You have seven $1 bills, three $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)?
There are a total of 20 bills in the wallet, so the probability of selecting a $1 bill first is 7/20. Since we do not replace the selected bill, there are now 19 bills in the wallet, of which 5 are $10 bills. Therefore, the probability of selecting a $10 bill second, given that a $1 bill was selected first, is 5/19.
To find the probability of selecting a $1 bill first and then a $10 bill, we multiply the probabilities:
P($1, then $10) = (7/20) x (5/19)
P($1, then $10) = 35/380
P($1, then $10) = 0.0921
Therefore, the probability of selecting a $1 bill first and then a $10 bill is approximately 0.0921 or 9.21%.
To find the probability of selecting a $1 bill and then a $10 bill without replacement, we need to calculate the probability of selecting a $1 bill first and then multiply it by the probability of selecting a $10 bill given that a $1 bill has already been selected.
Step 1: Calculate the probability of selecting a $1 bill first.
There are a total of 7 + 3 + 5 + 5 = 20 bills in the wallet.
Since there are 7 $1 bills, the probability of selecting a $1 bill first is equal to the number of $1 bills divided by the total number of bills:
P($1) = 7/20
Step 2: Calculate the probability of selecting a $10 bill given that a $1 bill has already been selected.
After selecting a $1 bill, there are 6 $1 bills remaining and a total of 19 bills left in the wallet.
Since there are 5 $10 bills, the probability of selecting a $10 bill given that a $1 bill has already been selected is equal to the number of $10 bills divided by the total number of bills remaining:
P($10 | $1) = 5/19
Step 3: Multiply the probabilities from step 1 and step 2.
P($1, then $10) = P($1) * P($10 | $1)
= (7/20) * (5/19)
= 35/380
= 7/76
Therefore, the probability of selecting a $1 bill and then a $10 bill without replacement is 7/76.