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

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

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

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

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

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Solution

The correct option is A $\frac{3 n^{2} - 3 n + 2}{(3 n - 1) (3 n - 2)}$
Let the consecutive numbers be
$x, x + 1, x + 2, . . ., x + 3 n - 1$
we arrange them in three rows, each consisting of $n$ numbers as follows:
$R_{1} : x, x + 3, x + 6, . . ., x + 3 n - 3 R_{2} : x + 1, x + 4, x + 7, . . . x + 3 n - 2 R_{3} : x + 2, x + 5, x + 8, . . ., x + 3 n - 1$
Sum of three numbers shall be divisible by $3$ is all three numbers are from the same row
as one each from $R_{1}, R_{2} {, R}_{3}$
The number of selection of 3 numbers from $R_{1}$ or $R_{2}$ or $R_{3}$ is
$^{n} C_{3} +^{n} C_{3} +^{n} C_{3} = 3^{n} C_{3} = \frac{1}{2} n (n - 1) (n - 1)$
The number of selection of 1 number from each of $R_{1}, R_{2} {, R}_{3}$ is
$^{n} C_{1} .^{n} C_{1} .^{n} C_{1} = n^{3}$
Thus. the total number as favorable cases
$n^{3} + \frac{1}{2} n (n - 1) (n - 1) = \frac{1}{2} n (2 n^{2} - 3 n + 2)$
The exhaustive number of cases = $^{3 n} C_{3} = \frac{1}{2} n (3 n - 1) (3 n - 1)$
Therefore required probability $= \frac{3 n^{2} - 3 n + 2}{9 n^{2} - 9 n + 2}$

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

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