Bill has a bag with 8 balls numbered 1 through 8 . He is playing a game of chance. This game is this: Bill chooses one ball from the bag at random. He wins \( \$ 1 \) if the number 1 is selected, \( \$ 2 \) if the number 2 is selected, \( \$ 3 \) if the number 3 is selected, \( \$ 4 \) if the number 4 is selected, and \( \$ 5 \) if the number 5 is selected. He loses \( \$ 7 \) if 6 , 7 , or 8 is selected. \( \begin{array}{l}\text { (a) Find the expected value of playing the game. } \\ \text { (b) What can Bill expect in the long run, after playing the game many times? } \\ \text { (He replaces the ball in the bag each time.) } \\ \text { Bill can expect to gain money, } \\ \text { He can expect to win } \square \text { dollars per selection. } \\ \text { Bill can expect to lose money. } \\ \text { He can expect to lose } \square \text { dollars per selection. } \\ \text { Bill can expect to break even (neither gain nor lose money). }\end{array} \)

To calculate the expected value, we first determine the outcomes and their probabilities: - Winning $1 (1)$: 1 ball (Probability = 1/8) - Winning $2 (2)$: 1 ball (Probability = 1/8) - Winning $3 (3)$: 1 ball (Probability = 1/8) - Winning $4 (4)$: 1 ball (Probability = 1/8) - Winning $5 (5)$: 1 ball (Probability = 1/8) - Losing $7 (6, 7, 8)$: 3 balls (Probability = 3/8) Now, we compute the expected value (EV): \[ \text{EV} = \left(\frac{1}{8} \times 1\right) + \left(\frac{1}{8} \times 2\right) + \left(\frac{1}{8} \times 3\right) + \left(\frac{1}{8} \times 4\right) + \left(\frac{1}{8} \times 5\right) + \left(\frac{3}{8} \times -7\right) \] Calculating this gives: \[ \text{EV} = \frac{1 + 2 + 3 + 4 + 5 - 21}{8} = \frac{-6}{8} = -0.75 \] So, Bill can expect to **lose** $0.75 per selection. In the long run, assuming Bill plays many times, he will consistently lose about $0.75 for each game. So while it might be fun to play, his wallet won’t be smiling! In the end, it’s a great reminder that games of chance can be as risky as they are entertaining!