A number is chosen at random from the set of consecutive natural numbers $\{1, 2, 3, \ldots, 24\}$. What is the probability that the number chosen is a factor of $4!$? Express your answer as a common fraction.
I want to find the fraction of numbers in the set that are factors of $4!$. To do that, I need to know what $4!$ is and what its factors are. I recall that $4!$ means $4 \times 3 \times 2 \times 1$, which is $24$. So $4! = 24$, and I can list its factors as $1, 2, 3, 4, 6, 8, 12, 24$. There are $8$ factors of $4!$ in the set, out of $24$ numbers in total. So the probability of choosing a factor of $4!$ at random is $\frac{8}{24}$. I can simplify this fraction by dividing the numerator and denominator by $8$. The simplified fraction is $\frac{1}{3}$. 