From 3 n consecutive integers three integers are selected at random. The probability that their sum is divisible by 3 is

Question

From 3 n consecutive integers three integers are selected at random. The probability that their sum is divisible by 3 is (A) (3 . nC3+n2/3 n C 3) (B) (2 . nC3+n3/3 n C 3) (C) (3 n2-3 n+2/(3 n-1)(3 n-2)) (D) (3 n2-3 n+2/(3 n+1)(3 n+2))

Tardigrade Admin · Accepted Answer

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 C3 Case 2: 3 k+(3 k+1)+(3 k+2)= n C1 × n C1 × n C1 Case 3:(3 k+1)+(3 k+1)+(3 k+1)= n C3 Case 4: (3 k+2)+(3 k+2)+(3 k+2)= n C3: Required probability =( C3+n3+ n C3+ n C3/ 3 n C3) =(n3+3 ⋅ n C3/ n C3) =(n3+(3 n(n-1)(n-2)/6)/(3 n(3 n-1)(3 n-2))6) =(6 n3+3 n3-9 n2+6 n/27 n3-27 n2+6 n) =(3 n2-3 n+2/9 n2-9 n+2)=(3 n2-3 n+2/(3 n-1)(3 n-2))

Q. From $3 n$ consecutive integers three integers are selected at random. The probability that their sum is divisible by 3 is

$\frac{3. n _{C_{3}} + n ^{2}}{3 n _{C_{3}}}$

$\frac{2. n _{C_{3}} + n ^{3}}{3 n _{C_{3}}}$

$\frac{3 n ^{2} - 3 n + 2}{( 3 n - 1 ) ( 3 n - 2 )}$

$\frac{3 n ^{2} - 3 n + 2}{( 3 n + 1 ) ( 3 n + 2 )}$

Q. From 3n consecutive integers three integers are selected at random. The probability that their sum is divisible by 3 is

3nC3​​3.nC3​​+n2​

3nC3​​2.nC3​​+n3​

(3n−1)(3n−2)3n2−3n+2​

(3n+1)(3n+2)3n2−3n+2​

Q. From $3 n$ consecutive integers three integers are selected at random. The probability that their sum is divisible by 3 is

$\frac{3. n _{C_{3}} + n ^{2}}{3 n _{C_{3}}}$

$\frac{2. n _{C_{3}} + n ^{3}}{3 n _{C_{3}}}$

$\frac{3 n ^{2} - 3 n + 2}{( 3 n - 1 ) ( 3 n - 2 )}$

$\frac{3 n ^{2} - 3 n + 2}{( 3 n + 1 ) ( 3 n + 2 )}$