Question:

Three numbers are chosen from 1 to 30. Find the probability that they are NOT 3 consecutive numbers.

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Consecutive triplets are of the form \( x, x+1, x+2 \). Count them carefully.
Updated On: Jun 4, 2025
  • \( \frac{1}{145} \)
  • \( \frac{142}{145} \)
  • \( \frac{143}{145} \)
  • \( \frac{144}{145} \)
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The Correct Option is D

Solution and Explanation

Total ways to choose 3 distinct numbers: \( {}^{30}C_3 = 4060 \) Favorable cases to avoid: 3 consecutive numbers: Possible such triplets: (1,2,3), (2,3,4), ..., (28,29,30) → total 28 \[ \Rightarrow \text{Required } = \frac{4060 - 28}{4060} = \frac{4032}{4060} = \frac{144}{145} \]
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