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?
\(\frac{23}{120}\)
The sample space is the integers \(1\) to \(120\), so there are \(120\) equally likely outcomes.
Count multiples of 3: \(\lfloor120/3\rfloor=40\). Count multiples of 5: \(\lfloor120/5\rfloor=24\). Their overlap is the multiples of \(\text{lcm}(3,5)=15\): \(\lfloor120/15\rfloor=8\).
By inclusion-exclusion, the count divisible by 3 or 5 is \(40+24-8=56\).
Now remove those also divisible by 24. Multiples of 24 up to 120 are \(24,48,72,96,120\), i.e. 5 numbers, and each is a multiple of 3, so all 5 lie inside the 56.
Favourable count \(=56-5=51\), giving probability \(\dfrac{51}{120}\).
The governing rule is inclusion-exclusion for the union, followed by deleting the forbidden multiples of 24.
This working gives \(\dfrac{51}{120}\), while the option marked correct in the official answer key is 23/120.
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