Question

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

• A. \( \tfrac{1}{2} \)
• B. \( \tfrac{2}{5} \)
• C. \( \tfrac{8}{15} \)
• D. \( \tfrac{9}{20} \)

Hint:

For “\(A\) or \(B\)” counts on multiples, always use inclusion–exclusion: \(|A\cup B|=|A|+|B|-|A\cap B|\).
If a question intends “either…but not both,” subtract the intersection once more.

Answer 1

The Correct Option is D

Step 1 (Count favourable numbers via inclusion–exclusion).
Multiples of \(3\) in \(1\)–\(20\): \( \lfloor 20/3 \rfloor = 6 \) \( \{3,6,9,12,15,18\} \).
Multiples of \(5\) in \(1\)–\(20\): \( \lfloor 20/5 \rfloor = 4 \) \( \{5,10,15,20\} \).
Common multiples (of \(15\)): \( \lfloor 20/15 \rfloor = 1 \{15\} \).
Step 2 (Apply inclusion–exclusion).
Favourable count \(= 6 + 4 - 1 = 9\).
Step 3 (Total outcomes and probability).
Total tickets \(= 20\).
Probability \(= \dfrac{9}{20} = 0.45\).
Step 4 (Note on options).
Since \( \dfrac{9}{20} \) does not match any given option, the options likely have a typo.
If the intention was “multiple of \(3\) or \(5\) but not both,” then favourable \(= 9-1=8\) and probability \(= \dfrac{8}{20}=\dfrac{2}{5}\) (which matches Option 2).
\[ \boxed{\tfrac{9}{20}\ \text{(not listed among options)}} \]