There are a total of 6 + 8 + 2 + 4 = 20 bills in the wallet.
The probability of selecting a $1 bill first is 6/20 or 3/10.
After selecting a $1 bill, there are now 19 bills left in the wallet, including two $10 bills. Therefore, the probability of selecting a $10 bill second, without replacing the first bill, is 2/19.
The probability of selecting a $1 bill first and a $10 bill second is the product of the probabilities: (3/10) x (2/19) = 6/190 or 3/95.
Therefore, the probability of selecting a $1 bill first and a $10 bill second is 3/95.
You have six $1 bill’s eight $5 bills two $10 bills and four $20 bills in your wallet. You select a bill at random. Without replacing the bill you chose a second bill. What is p($1, then $10)?
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