Emilio449
Answered

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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

Sagot :

AL2006

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 !


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