One of the positive factors of 60 is chosen at random. What is the probability that it is an even number?

Since $60=2^2\times 3\times 5$, it has $(2+1)(1+1)(1+1)=12$ factors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. The odd ones are exactly the factors of 15 — namely 1, 3, 5 and 15 — so 8 factors are even and the probability is 8/12 = 2/3. Answering 1/2 assumes even and odd factors must balance, which is false for any number with a factor of 2; 1/3 is the probability of an ODD factor; and 5/6 = 10/12 comes from spotting only 1 and 15 as odd and overlooking 3 and 5.