)\cdot 1$
$\Rightarrow 130=40+\mathrm{n}-1$
$\Rightarrow \mathrm{n}=91$
Now,
$\mathrm{LCM}$ of $2,3,4=12$
So, numbers which are divisible by 12 between 40 to $130=48,60,72,84,96$, 108,120
So, total of 7 numbers are there
So,
Probability $=7∕91$
$\therefore$ The required probability is $7∕91$.