3n Consecutive integers n integers will be of the form 3k n integers will be of the form 3k+1 n integers will be of the form 3k+2 If the sum of 3 chosen integers is also divisible by 3 , then Case 1:3k+3k+3k=‌nC3 Case 2:3k+(3k+1)+(3k+2)=‌nC1×‌nC1×‌nC1 Case 3:(3k+1)+(3k+1)+(3k+1)=‌nC3 Case 4:(3k+2)+(3k+2)+(3k+2)=‌nC3 ∴ Required probability ‌‌=‌