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)?
9/39
5/64
3/80
1/12
my answer: 3/80
your answer is incorrect
the correct answer is D
are you sureeee
yes I am sure
If I wasn't I wouldn't intentionally give you false information
yes
p(1) = 4/16 = 1/4
p(10) = 5/15 = 1/3
1/4 * 1/3 = 1/12
To find the probability of selecting a $1 bill followed by a $10 bill without replacement, we need to determine the total number of possible outcomes and the number of favorable outcomes.
First, let's find the total number of bills. We have:
4 + 2 + 5 + 5 = 16 bills in total.
Now, let's determine the number of favorable outcomes, which is selecting a $1 bill followed by a $10 bill.
For the first selection, the probability of selecting a $1 bill is 4 out of 16 bills. After selecting a $1 bill, we are left with 3 $1 bills in our wallet.
For the second selection, the probability of selecting a $10 bill is 5 out of the remaining 15 bills.
So the probability of selecting a $1 bill followed by a $10 bill is:
(4/16) * (5/15) = 20/240 = 1/12
Therefore, the correct answer is 1/12, not 3/80.