Mathematics - Probability Question with Solution | TestHub

$x^3+y^3$ is divisible by $3\Rightarrow x+y$ is divisible by $3\Rightarrow x,y$ are multiples of 3 or one leaves remainder 1 and the other 2 when divided by 3 .

$3,6,9\dots .30$ are multiple of $3;1,4,7,\dots \dots ,28$ leave remainder 1

$2,5,8,\dots ,29$ leave remainder 2

Probability $=\frac{\left(\genfrac{}{}{0ex}{}{10}{1}\right)\left(\genfrac{}{}{0ex}{}{10}{1}\right)+\left(\genfrac{}{}{0ex}{}{10}{2}\right)}{\left(\genfrac{}{}{0ex}{}{30}{2}\right)}$ $=\frac{145}{15\times 29}=\frac{1}{3}$