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

Introduction / Context:
This question uses basic probability with divisibility rules for integers. The tickets are numbered from 1 to 20, and we want the probability that a randomly selected ticket has a number divisible by 3 or divisible by 5. The main task is to count how many numbers meet this condition and compare that count with the total number of tickets.

Given Data / Assumptions:

• Tickets are numbered 1, 2, 3, up to 20.
• One ticket is drawn at random, and each ticket has equal chance of selection.
• Favourable tickets are those with numbers that are multiples of 3 or multiples of 5.

Concept / Approach:
We apply the union principle for counting. First count the numbers that are multiples of 3, then count those that are multiples of 5, and finally subtract the numbers that are multiples of both 3 and 5, because those have been counted twice. The probability is the ratio of this union count to the total number of tickets, which is 20.

Step-by-Step Solution:
Total tickets = 20. Multiples of 3 between 1 and 20: 3, 6, 9, 12, 15, 18, which is 6 numbers. Multiples of 5 between 1 and 20: 5, 10, 15, 20, which is 4 numbers. Multiples of both 3 and 5 are multiples of 15: 15 only, which is 1 number. Total favourable numbers = 6 + 4 - 1 = 9. Required probability = 9 / 20.

Verification / Alternative check:
List all numbers from 1 to 20 and mark those divisible by 3 or 5. The marked numbers are 3, 5, 6, 9, 10, 12, 15, 18 and 20. Counting them gives 9 favourable tickets. Since there are 20 tickets in all, the probability is 9 / 20. The fraction is already in simplest form, so this confirms our earlier counting and ensures there is no arithmetic mistake.

Why Other Options Are Wrong:
1/2 would require 10 favourable numbers out of 20, but only 9 numbers meet the condition. 3/5 corresponds to 12 favourable numbers, which is more than we actually have. 2/5 equals 8 favourable numbers, which is slightly less than the correct count of 9.

Common Pitfalls:
Some learners forget to subtract the overlap of numbers that are multiples of both 3 and 5, leading to 10 instead of 9 favourable numbers. Others may miscount by skipping a multiple or mistakenly including a non multiple. When the range is small, writing out all numbers and checking divisibility carefully is a very effective strategy to avoid these errors.

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