The tickets numbered from 1 to 20 are mixed up and then a ticket is drawn at…
2025
The tickets numbered from 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket has a number which is a multiple of 3 or 5?
- A.
9/20
- B.
9/24
- C.
9/27
- D.
9/30
Show answer & explanation
Correct answer: A
Concept: For equally likely outcomes, probability P(event) = (number of favourable outcomes) / (total number of outcomes). When the event is "a multiple of 3 or a multiple of 5", use the inclusion-exclusion rule: n(A or B) = n(A) + n(B) - n(A and B), so outcomes common to both conditions are not counted twice.
Sample space: the tickets are numbered 1 to 20, so the total number of outcomes n(S) = 20.
Multiples of 3 up to 20: 3, 6, 9, 12, 15, 18 -- so n(A) = 6.
Multiples of 5 up to 20: 5, 10, 15, 20 -- so n(B) = 4.
Multiples of both 3 and 5 (i.e. multiples of 15) up to 20: 15 -- so n(A and B) = 1.
Favourable outcomes by inclusion-exclusion: n(A or B) = n(A) + n(B) - n(A and B) = 6 + 4 - 1 = 9.
Probability: P(multiple of 3 or 5) = n(A or B) / n(S) = 9/20.
Cross-check: listing every favourable ticket directly -- 3, 5, 6, 9, 10, 12, 15, 18, 20 -- gives exactly 9 distinct numbers out of 20, confirming the inclusion-exclusion result.