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From a sequence of 3n consecutive integers, three numbers are picked randomly. What is the probability that their total is a multiple of 3?
- (3n² - 3n + 2) / [n(3n - 1)(3n - 2)]
- (3n² - 3n + 2) / [n(3n - 1)(3n - 2)]
- n³ / [(3n - 1)(n - 2)(n - 3)]
- None of the above
Correct answer: (3n² - 3n + 2) / [n(3n - 1)(3n - 2)]
Solution
The probability that the total of three randomly picked numbers from a sequence of 3n consecutive integers is a multiple of 3 can be found by considering the arrangement of these integers in rows such that the sum of three numbers is divisible by 3 if they are either all in the same row or all in different rows, leading to the given probability formula.
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