Number from 1 to 30 divisible by none of 2, 3, 5

Question

An integer is chosen uniformly at random from 1 to 30. What is the probability that it is divisible by none of 2, 3, and 5?

Accepted Answer

First count how many numbers are divisible by at least one of 2, 3, 5, using inclusion-exclusion, then take the complement. Among 1 to 30: - singles: multiples of 2 give 15, of 3 give 10, of 5 give 6, - pairs: multiples of 6 give 5, of 10 give 3, of 15 give 2, - triple: multiples of 30 give 1. Add singles, subtract pairs, add the triple: \(\text{at least one}) = (15 + 10 + 6) - (5 + 3 + 2) + 1 = 31 - 10 + 1 = 22. So 30 - 22 = 8 numbers avoid all three factors. The probability is \frac{8}{30} = \frac{4}{15} \approx 0.2667. Those 8 numbers are 1, 7, 11, 13, 17, 19, 23, 29, exactly the ones shar…

Number from 1 to 30 divisible by none of 2, 3, 5

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