You have six $1 bills, eight $5 bills, two $10 bills, and four$20 dollar 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. 77/190
B. 3/100
C.3/95
D.2/5

see Reiny's solution at

http://www.jiskha.com/display.cgi?id=1394074310

To find the probability of selecting a $1 bill and then a $10 bill without replacement, we need to calculate the probability of each step.

Step 1: Selecting a $1 bill
There are a total of 6 $1 bills out of a total of 20 bills in the wallet.
So, the probability of selecting a $1 bill on the first attempt is 6/20.

Step 2: Selecting a $10 bill
After selecting a $1 bill, we have 19 bills left in the wallet.
Out of these, there are 2 $10 bills.
So, the probability of selecting a $10 bill on the second attempt is 2/19.

Since we are choosing without replacement, we multiply the probabilities of each step to get the overall probability.

P($1, then $10) = (6/20) * (2/19)
P($1, then $10) = 12/380
Simplifying the fraction, we get:
P($1, then $10) = 6/190

Therefore, the correct answer is:
A. 6/190