1. **Problem Statement:** We have tickets numbered from 1 to 20. We want to find the probability that a randomly drawn ticket has a number that is a multiple of 3 or 5. 2. **Formula and Rules:** The probability of an event is given by: $$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$$ We use the principle of inclusion-exclusion to count numbers that are multiples of 3 or 5: $$|A \cup B| = |A| + |B| - |A \cap B|$$ where: - $A$ is the set of multiples of 3 - $B$ is the set of multiples of 5 3. **Step-by-step Calculation:** - Total tickets: 20 - Multiples of 3 between 1 and 20: 3, 6, 9, 12, 15, 18 (6 numbers) - Multiples of 5 between 1 and 20: 5, 10, 15, 20 (4 numbers) - Multiples of both 3 and 5 (i.e., multiples of 15): 15 (1 number) Using inclusion-exclusion: $$|A \cup B| = 6 + 4 - 1 = 9$$ 4. **Calculate Probability:** $$\text{Probability} = \frac{9}{20}$$ 5. **Answer:** The probability that the ticket drawn has a number which is a multiple of 3 or 5 is **$\frac{9}{20}$**.