Probability a number from 1 to 30 is divisible by 2, 3, or 5

Question

An integer is chosen uniformly at random from \{1, 2, \dots, 30\}. What is the probability it is divisible by 2, 3, or 5?

Accepted Answer

With three overlapping sets you must alternate signs: add the singles, subtract the pairwise overlaps, add back the triple overlap. Count multiples up to 30: - Singles: 2 = 15, 3 = 10, 5 = 6. - Pairs (divisible by the product): 6 = 5, 10 = 3, 15 = 2. - Triple: 30 = 1. A \cup B \cup C = (15 + 10 + 6) - (5 + 3 + 2) + 1 = 31 - 10 + 1 = 22. So 22 of the 30 numbers qualify, and P = \frac{22}{30} = \frac{11}{15} \approx 0.733. The only numbers left out are those coprime to 2, 3, 5: namely \{1, 7, 11, 13, 17, 19, 23, 29\}, exactly 8, and 30 - 8 = 22 confirms the count. Takeaway. For three or more ev…

Probability a number from 1 to 30 is divisible by 2, 3, or 5

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