# Combination Probability

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? • 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? – shardulc says Reinstate Monica Mar 14 '16 at 11:52 • 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? – N. F. Taussig Mar 14 '16 at 12:03 • 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 ? – true blue anil Mar 14 '16 at 12:08 • 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. – labyrinthdeux Mar 14 '16 at 12:30 ## 1 Answer What is the probability that her friends will win exactly$\$35$?

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 $$\frac{\dbinom{4}{3}}{\dbinom{16}{3}}$$