A number is chosen at random from the set 1, 2, 3, . . . , 120. What is the probability that the number chosen is divisible by 6…

Question

A number is chosen at random from the set 1, 2, 3, . . . , 120. What is the probability that the number chosen is divisible by 6 or 8 but not divisible by 24?

1. A. 23/120
2. B. 25/120 (Correct answer)
3. C. 21/120
4. D. 19/120

Correct Answer

Option B — 25/120

Detailed Solution & Explanation

The correct answer is 25/120.

Key Points

• Count each set within 1 to 120:
  • divisible by 6: 120/6 = 20
  • divisible by 8: 120/8 = 15
  • divisible by both, i.e. by LCM(6, 8) = 24: 120/24 = 5
• By inclusion-exclusion, divisible by 6 or 8 = 20 + 15 − 5 = 30.
• The question excludes multiples of 24, and every one of those 5 is inside that 30, so favourable = 30 − 5 = 25.
• Probability = 25/120.

Additional Information

• The overlap of "multiples of 6" and "multiples of 8" is the multiples of their LCM, not their product — 24, not 48. That distinction is the crux here.
• Inclusion-exclusion: n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
• Note the phrasing "or … but not …", which asks for the symmetric part of the union: everything in A ∪ B minus the intersection.
• The answer is deliberately left unsimplified as 25/120 to match the options; in lowest terms it is 5/24.