a) The probability distribution is:
[tex]P(X = -1) = \frac{8999}{9000}[/tex]
[tex]P(X = 2699) = \frac{1}{9000}[/tex]
b) The expected winnings are of -$0.7.
c) The interpretation is that for each game you play, you expect to lost $0.7.
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Item a:
Thus, the probability of you choosing the correct number, and thus earning $2699(discount of the price of $1), is:
[tex]P(X = 2699) = \frac{1}{9000}[/tex]
An the probability of you choosing the wrong number, and thus paying $1, is:
[tex]P(X = -1) = \frac{8999}{9000}[/tex]
Thus, the distribution is:
[tex]P(X = -1) = \frac{8999}{9000}[/tex]
[tex]P(X = 2699) = \frac{1}{9000}[/tex]
Item b:
[tex]E = -\frac{8999}{9000} + 2700\frac{1}{9000} = \frac{-8999 + 2700}{9000} = -0.7[/tex]
The expected winnings are of -$0.7.
Item c:
The interpretation is that for each game you play, you expect to lost $0.7.
A similar problem is given at https://brainly.com/question/24855677