Abdul is playing a game of chance in which he tosses a dart into a rotating dartboard with 8 equal-sized slices numbered 1 through 8 . The dart lands on a numbered slice at random. This game is this: Abdul tosses the dart once. He wins \( \$ 1 \) if the dart lands in slice \( 1, \$ 2 \) if the dart lands in slice \( 2, \$ 5 \) if the dart lands in slice 3 , and \( \$ 8 \) if dart lands in slice 4 . He loses \( \$ 2.50 \) if the dart lands in slices \( 5,6,7 \), or 8 . (a) Find the expected value of playing the game. \( \square \) dollars (b) What can Abdul expect in the long run, after playing the game many times? Abdul can expect to gain money. He can expect to win \( \square \) dollars per toss. Abdul can expect to lose money. He can expect to lose \( \square \) dollars per toss. Abdul can expect to break even (neither gain nor lose money).

To find the expected value of playing the game, we can calculate the outcomes based on the probabilities of landing on each slice. Each slice has a probability of \( \frac{1}{8} \). The expected value (EV) can be computed as follows: EV = \( \left( \frac{1}{8} \times 1 \right) + \left( \frac{1}{8} \times 2 \right) + \left( \frac{1}{8} \times 5 \right) + \left( \frac{1}{8} \times 8 \right) + \left( \frac{1}{8} \times (-2.50) \right) \times 4 \) = \( \frac{1 + 2 + 5 + 8 - 10}{8} \) = \( \frac{6}{8} = \$0.75. \) So, the expected value of playing the game is \( 0.75 \) dollars per toss. In the long run, after playing many times, Abdul can expect to gain money, winning an average of \( 0.75 \) dollars per toss. This means that if he plays the game repeatedly, he’s likely to accumulate a profit over time, making this a fun game with positive expectations!