A number is chosen at random among the first 120 natural numbers. The…

Concept: When all outcomes are equally likely, probability = (number of favourable outcomes) / (total number of outcomes). When two conditions are joined by "or", the favourable set is the union of the two sets — and if one condition’s set is entirely contained within the other’s, that union simply equals the larger set.

1. The first 120 natural numbers form the sample space, so the total number of outcomes is 120.

2. Multiples of 5 up to 120 are 5, 10, 15, 20, ..., 120 — an arithmetic sequence with common difference 5, giving 120/5 = 24 such numbers.

3. Since 15 = 3 × 5, every multiple of 15 is automatically a multiple of 5. So the set of numbers that are "a multiple of 5 or 15" is exactly the set of multiples of 5 — the "15" condition contributes no additional numbers.

4. Favourable outcomes = 24, so the probability = 24/120.

5. Reducing 24/120 by their HCF (24) gives 1/5.

Cross-check: By the inclusion–exclusion rule, |A ∪ B| = |A| + |B| − |A ∩ B|, where A is the set of multiples of 5 (24 numbers) and B is the set of multiples of 15 (120/15 = 8 numbers). Because every element of B already lies in A, A ∩ B = B, so |A ∪ B| = 24 + 8 − 8 = 24 — the same favourable count, confirming the probability is 1/5.