Question

Out of $3n$ consecutive integers, three are selected at random, then the chance that their sum is divisible by $3$ is

• A. $\frac{3n^2-3n+2}{(3n-1)(3n-2)}$
• B. $\frac{1}{(3n-1)(3n-2)}$
• C. $\frac{3n^2}{(3n-1)(3n-2)}$
• D. $\frac{n^2}{(3n-1)(3n-2)}$

Answer 1

The correct option is A $\frac{3n^2-3n+2}{(3n-1)(3n-2)}$

Three integers from a total of $3n$ consecutive integers can be selected in ${}^{3n}{C}_3$ ways.
Now, we have to find all such combination of three such that their sum is divisible by $3$.
There are four possible cases:
$1$. All three numbers are themselves divisible by $3$.
Out of $3n$ consecutive integers, exactly $n$ will be divisible by $3$. So, any three of them can be selected in ${}^n{C}_3$ ways.
$2$. All three numbers give a remainder of $1$ when divided by $3$.
Out of $3n$ consecutive integers, exactly $n$ will give a remainder of $1$, when divided by $3$. So, any three of them can be selected in ${}^n{C}_3$ ways.
$3$. All three numbers give a remainder of $2$ when divided by $3$.
Out of $3n$ consecutive integers, exactly $n$ will give a remainder of $2$, when divided by $3$. So, any three of them can be selected in ${}^n{C}_3$ ways.
$4.$ Out of three selected , one is divisible by $3$, one gives a remainder of $1$ when divided by $3$, the other one gives a remainder of $2$ when divided by $3$. There are exactly $n$ numbers from each category. So, three of them can be chosen in $n^3$ number of ways.
Hence, the number of ways in which $3$ numbers can be selected so that sum is divisible by $3$ is equal to $3\left({}^n{C}_3\right)+n^3$.
Probability$=\frac{3\left({}_3^{n}C\right)+n^3}{{}_3^{3n}C}$
Simplifying, we get the probability$=\frac{{3n}^2-3n+2}{(3n-1)(3n-2)}$