Amanda has a deck of 10 cards numbered 1 through 10 . She is playing a game of chance. This game is this: Amanda chooses one card from the deck at random. She wins an amount of money equal to the value of the card if an odd numbered ca drawn. She loses \( \$ 6 \) if an even numbered card is drawn. (a) Find the expected value of playing the game. dollars (b) What can Amanda expect in the long run, after playing the game many times? (She replaces the card in the deck each time.) She can expect to win \( \square \) dollars per draw. Amanda can expect to lose money. She can expect to lose \( \square \) dollars per draw. Amanda can expect to break even (neither gain nor lose money).

To calculate the expected value of playing Amanda's game, we first need to understand the outcomes based on the cards drawn. **(a)** To find the expected value, we consider the winnings and losses for each card: - Odd cards: 1, 3, 5, 7, 9 (Total winnings: \(1 + 3 + 5 + 7 + 9 = 25\)) - Even cards: 2, 4, 6, 8, 10 (Total loss per each is -\$6) Each card has an equal probability of \( \frac{1}{10} \). Now we calculate the expected winnings from odd cards: \[ \text{Odds Winning} = \frac{25}{10} = 2.5 \text{ (average from odd cards)} \] Total probability for odd draws = 5 cards = \( \frac{5}{10} = 0.5 \). Expected loss from even cards: \[ \text{Even Loss} = 5 \times -6 = -30 \] Total probability for even draws = 5 cards = \( \frac{5}{10} = 0.5 \). Combining everything, the expected value (EV) can be calculated as follows: \[ \text{EV} = (0.5 \times 2.5) + (0.5 \times -6) = 1.25 - 3 = -1.75 \] So the expected value of playing the game is \(-\$1.75\). **(b)** In the long run, after playing the game many times, Amanda can expect to lose money. Since the expected value is negative, she can expect to lose \( \$1.75 \) per draw. This consistent loss highlights that over time, the nature of this game won't be in her favor, making it a gamble where she should tread carefully if she wishes to keep her money!