TS EAMCET · Maths · Probability

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

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

Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

Three integers are selected at random from 3n consecutive integers. Let consecutive integers are x, x+1, x+2, x+3, x+4, x+5 .... x+3n, x+3n-1 Now, three integers are selected at random in such a way that their sum is divisible by 3. Let us…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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

Step-by-step Solution

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