A large hat contains five $10 bills,seven $5 bills and ten $1 bills.In a game,you are allowed to reach in and pull two bills at random from the hat.If both bills are of the same denomination,you get to keep both of them.In order to make this a fair game,how much should you pay to participate in this game?
A)$2.46
B)$2.16
C)$3.56
D)$5.08

Question

Quizplus · Accepted Answer

The total value in the hat is:
$$5(10)+7(5)+10(1)= 85$$
The probability of drawing two bills of the same denomination is:
$$\frac{5}{22}\cdot\frac{4}{21}+\frac{7}{22}\cdot\frac{6}{21}+\frac{10}{22}\cdot\frac{9}{21}=\frac{127}{462}$$
The expected value of winning is then:
$$10\cdot\frac{5}{22}\cdot\frac{4}{21}+5\cdot\frac{7}{22}\cdot\frac{6}{21}+1\cdot\frac{10}{22}\cdot\frac{9}{21}=\frac{65}{231}$$
Therefore, in order for the game to be fair, the cost to participate must equal the expected value of winning. So, the answer is:
$$\frac{65}{231}=\boxed{\mathrm{(B)\ } 2.16}$$

A Large Hat Contains Five $10 Bills,seven $5 Bills and Ten