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 !