Two numbers a and b are chosen at random from the set {1, 2, 3, ......, 3n}. The probability that $a^2 -b^2$ is divisible by 3, is
Answer & explanation
Correct answer: option 2
The number of ways of choosing two numbers from the given set is ${^{3n}C}_2.$
Let us divide given 3 numbers into three groups $G_1, G_2 $ and $G_3 $ as ffollows :
$G_1 : 3, 6, 9, .....3n $
$G_2 : 1 , 4, 7, 10, ....., 3n-1$
$G_3 : 2, 5, 8, 11, ....., 3n -2 $
We have, $ a^2 - b^2 = (a-b) (a + b)$
Therefore, $ a^2-b^2$ will be divisible by 3 if either a and b are chosen from the same group or one of them is chosen from group $G_2$ and the other from group $G_3 $. Therefor, the number of favourable elementary events is
$({^nC}_2 + {^nC}_2 + {^nC}_2) + {^nC}_1 × {^nC}_1 = 3 {^nC}_2 + n^2 $
∴ Required probability $= \frac{3×{^nC}_2 +n^2 }{^{3n}C_2}= \frac{5n-3}{3(3n-1)}$