Four slips of paper labeled A, B, C and D are drawn from a hat at random, one at a time, without replacement. What is the probability of drawing C first and D last? Express your answer as a common fraction.
There are 4!=24 ways to draw the slips out of the hat.
There are only 2 ways to draw C first and D last:
CABD
CBAD
because once you decide when you draw C and D, the only two papers you have to worry about ordering are A and B in the middle and there are only 2 ways to do that.
So the probability is \(\frac{2}{24}=\boxed{\frac{1}{12}}\)
There are 4!=24 ways to draw the slips out of the hat.
There are only 2 ways to draw C first and D last:
CABD
CBAD
because once you decide when you draw C and D, the only two papers you have to worry about ordering are A and B in the middle and there are only 2 ways to do that.
So the probability is \(\frac{2}{24}=\boxed{\frac{1}{12}}\)