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There are 1010 cards in a hat, numbered 11 to 1010. The game is to draw one card out of the hat. If the number you draw is even, you win $20\$20. If the number you draw is odd, you win nothing. If you play the game, what is the expected payoff?\newline$\$____

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Q. There are 1010 cards in a hat, numbered 11 to 1010. The game is to draw one card out of the hat. If the number you draw is even, you win $20\$20. If the number you draw is odd, you win nothing. If you play the game, what is the expected payoff?\newline$\$____
  1. Prompt: question_prompt: What's the expected payoff for drawing an even number from a hat with cards numbered 11 to 1010?
  2. Even Numbers: There are 55 even numbers (22, 44, 66, 88, 1010) and 55 odd numbers (11, 33, 55, 2200, 2211) in the hat.
  3. Probability: The probability of drawing an even number is 55 out of 1010, or 12\frac{1}{2}.
  4. Expected Value - Even: The probability of drawing an odd number is also 55 out of 1010, or 12\frac{1}{2}.
  5. Calculation - Even: If you draw an even number, you win $20\$20. So the expected value for drawing an even number is 12×$20\frac{1}{2} \times \$20.
  6. Expected Value - Odd: Calculate the expected value for drawing an even number: 12×($20)=($10)\frac{1}{2} \times (\$20) = (\$10).
  7. Calculation - Odd: If you draw an odd number, you win $0\$0. So the expected value for drawing an odd number is 12×$0\frac{1}{2} \times \$0.
  8. Total Expected Payoff: Calculate the expected value for drawing an odd number: 12×($0)=($0)\frac{1}{2} \times (\$0) = (\$0).
  9. Total Expected Payoff: Calculate the expected value for drawing an odd number: 12×($0)=($0)\frac{1}{2} \times (\$0) = (\$0). Add the expected values together to get the total expected payoff: ($10)+($0)=($10)(\$10) + (\$0) = (\$10).

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