Assume that it costs $\$ 5$ to play a state's daily number. The player chooses a three-digit number between 000 and 999 , inclusive, and if the number is selected that day, then the player wins $\$ 310$ (this means the player's profit is $\$ 310-\$ 5=\$ 305$ ). If the player buys 9 chances on 9 different numbers, what is the expected value for the player?

Step 1 ：Assume that it costs $5 to play a state's daily number. The player chooses a three-digit number between 000 and 999, inclusive, and if the number is selected that day, then the player wins $310. This means the player's profit is $310 - $5 = $305. If the player buys 9 chances on 9 different numbers, what is the expected value for the player?

Step 2 ：The expected value is the sum of the products of the possible outcomes and their respective probabilities. In this case, there are two possible outcomes: winning $305 or losing $5.

Step 3 ：The probability of winning is \(\frac{9}{1000}\) (since the player buys 9 chances out of 1000 possible numbers) and the probability of losing is \(\frac{991}{1000}\) (since there are 991 numbers that the player did not choose).

Step 4 ：Therefore, the expected value can be calculated as follows: Expected Value = \(305 \times \frac{9}{1000}\) - \(5 \times \frac{991}{1000}\)

Step 5 ：The expected value is approximately -2.21.

Step 6 ：Final Answer: The expected value for the player is \(\boxed{-2.21}\). This means that on average, the player will lose about $2.21 each time they play.