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,putting the first bill back in the hat before choosing the second bill.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)$3.56
C)$4.32
D)$8.64

Question

Quizplus · Accepted Answer

The expected value of the game can be calculated by finding the probability of drawing two bills of the same denomination and multiplying those probabilities by the value of the bills you get to keep. - For the $10 bills, the probability of drawing two in a row is $\frac{5}{22} \times \frac{5}{22}$ because there are 5 $10 bills out of 22 total bills, and you replace the first bill before drawing the second. The value of keeping both bills is $20, so the expected value from $10 bills is $\frac{5}{22} \times \frac{5}{22} \times 20$.- For the $5 bills, the probability of drawing two in a row is $\frac{7}{22} \times \frac{7}{22}$, with a value of $10 for keeping both, so the expected value from $5 bills is $\frac{7}{22} \times \frac{7}{22} \times 10$.- For the $1 bills, the probability of drawing two in a row is $\frac{10}{22} \times \frac{10}{22}$, with a value of $2 for keeping both, so the expected value from $1 bills is $\frac{10}{22} \times \frac{10}{22} \times 2$.Adding these expected values together gives the total expected value of the game:$$\left(\frac{5}{22} \times \frac{5}{22} \times 20\right) + \left(\frac{7}{22} \times \frac{7}{22} \times 10\right) + \left(\frac{10}{22} \times \frac{10}{22} \times 2\right) = \frac{25}{484} \times 20 + \frac{49}{484} \times 10 + \frac{100}{484} \times 2 = \frac{500 + 490 + 200}{484} = \frac{1190}{484} = 2.46$$Therefore, to make the game fair, you should pay $2.46 to participate.

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