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Q.
Two numbers x and y are chosen at random (without replacement) from amongst the numbers $1, 2, 3, .....2004$. The probability that $x^3 + y^3$ is divisible by 3 is
Probability
Solution:
We divide the number in three groups
$3k+1$ type $\left\{1, 4, 7, .................., 2005\right\}$
$3k+2$ type $\left\{2, 5, 8, .................., 2006\right\}$
$3k + 3$ type $\left\{3, 6, 9, .................., 2007\right\}$
$x^3 + y^3$ is divisible by 3 if x and y both belong to $3^{rd}$ group or one of them belongs to the first group and the other to the second group. So favourable number of cases
$=^{669}C_2 +669 \times 669$
Total number of cases $^{2007}C_2$
$\therefore $ Desired probability $=\frac{\frac{669\times668}{2}+669\times 668}{\frac{2007\times2006}{2}}$
$=\frac{669\times2006}{2007\times2006}=\frac{1}{3}$