Tickets numbered from 1 to 20 are mixed thoroughly and then one ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or a multiple of 5?

Introduction / Context:
This question tests basic probability with simple counting and the idea of numbers that are multiples of 3 or 5. It also illustrates how to handle overlap when a number satisfies both conditions at the same time, using the addition principle for events that are not mutually exclusive.

Given Data / Assumptions:

• Tickets are numbered from 1 to 20.
• Exactly one ticket is drawn at random.
• Each ticket is equally likely to be drawn.
• We are interested in numbers that are multiples of 3 or multiples of 5.

Concept / Approach:
We use classical probability: P(event) = favourable outcomes / total outcomes. Here the event is that the ticket number is a multiple of 3 or 5. We count multiples of 3, multiples of 5, and then subtract the overlap (multiples of both 3 and 5, that is multiples of 15) to avoid double counting.

Step-by-Step Solution:
Total tickets = 20, so total outcomes = 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). Number 15 is common to both sets, so it is counted twice above. Favourable outcomes = 6 + 4 - 1 = 9 (inclusion exclusion principle). Probability = favourable / total = 9 / 20.

Verification / Alternative check:
We can also list all numbers from 1 to 20 and highlight those divisible by 3 or 5. Counting them directly again gives 9 numbers. Since 9 and 20 have no common factor other than 1, the fraction 9/20 is already in simplest form, which confirms our result.

Why Other Options Are Wrong:
1/2: This would require 10 favourable numbers, but there are only 9. 3/5: This equals 12/20, which is larger than 9/20 and overestimates the probability. 2/5: This equals 8/20, which is slightly less than the correct count of 9 favourable numbers.

Common Pitfalls:
Students often forget to subtract the overlap when using the idea of multiples of 3 or 5. Counting 6 multiples of 3 and 4 multiples of 5 and simply adding them gives 10, which is wrong because 15 is then counted twice. Always check for numbers that satisfy both conditions when using the addition principle for probability problems.

Final Answer:
Therefore, the required probability that the drawn ticket shows a number that is a multiple of 3 or 5 is 9/20.