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**Hint:**Start by preparing the set as per the statement given in the question. Assume two elements from this set and form the equation as per the condition given. After forming the equation, look for conditions for divisibility by 3 and proceed. Apply the probability formula to get the answer.

**Complete step-by-step answer:**

Given, set of first 18 natural numbers

So, the set be $P = \left( {x,y} \right) \in \left\{ {1,2,3,4,5........18} \right\}$

Now, according to the question, the sum of cubes of the number must be divisible by 3

Which means ${x^3} + {y^3}$ must be divisible by 3

And we know ${x^3} + {y^3} = (x + y)({x^2} - xy + {y^2})$

$\therefore {x^3} + {y^3} = (x + y)({x^2} - xy + {y^2})$ will be divisible by 3

Now the number, $(x + y)$ or $({x^2} - xy + {y^2})$ should be divisible by 3

Now, let us see how it will be divisible by 3.

Number (x + y) will be divisible by 3 if x and y both are multiple of 3 or among x, y anyone leaves the remainder 2 when divided by 3.

Now 3, 6, 9, 12, 15, 18 total of 6 numbers are multiple of 3 and 1, 4, 7, 10, 13, 16 total of 6 numbers are those who leaves remainder 1 when divided by 3 and 2, 5, 8, 11, 14, 17 total of 6 numbers are those who leaves remainder 2 when divided by 3

Now, for the selection of such number we can use combination concept

$n(A) = {}^6{C_1} \times {}^6{C_1} + {}^6{C_2}$

And we know ,${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

$n(A) = 36 + 15 = 51$

Similarly, we have to choose 2 numbers out of 18, which can be done in the following number of ways.

$n(S) = {}^{18}{C_2} = 153$

$P(A) = \dfrac{{n(A)}}{{n(S)}} = \dfrac{{51}}{{153}} = \dfrac{{17}}{{51}}$

Therefore, the probability of the required condition is $\dfrac{{17}}{{51}}$.

**So, the correct answer is “Option B”.**

**Note:**Similar questions can be solved using the same procedure as followed above. Attention must be given while forming the equations or splitting the terms as any wrong step might lead to wrong or absurd value. Also Probability of any event can never be greater than 1.

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