1. **Problem statement:** Find the probability that a randomly selected positive integer from 1 to 100 is divisible by 2 or 5. 2. **Formula and rules:** Use the principle of inclusion-exclusion for probabilities: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ where: - $A$ is the event "divisible by 2" - $B$ is the event "divisible by 5" 3. **Calculate each probability:** - Numbers divisible by 2 from 1 to 100: $\lfloor \frac{100}{2} \rfloor = 50$ - Numbers divisible by 5 from 1 to 100: $\lfloor \frac{100}{5} \rfloor = 20$ - Numbers divisible by both 2 and 5 (i.e., divisible by 10): $\lfloor \frac{100}{10} \rfloor = 10$ 4. **Calculate probabilities:** - $P(A) = \frac{50}{100} = 0.5$ - $P(B) = \frac{20}{100} = 0.2$ - $P(A \cap B) = \frac{10}{100} = 0.1$ 5. **Apply inclusion-exclusion:** $$P(A \cup B) = 0.5 + 0.2 - 0.1 = 0.6$$ 6. **Final answer:** The probability that a randomly selected positive integer from 1 to 100 is divisible by 2 or 5 is **0.6** or **60%**.