Emilio449
Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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 !


We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.