Step 1 :For part a, since each digit can be any number from 0 to 9 inclusive, there are 10 possibilities for each digit. Since there are three digits, the total number of different selections is \(10^3\).
Step 2 :For part b, the probability of winning is the number of winning outcomes divided by the total number of outcomes. Since there is only one winning outcome (the three numbers you selected), the probability of winning is \(\frac{1}{10^3}\).
Step 3 :For part c, if you win, you collect $400. Since you paid $4 to play, your net profit is $400 - $4 = $396.
Step 4 :For part d, the expected value is the sum of the possible outcomes multiplied by their probabilities. In this case, there are two possible outcomes: winning $396 with probability \(\frac{1}{10^3}\), and losing $4 with probability \(1 - \frac{1}{10^3}\).
Step 5 :Final Answer: a. The total number of different selections possible is \(\boxed{1000}\). b. The probability of winning is \(\boxed{0.001}\). c. The net profit if you win is \(\boxed{396}\). d. The expected value for a $4 bet is \(\boxed{-3.6}\).