Question

Two distinct numbers are chosen at random from the set $\{1,2,3,....,3n\}$. The probability that $x^2-y^2$ is divisible by $3$ is $\frac{pn+q}{r(3n-1)}$, then $p+q+r=$____

Answer 1

Given: set of $\{1,2,3,......3n\}$ numbers. The probability of $x^2-y^2$ is divisible by 3 = $\frac{pn+q}{r(3n-1)}$

To find: $p+q+r$

Sol: For $x^2-y^2$ to be divisible by $3$

$(x^2-y^2)=(x-y)(x+y)$

Thus, $(x-y)$ or $(x+y)$ should be divisible by $3$

$\therefore x$ and $y$ will be divisible by 3

$x$ and $y$ can be $3,6,9,\mathrm{12....}$

We know the probability that $x^2-y^2$ is divisible by 3$=\frac{5n-3}{9n-3}$
Comparing this with given condition we get,

$\frac{5n-3}{3(3n-1)}=\frac{pn+q}{r(3n-1)}$

We can see that, $p=5,q=(-3),r=r=3$
Hence, $p+q+r=5-3+3$
$=5$ is the answer