You have four $1 bills, two $5 bills, five $10 bills, and five $20 bills in your wallet. You select a bill at random. Without replacing the bill, you choose a second bill. What is P($1, then $10)?
A. start fraction 9 over 39 end fraction
B. The fraction is 5 over 64.
C. start fraction 3 over 80 end fraction
D. start fraction 1 over 12 end fraction
There are a total of 16 bills in the wallet, so there are 16 choices for the first bill selected.
If the first bill selected is a $1 bill, there are 15 bills left, and 5 of them are $10 bills. Therefore, the probability of selecting a $1 bill followed by a $10 bill is $\dfrac{4}{16} \cdot \dfrac{5}{15} = \dfrac{1}{4} \cdot \dfrac{1}{3} = \dfrac{1}{12}$.
If the first bill selected is a $10 bill, there are only 4 other $10 bills left, so the probability of selecting a $10 bill followed by another $10 bill is $\dfrac{5}{16} \cdot \dfrac{4}{15} = \dfrac{1}{8}$.
For all other first bills selected, there are 5 $10 bills left and a total of 15 bills left, so the probability of selecting a $10 bill second is $\dfrac{10}{16} \cdot \dfrac{5}{15} = \dfrac{1}{6}$.
Therefore, the overall probability of selecting a $1 bill followed by a $10 bill is the sum of the probabilities of the two possible outcomes: $\dfrac{1}{12} + \dfrac{1}{6} = \dfrac{1}{8}$.
The closest answer choice is $\boxed{\textbf{(D) }\frac{1}{12}}$.