Ilike really need ! 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

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 !

6inches by 8 inches