Tickets numbered 1 to 20 are mixed and one ticket is drawn at random. What is the probability that the number on the ticket is a multiple of 3 or 7?

Introduction / Context:
We need the probability that the chosen number is in the union of two sets: multiples of 3 and multiples of 7. Inclusion–exclusion prevents double counting when overlaps exist.

Given Data / Assumptions:

• Range: 1..20 inclusive.
• Multiples of 3: ⌊20/3⌋ = 6.
• Multiples of 7: ⌊20/7⌋ = 2.
• LCM(3,7) = 21 → overlap count in 1..20 is 0.

Concept / Approach:
Count = count(3) + count(7) − count(21). Probability = count / 20.

Step-by-Step Solution:
Count(3) = 6 (3,6,9,12,15,18).Count(7) = 2 (7,14).Overlap = 0; total favourable = 6 + 2 = 8.Probability = 8/20 = 2/5.

Verification / Alternative check:
Listing confirms 8 valid numbers; fraction reduces to 2/5.

Why Other Options Are Wrong:
1/2 would require 10 favourable numbers; 7/20 equals 7/20 = 0.35 (undercount).

Common Pitfalls:
Forgetting to apply inclusion–exclusion when overlaps are possible; here the overlap happens at 21 which is outside the range, so it is 0.

Final Answer:
2/5