What is the probability that a randomly drawn positive factor of $60$ is less than $7$?
$\mathrm{(A) \ } \frac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{4}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2}$
Solution 1
For a positive number $n$ which is not a perfect square, exactly half of the positive factors will be less than $\sqrt{n}$. 
Since $60$ is not a perfect square, half of the positive factors of $60$ will be less than $\sqrt{60}\approx 7.746$. 
Clearly, there are no positive factors of $60$ between $7$ and $\sqrt{60}$. 
Therefore half of the positive factors will be less than $7$. 
So the answer is $\boxed{\mathrm{(E)}\ \frac{1}{2}}$.

Solution 2
Testing all numbers less than $7$, numbers $1, 2, 3, 4, 5$, and $6$ divide $60$. The prime factorization of $60$ is $2^2\cdot 3 \cdot 5$. Using the formula for the number of divisors, the total number of divisors of $60$ is $(3)(2)(2) = 12$. Therefore, our desired probability is $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$

Solution 3
This is not too bad with casework. Notice that $1*60=2*30=3*20=4*15=5*12=6*10=60$. Hence, $60$ has $12$ factors, of which $6$ are less than $7$. Thus, the answer is $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$.
Solution by franzliszt

Solution 4
This following solution isn't recommended,  but just list out all the divisors of $60$ to find $12$ divisors.  $6$ of these are less than $7,$ so the answer is $\frac{6}{12} = \boxed{\mathrm{(E)}\ \frac{1}{2}}$. ~Sophia866