Question

Two numbers X and Y are chosen at random (without replacement) from among the numbers 1, 2,3, ... 3n. The probability that $X^3+Y^3$ is divisible by 3 is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{9}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $\frac{1}{3}$

We divide the numbers into $3$ groups each with $n$ terms, the first group A having numbers of the form $3n+1$, the second group B having numbers of the form $3n+2$ and the third group C having numbers of the form $3n$.
To satisfy the condition, we need to select both numbers from group C OR select the numbers one from each group A and B.
Thus can be done in ${}_2^{n}C+n^2$ ways.

Hence, probability = $\frac{{}_2^{n}C+n^2}{{}_2^{3n}C}=\frac{\frac{n(n-1)}{2}+n^2}{\frac{3n(3n-1)}{2}}=\frac{3n-1}{3(3n-1)}=\frac{1}{3}$