lottery probability question

In a certain state lottery, a bin contains 40 white balls numbered 1-40, and a separate bin contains 24 red balls numbered 1-24. In the morning, you buy a card for $1, and mark that card with 3 "white numbers" (between 1 and 40) and 1 "red number" (between 1 and 24). Your red number can be the same as one of your white numbers if you want. In the evening, 3 white balls are drawn at random from the first bin (without replacement), and one red ball is drawn at random from the second bin. There are a variety of possible payoffs:

Payoff #1: If your white numbers match all 3 white balls, and your red number matches the red ball, you win $50,000

Payoff #2: If your white numbers match all 3 white balls, but your red number does not match the red ball, you win $2,000.

Payoff #3: If exactly 2 of your white numbers match numbers on the white balls and your red number matches the red ball, you win $500.

Payoff #4: If exactly 2 of your white numbers match a number on the white balls and your red number matches the red ball you win $10.

Payoff #5: If exactly 1 of your white numbers match any of the white balls and your red number matches the red ball, you win $3.

What is your EXPECTED PROFIT or LOSS if you play this lottery?

HOW CAN I APPROACH THIS CORRECTLY?

. . .>> HOW TO I APPROACH THIS PROBLEM CORRECTLY?

The expected profit or loss is computed by finding the probability of each option (with the sum of the various options' probabilities equalling 1, of course), and multiplying each probability by its payout (or deduction).

. . .* Google results for 'expected value probability'
. . .* http://www.google.com/search?hl=en&q=expected+value+probability

For instance, suppose a game costs $2 to play, and has two outcomes: you win and get $4 (for a net value of $2), or you lose and get nothing (for a net value of -$2). Suppose there is a two-thirds chance of losing. Then the expected value of one round of the game is:

. . .(2/3)(-$2) + (1/3)($2) = -$(4/3) + $(2/3) = -$(2/3)

In other words, this would not be a great game to play. On the other hand, suppose winning earned you $9, with the same probabilities. Then:

. . .(2/3)(-$2) + (1/3)($7) = -$(4/3) + $(7/3) = $(3/3) = $1

This game might be worth playing. *smile*

. . .>> Do i use combinatiions?

The order in which you listed your numbers should not matter so, yes, combinations should be fine.