Two numbers are selected at random from a set of first 120 natural numbers, then the probability that product of selected number is divisible by 3 is:
(a) $\dfrac{13}{357}$
(b) $\dfrac{199}{357}$
(c) $\dfrac{158}{357}$
(d) $\dfrac{160}{357}$

Hint: First, before proceeding for this, we must know how many numbers from 120 natural numbers are divisible by 3. Then, by using the $nth$ term formula of AP where a is first term, d is common difference, n is the number of terms and ${{a}_{n}}$is the nth term to get the count of numbers divisible by 3 as ${{a}_{n}}=a+\left( n-1 \right)d$. Then, by using the concept of probability to get any probability we can also use the concept of subtracting it from 1 which is not required probability, we get the desired result.

Complete step-by-step answer:
In this question, we are supposed to find the probability that a product of selected numbers is divisible by 3 when two numbers are selected at random from a set of first 120 natural numbers.
So, before proceeding for this, we must know how many numbers from 120 natural numbers are divisible by 3.
So, we know the concept of AP where the divisible by 3 numbers start from 3 and the last number 120 is also divisible by 3.
So, we get the AP for divisible 3 from the first 120 natural numbers as 3, 6, 9, ....., 120.
Now, by using the $nth$ term formula of AP where a is first term, d is common difference, n is the number of terms and ${{a}_{n}}$is the nth term to get the count of numbers divisible by 3 as:
${{a}_{n}}=a+\left( n-1 \right)d$
No, by substituting the value of a as 3, d as 3 and ${{a}_{n}}$as 120, we get the total count of terms divisible by 3 which is n as:
$\begin{align}
  & 120=3+\left( n-1 \right)3 \\
 & \Rightarrow 117=3n-3 \\
 & \Rightarrow 120=3n \\
 & \Rightarrow n=\dfrac{120}{3} \\
 & \Rightarrow n=40 \\
\end{align}$
So, we get 40 terms from the first 120 natural numbers that are divisible 3.
Now, we can also say that 80 terms are not divisible by 3 from the first 120 natural numbers.
Then, by using the concept of combination to get the total number of cases that two numbers are selected at random from 120 numbers as:
${}^{120}{{C}_{2}}$
Then, by using the concept of probability to get any probability we can also use the concept of subtracting it from 1 which is not required probability.
So, we get the probability of numbers selected from 80 numbers that are not divisible by 3 is given by:
${}^{80}{{C}_{2}}$
Then, we get the probability by taking the ratio of favourable outcomes to the total outcomes for the probability of numbers not divisible by 3 as:
$\dfrac{{}^{80}{{C}_{2}}}{{}^{120}{{C}_{2}}}$
Then, by using the concept mentioned above , we get the probability of the two numbers being selected is divisible by 3 is given by:
$1-\dfrac{{}^{80}{{C}_{2}}}{{}^{120}{{C}_{2}}}$
Then, by solving the above expression, we get the required probability as:
$\begin{align}
  & 1-\dfrac{80!\times 2!\times \left( 120-2 \right)!}{2!\times \left( 80-2 \right)!\times 120!} \\
 & \Rightarrow 1-\dfrac{80\times 79\times 78!\times 118!}{78!\times 120\times 119\times 118!} \\
 & \Rightarrow 1-\dfrac{80\times 79}{120\times 119} \\
 & \Rightarrow 1-\dfrac{158}{357} \\
 & \Rightarrow \dfrac{357-158}{357} \\
 & \Rightarrow \dfrac{199}{357} \\
\end{align}$
So, we get the probability of two numbers selected from the first 120 natural numbers divisible by 3 as $\dfrac{199}{357}$.

So, the correct answer is “Option (b)”.

Note: Now, to solve these type of the questions we need to know some of the basics to get the value of ${}^{n}{{C}_{r}}$ is given by the formula of the combination:
$^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
To find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4.
$\begin{align}
  & 4!=4\times 3\times 2\times 1 \\
 & \Rightarrow 24 \\
\end{align}$