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During lunch, Charlotte and her friends decided to play a game while waiting for their order. She told her friends that one of them will get a chance to win an amount of money based on the three randomly selected bills to be taken from her allowance. Charlotte has two ten-dollar bills, four five-dollar bills, and ten one-dollar bills. What is the probability that one of her friends will:

  1. win exactly $35$ dollars?
  2. select two ten-dollar bills?
  3. win exactly $15$ dollars? *She has exactly 7 friends I'm stuck on solving this one.

Since there are only three trials: _ _ _ the maximum amount that his friend could get is only \$25 (2 \$10 and 1 \$5), the probability of part 1 is $0\%$.

On part 2, it seems that the order is not important so, I used combination. (nCr) There are a total of 16 dollar bills. Thus, $$\frac{\binom{16}{2}}{\binom{16}{3}} \approx 0.21$$

On part 3, there is also $0\%$ probability of winning exactly $15$ dollars.

I am quite hesitant of my answers. Can someone please check out if what am I doing is correct?

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  • $\begingroup$ Could you tell us more about how the game is played? Do Charlotte's friends say that such-and-such a bill will be drawn, or something else? $\endgroup$ – shardulc says Reinstate Monica Mar 14 '16 at 11:52
  • $\begingroup$ Do you mean that each of her seven friends makes a selection? Are we interested in the probability that at least one of them will select two ten dollar bills? $\endgroup$ – N. F. Taussig Mar 14 '16 at 12:03
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    $\begingroup$ What is the role of exactly 7 friends ? Do they draw turn by turn or what ? Since there are a total of 16 bills, more than one (but not all) can draw. How is it decided who gets the prize ? The question seems very vague. Have you made it up ? $\endgroup$ – true blue anil Mar 14 '16 at 12:08
  • $\begingroup$ Edit: The role of "exactly 7 friends" is from another problem that used the same universe. I thought at first that it will be helpful. Then I realized that it would not be. So, I removed it. This is from our work text. The problem itself is vague so I tried to answer it as simple as possible. $\endgroup$ – labyrinthdeux Mar 14 '16 at 12:30
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What is the probability that her friends will win exactly $\$ 35$?

Your answer is correct.

What is the probability that her friends will select two ten dollar bills?

Assuming that her friends cannot see the bills that are being selected (so that the selection of any particular bill is equally likely), the probability that a particular friend will select two ten dollar bills is $$\frac{\dbinom{2}{2}\dbinom{14}{1}}{\dbinom{16}{3}}$$ since the friend must select both ten dollar bills and one of the other $14$ available bills when selecting three of the $16$ bills.

What is the probability that her friends will win exactly $\$ 15$.

To win $\$15$, her friends have to select three of the four five dollar bills. Thus, under the same assumptions as above, the probability that a particular friend will select exactly $\$15$ is $$\frac{\dbinom{4}{3}}{\dbinom{16}{3}}$$

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