Two different numbers are chosen at random from the set {1, 2, 3, ..., 5}. What is the probability that the product of the numbers is even?
The only way to not get an even product is to have two odd numbers. There are \({5 \choose 2} = \frac{5!}{2!(5-2)!} = 10\) possible pairs you can choose, and we have three pairs (out of the ten) of odd numbers: (1,3), (1,5), and (3,5). Therefore \(\frac{10}{10}-\frac{3}{10} = \boxed{\frac{7}{10}}\).