# Step-By-Step Solution with Explanations ## Given Information - James bets $25 on red and $5 on number 33 (black) in roulette. - Roulette wheel has 38 numbers (1-36, , 00). - If red hits (not 33): Wins $25 (red) - loses $5 (33) = +$20. - If 33 hits (black): Wins $25 (red) + $175 (33 pays 35:1) - loses $30 (both bets) = +$150. - If black (not 33) or green: loses both bets = -$30. ### Probabilities: - Wins $150 (33 hits): \( P = \frac{1}{38} = .0263 \) - Wins $20 (red not 33): \( P = \frac{18}{38} = .4737 \) - Loses $30 (other): \( P = \frac{19}{38} = .500 \) --- ## Definitions Used - **Expected Value (\(\mu\)):** The average winnings per spin, calculated as \(\sum x P(x)\). - **Standard Deviation (\(\sigma\)):** Measure of spread in winnings per spin, calculated as \(\sqrt{E(X^2) - (E(X))^2}\). - **Central Limit Theorem:** For large \( n \), the mean of independently repeated random samples will be approximately normally distributed. --- ## Step-by-Step Solution ### 1. Find the Expected Value (\(\mu\)) \[ \mu = \sum x P(x) \] \[ = 150 \times .0263 + 20 \times .4737 + (-30) \times .5 \] \[ = 3.945 + 9.474 - 15 \] \[ = -1.581 \] --- ### 2. Find the Standard Deviation (\(\sigma\)) First, calculate \( E(X^2) \): \[ E(X^2) = \sum x^2 P(x) \] \[ = (150^2) \times .0263 + (20^2) \times .4737 + ((-30)^2) \times .5 \] \[ = 22500 \times .0263 + 400 \times .4737 + 900 \times .5 \] \[ = 591.75 + 189.48 + 450 \] \[ = 1231.23 \] Now the variance: \[ \sigma^2 = E(X^2) - (E(X))^2 = 1231.23 - (-1.581)^2 = 1231.23 - 2.499 = 1228.73 \] \[ \sigma = \sqrt{1228.73} \approx 35.05 \] --- ### 3. For 250 Games Let \( \bar{x} \) = mean winnings per game. #### a. Expected Value of \( \bar{x} \), \(\mu_{\bar{x}}\): \[ \mu_{\bar{x}} = \mu = -1.581 \] #### b. Standard Deviation of \( \bar{x} \): \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{35.05}{\sqrt{250}} = \frac{35.05}{50} = .701 \] #### c. Total Expected Loss for 250 games: \[ \text{Total Expected Loss} = 250 \times (-1.581) = -3952.5 \] #### d. Standard Deviation for 250 games: \[ \text{Total Standard Deviation} = \sigma \times \sqrt{n} = 35.05 \times 50 = 1752.5 \] --- ### 4. Central Limit Theorem The **Central Limit Theorem** tells us the sum/mean of many independent bets will be approximately normally distributed. --- ### 5. Probability Calculations #### a. Probability of Making Money (Profit > in 250 games) \[ Z = \frac{ - (-3952.5)}{1752.5} = \frac{3952.5}{1752.5} \approx 2.256 \] From Z-tables, \( P(Z > 2.256) \approx .012 \) (about 1.2%). --- #### b. Probability of Profit over $100 \[ Z = \frac{100 - (-3952.5)}{1752.5} = \frac{4952.5}{1752.5} \approx 2.828 \] \( P(Z > 2.828) \approx .0023 \) (about .23%). --- #### c. Probability of Losing over $30,000 \[ Z = \frac{-30000 - (-3952.5)}{1752.5} = \frac{-26047.5}{1752.5} \approx -14.87 \] This Z is extremely negative; probability is virtually . --- #### d. Probability of Losing over $600 \[ Z = \frac{-600 - (-3952.5)}{1752.5} = \frac{-2047.5}{1752.5} \approx -1.168 \] Probability: \( P(Z < -1.168) \approx .121 \) (about 12.1%). --- ## Final Answers Listed 1. **Expected value per game:** \( -1.581 \) 2. **Standard deviation per game:** \( 35.05 \) 3. **Expected loss for 250 games:** \( -3952.5 \) 4. **Standard deviation for 250 games:** \( 1752.5 \) 5. **Central Limit Theorem** applies. 6. **Probability of making money (profit > ):** \( .012 \) (1.2%) 7. **Probability of profit > \$100:** \( .0023 \) (.23%) 8. **Probability of losing over \$30,000:** \( \approx \) 9. **Probability of losing over \$600:** \( .121 \) (12.1%) --- ### All Final Answers Together - **Expected value per game:** -1.581 - **Standard deviation per game:** 35.05 - **Expected loss for 250 games:** -3952.5 - **Standard deviation for 250 games:** 1752.5 - **Central Limit Theorem** applies. - **Probability of making money (profit > ):** .012 (1.2%) - **Probability of profit > $100:** .0023 (.23%) - **Probability of losing over $30,000:** ~ - **Probability of losing over $600:** .121 (12.1%)