Probability of distribution

Hi, I am having difficulty with this problem.

New Jersy has a "Pick 3" game in which you pick one of the 1000 3-digit numbers between 000 and 999, recieving $275 for a winning $1 bet and nothing otherwise.

A) Construct the probability distribution of the random variable X = winnings.

For this one, I thought it was simply P(000)=P(001)=P(002)...P(997)=P(998)=P(999)=.001

B) Find the mean of the probability distribution.

I understand what probability distribution is but don't understand how to apply it to this problem. I believe you normally take the number of trials multiplied by the probablility of it happening divided by the overall number. In this case, I know you would divide by a 1000 but how do you figure out the first part to it? :?

C) Based on the mean in (b) and the $1 cost to play the game, on average, how much can you expect to lose each time you play this lottery?

I'm not sure I fully understand this question...

Probability of Winnings, not of seleting a certain nnumber.

Lose $1 -- P(-1) = 999/1000
Win $275 - $1 -- P(274) = 1/1000

Thank you! That makes sense now for the first part. I'm still confused though on parts b and c. For part b, it's unclear to me if they are asking about the chances of winning based on the amount of money ($275) that the person won or the average chance of a person winning in general. If it's the chance of a person winning, then would it be 999/1000? And then how can it be attributed to part c?

Calculate the mean by the standard definition.

-1 * (999/1000) + 274 * (1/1000)