Tickets are numbered from 1 to 18 and mixed together. One ticket is drawn at random. What is the probability that the number on the drawn ticket is a multiple of 2 or a multiple of 3?

Introduction / Context:
This question uses basic probability with divisibility conditions on integers. The tickets are numbered from 1 to 18, and we want the probability that a randomly selected ticket shows a number that is divisible by 2 or divisible by 3. The task is to count such numbers and divide by the total number of tickets.

Given Data / Assumptions:

• Tickets are numbered 1, 2, 3, up to 18.
• One ticket is drawn at random from these 18 tickets.
• Each ticket has equal chance of being selected.
• The favourable numbers are those that are multiples of 2 or multiples of 3.

Concept / Approach:
Use the union principle for counting. Count all multiples of 2, count all multiples of 3, and then subtract the numbers that are multiples of both 2 and 3, because those numbers have been counted twice. Once the count of favourable numbers is known, probability equals favourable count divided by total count, which here is 18.

Step-by-Step Solution:
Total tickets = 18. Multiples of 2 from 1 to 18: 2, 4, 6, 8, 10, 12, 14, 16, 18, which is 9 numbers. Multiples of 3 from 1 to 18: 3, 6, 9, 12, 15, 18, which is 6 numbers. Multiples of both 2 and 3 are multiples of 6: 6, 12, 18, which is 3 numbers. Total favourable numbers = 9 + 6 - 3 = 12. Required probability = 12 / 18. Simplify 12 / 18 by dividing numerator and denominator by 6 to get 2 / 3.

Verification / Alternative check:
We can check by writing all numbers from 1 to 18 and marking those that are multiples of 2 or 3. The marked numbers will be 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18, which is exactly 12 numbers. Since there are 18 numbers in total, the probability is 12 / 18, which simplifies to 2 / 3, confirming the result.

Why Other Options Are Wrong:
1/3 would correspond to only 6 favourable numbers, which is too small. 3/5 equals 10 out of 18, but we found 12 favourable numbers. 5/6 would require 15 favourable numbers, which is more than actually satisfy the condition.

Common Pitfalls:
Learners sometimes forget to subtract the overlap, that is the multiples of 6, and obtain 15 instead of 12 favourable numbers. Others may miss one or two numbers when listing multiples by hand. Using the inclusion exclusion idea and checking the list carefully helps avoid such counting errors.

Final Answer:
The probability that the number on the ticket is a multiple of 2 or 3 is 2/3.