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Jina is playing a game in which she spins a spinner with 6 equal-sized slices numbered 1 through 6 . The spinner stops on a numbered slice at random. This game is this: Jina spins the spinner once. She wins \( \$ 1 \) if the spinner stops on the number \( 1, \$ 3 \) if the spinner stops on the number 2 , \( \$ 5 \) if the spinner stops on the number 3 , and \( \$ 7 \) if the spinner stops on the number 4 . She loses \( \$ 8 \) if the spinner stops on 5 or 6 . (a) Find the expected value of playing the game. dollars (b) What can Jina expect in the long run, after playing the game many times? She can expect to win \( \square \) dollars per spin. Jina can expect to lose money. She can expect to lose \( \square \) dollars per spin. Jina can expect to break even (neither gain nor lose money).

Ask by Guzman Sullivan. in the United States
Feb 28,2025

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Tutor-Verified Answer

Answer

(a) The expected value of playing the game is 0 dollars. (b) Jina can expect to break even; she neither gains nor loses money per spin.

Solution

We first find the probabilities and payoffs. Since there are 6 equal-sized slices on the spinner, each outcome has a probability of 1/6. The payoffs are: • Spin 1: win $1 • Spin 2: win $3 • Spin 3: win $5 • Spin 4: win $7 • Spin 5: lose $8 • Spin 6: lose $8 (a) To calculate the expected value (E), multiply each outcome by its probability and add them up:   E = (1/6)(1) + (1/6)(3) + (1/6)(5) + (1/6)(7) + (1/6)(–8) + (1/6)(–8)     = (1 + 3 + 5 + 7 – 8 – 8) / 6     = (16 – 16) / 6     = 0 / 6     = 0 dollars (b) Since the expected value is 0 dollars per spin, after playing many spins, Jina can expect, on average, neither to win nor lose money. In other words, she can expect to break even. Final Answers: (a) The expected value of playing the game is 0 dollars. (b) Jina can expect to break even; she neither gains nor loses money per spin.

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To find the expected value of Jina’s game, we need to calculate the payouts for each outcome, weight them by their probabilities, and sum them up. The spinner is fair, which means each slice has a probability of \( \frac{1}{6} \). - Winning \( \$ 1 \) for landing on 1: \( \frac{1}{6} \times 1 = \frac{1}{6} \) - Winning \( \$ 3 \) for landing on 2: \( \frac{1}{6} \times 3 = \frac{3}{6} \) - Winning \( \$ 5 \) for landing on 3: \( \frac{1}{6} \times 5 = \frac{5}{6} \) - Winning \( \$ 7 \) for landing on 4: \( \frac{1}{6} \times 7 = \frac{7}{6} \) - Losing \( \$ 8 \) for landing on 5: \( \frac{1}{6} \times (-8) = -\frac{8}{6} \) - Losing \( \$ 8 \) for landing on 6: \( \frac{1}{6} \times (-8) = -\frac{8}{6} \) Now, let's sum these values together: \[ E(X) = \frac{1}{6} + \frac{3}{6} + \frac{5}{6} + \frac{7}{6} - \frac{8}{6} - \frac{8}{6} \] Combining these gives: \[ E(X) = \frac{1 + 3 + 5 + 7 - 8 - 8}{6} = \frac{1 + 3 + 5 + 7 - 16}{6} = \frac{-0}{6} = 0 \] So, the expected value of playing the game is \( 0 \) dollars. (b) In the long run, Jina can expect to break even (neither gain nor lose money). Since the expected value is \( 0 \), it indicates a fair game without any tilt toward winnings or losses. Thus, she can expect to win \( 0 \) dollars per spin or potentially neither winning nor losing money after many plays.

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