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An online gambling site offers a first prize of \( \$ 50,000 \) and two second prizes of \( \$ 10,000 \) each for registered users when they place a bet. A random bet will be selected over a 24 -hour period. Three million bets are received in the contest. Find the expected winnings if you can place one registered bet of \( \$ 3 \) in the given period. Let \( x \) be the amount of the net winnings. What are the possible values of \( x \) ? \( \square \) (Use a comma to separate answers as needed. Do not use commas in the individual values. Do not include the \( \$ \) symbol in your answer.) The expected payback for a person who places one registered bet is \( \$ \square \). (Round to two decimal places as needed.)

Ask by Burgess Wyatt. in the United States
Mar 09,2025

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Answer

Possible values of \( x \): 49997, 9997, -3 Expected payback: -2.98

Solution

Let \( x \) be your net winnings (the prize minus the \$3 bet). There are three cases: 1. If you win the first prize: The prize is \$50,000 so the net winning is \[ x = 50000 - 3 = 49997. \] 2. If you win a second prize: The prize is \$10,000 so the net winning is \[ x = 10000 - 3 = 9997. \] 3. If you do not win a prize: You receive \$0 and lose your \$3 bet, so \[ x = 0 - 3 = -3. \] Thus, the possible values of \( x \) are: \[ 49997, 9997, -3. \] Now, given that there are 3,000,000 bets in total, the probabilities are: - First prize: \( \frac{1}{3000000} \). - Each second prize: \( \frac{1}{3000000} \) (and there are 2 of these, so a combined probability of \( \frac{2}{3000000} \)). - No prize: \( 1 - \frac{1+2}{3000000} = \frac{2999997}{3000000} \). The expected net winnings are computed as follows: \[ \text{Expected Payback} = \left(\frac{1}{3000000}\right)(49997) + \left(\frac{2}{3000000}\right)(9997) + \left(\frac{2999997}{3000000}\right)(-3). \] Let's calculate each term: - First term: \[ \frac{49997}{3000000} \approx 0.0166657. \] - Second term: \[ \frac{2 \times 9997}{3000000} = \frac{19994}{3000000} \approx 0.0066647. \] - Third term: \[ \frac{2999997}{3000000} \times (-3) \approx -2.999997. \] Adding these: \[ \text{Expected Payback} \approx 0.0166657 + 0.0066647 - 2.999997 \approx -2.9766666. \] Rounded to two decimal places, the expected payback is: \[ -2.98. \] Thus, the answers are: Possible values of \( x \): \( 49997, 9997, -3 \) Expected payback: \( -2.98 \)

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To find the possible values of \( x \), we consider the outcomes of placing a single bet. The possible values of winnings \( x \) based on the contest rules are: 1. Win the first prize: \( 50000 - 3 = 49997 \) 2. Win one of the second prizes: \( 10000 - 3 = 9997 \) (since there are two second prizes, this value counts twice) 3. Win nothing: \( 0 - 3 = -3 \) Thus, the possible values of \( x \) are: \( 49997, 9997, -3 \). Now, to find the expected payback, we first compute the probabilities: - Probability of winning the first prize: \( \frac{1}{3000000} \) - Probability of winning a second prize: \( \frac{2}{3000000} \) - Probability of winning nothing: \( 1 - \left( \frac{1 + 2}{3000000} \right) = \frac{2999997}{3000000} \) Now, we can calculate the expected winnings \( E[x] \): \[ E[x] = 49997 \times \frac{1}{3000000} + 9997 \times \frac{2}{3000000} + (-3) \times \frac{2999997}{3000000} \] Calculating each term: - \( 49997 \times \frac{1}{3000000} \approx 0.01666633 \) - \( 9997 \times \frac{2}{3000000} \approx 0.00666467 \) - \( -3 \times \frac{2999997}{3000000} \approx -2.99929067 \) Now summing these values gives: \[ E[x] \approx 0.01666633 + 0.00666467 - 2.99929067 \approx -2.97595967 \] Rounding to two decimal places, the expected payback for a person who places one registered bet is approximately \( -2.98 \). So the final answers are: Possible values of \( x \): 49997, 9997, -3 Expected payback: -2.98

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