Question

Two numbers $x$ and $y$ are chosen at random (without replacement) from amongst the numbers $1,2,\mathrm{3......3}n$. Then find the probability that $x^3+y^3$ is divisible by $3$.

• A. Required probability $P\left(E\right)=\frac{1}{6}$
• B. Required probability $P\left(E\right)=\frac{5}{6}$
• C. Required probability $P\left(E\right)=\frac{2}{3}$
• D. Required probability $P\left(E\right)=\frac{1}{3}$

Answer 1

The correct option is D Required probability $P\left(E\right)=\frac{1}{3}$

Total number of ways of choosing $x$ and $y$ is ${}^{3n}{C}_2=\frac{3n(3n-1)}{2}$
$\therefore$ If the sample space, then $n\left(S\right)=\frac{3n(3n-1)}{2}$
Now, arrange the given numbers as below,
$1,4,7,...,3n-2$
$2,5,8,...,3n-1$
$3,6,9,...3n$
We see that $x^3+y^3$ will be divisible by $3$ in the fol-lowing cases:
one of the two numbers belongs to the first row and the other one (of the two numbers)
belongs to the second row
Both numbers occur in third row, If $E$ be the event for favourable cases, then
$n\left(E\right)-\left(n\right)\left(n\right){+}^n{C}_2=n^2+\frac{n(n+1)}{2}=\frac{n}{2}(3n-1)$
$\therefore P\left(E\right)=\frac{\frac{1}{2}(3n-1)}{\frac{3n(3n-1)}{2}}=\frac{1}{3}$