Originally posted by Broken8ball
Without giving the answer, let me see if I can help some of you still trying.
Most people are going to answer with 50% (2 in 4 on the first pull) multiplied by 33% (1 in 3 on the second pull). This yields around 16.5%. The problem with this solution is that besides allowing for 2 reds, it also allows for 2 non reds. The problem states that 1 has to be red, you just don't know on which pull you got the red. That option has to be removed from the equation (just not going to tell you how to do it).