Question

I like really need help! you have six $1 bills, two $10 bills, and four $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)77/190 (B)3/100 (C)3/95 (D)2/5

Answer 1

Before you do anything to it, there are 12 bills in your wallet.
Six of them are singles, so the probability of selecting a single
on the first draw is
6 / 12 .

Now there are 11 bills in the wallet, and two of them are tens,
so the probability of selecting one of those is
2 / 11 .

The probability of both events unfolding just like that is

( 6 / 12 ) x ( 2 / 11 ) = 12/132 = 1/11 = 9.09 % (rounded)

Sadly, I don't find this solution listed among the given choices.
Still, my confidence in the methods I've described is unshakable.
1/11 or 9.09% is my answer and I'm stickin to it !