Thank you for reporting, we will resolve it shortly
Q.
From $3 n$ consecutive integers three integers are selected at random. The probability that their sum is divisible by 3 is
TS EAMCET 2021
Solution:
$3 n$ Consecutive integers
$n$ integers will be of the form $3 k$
$n$ integers will be of the form $3 k+1$
$n$ integers will be of the form $3 k+2$
If the sum of 3 chosen integers is also divisible by 3 , then
Case $1: 3 k+3 k+3 k={ }^n C_3$
Case $2: 3 k+(3 k+1)+(3 k+2)={ }^n C_1 \times{ }^n C_1 \times{ }^n C_1$
Case $3:(3 k+1)+(3 k+1)+(3 k+1)={ }^n C_3$
Case 4: $(3 k+2)+(3 k+2)+(3 k+2)={ }^n C_3$
: Required probability
$ =\frac{{ }^{C_3}+n^3+{ }^n C_3+{ }^n C_3}{{ }^{3 n} C_3} $
$ =\frac{n^3+3 \cdot{ }^n C_3}{{ }^n C_3} $
$ =\frac{n^3+\frac{3 n(n-1)(n-2)}{6}}{\frac{3 n(3 n-1)(3 n-2)}{6}} $
$ =\frac{6 n^3+3 n^3-9 n^2+6 n}{27 n^3-27 n^2+6 n} $
$ =\frac{3 n^2-3 n+2}{9 n^2-9 n+2}=\frac{3 n^2-3 n+2}{(3 n-1)(3 n-2)}$