Place the solution steps in the correct order: In American roulette, the wheel contains the numbers 1 through 36 , alternating between black and red. There are two green spaces numbered 0 and 00 . A player places a bet of \( \$ 1.75 \) on red to play the game. If the ball lands on red, the player get a \( \$ 1.75 \) for winning and receives the money back. If the ball does not land on red, then the player simply loses the \( \$ 1.75 \) placed on the bet. If the player places the same bet on red 7 .times, what is the player's expected winnings?

To find the player's expected winnings when placing a bet of $1.75 on red 7 times in American roulette, we need to consider the probability of winning and the probability of losing. In American roulette, there are 38 numbers on the wheel (1-36, alternating between black and red, and 0 and 00). The probability of the ball landing on red is 18/38, since there are 18 red numbers out of the total 38 numbers. The probability of winning is the probability of the ball landing on red, which is 18/38. The probability of losing is the probability of the ball not landing on red, which is 1 - 18/38 = 20/38. The expected value of a single bet is calculated as the probability of winning multiplied by the amount won, minus the probability of losing multiplied by the amount lost. In this case, the amount won is $1.75 and the amount lost is $1.75. Let's calculate the expected value of a single bet: \[ \text{Expected Value} = \left(\frac{18}{38}\right) \times \$1.75 - \left(\frac{20}{38}\right) \times \$1.75 \] Now, we need to calculate the expected value of placing the same bet 7 times. The expected value of placing the bet 7 times is the expected value of a single bet multiplied by 7. Let's calculate the expected value of placing the bet 7 times: \[ \text{Expected Value of 7 Bets} = 7 \times \text{Expected Value of a Single Bet} \] Now, we can calculate the expected value of placing the bet 7 times. Calculate the value by following steps: - step0: Calculate: \(\left(\frac{18}{38}\right)\times 1.75-\left(\frac{20}{38}\right)\times 1.75\) - step1: Reduce the fraction: \(\frac{9}{19}\times 1.75-\left(\frac{20}{38}\right)\times 1.75\) - step2: Reduce the fraction: \(\frac{9}{19}\times 1.75-\frac{10}{19}\times 1.75\) - step3: Multiply the numbers: \(\frac{63}{76}-\frac{10}{19}\times 1.75\) - step4: Multiply the numbers: \(\frac{63}{76}-\frac{35}{38}\) - step5: Reduce fractions to a common denominator: \(\frac{63}{76}-\frac{35\times 2}{38\times 2}\) - step6: Multiply the numbers: \(\frac{63}{76}-\frac{35\times 2}{76}\) - step7: Transform the expression: \(\frac{63-35\times 2}{76}\) - step8: Multiply the numbers: \(\frac{63-70}{76}\) - step9: Subtract the numbers: \(\frac{-7}{76}\) - step10: Rewrite the fraction: \(-\frac{7}{76}\) The expected value of a single bet is approximately -0.092105. Now, let's calculate the expected value of placing the bet 7 times: \[ \text{Expected Value of 7 Bets} = 7 \times \text{Expected Value of a Single Bet} \] \[ \text{Expected Value of 7 Bets} = 7 \times -0.092105 \] \[ \text{Expected Value of 7 Bets} = -0.643735 \] Therefore, the player's expected winnings when placing a bet of $1.75 on red 7 times in American roulette is approximately -$0.643735.