Juan is playing a game in which he 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: Juan spins the spinner once. He wins \( \$ 1 \) if the spinner stops on the number \( 1, \$ 4 \) if the spinner stops on the number 2 , \( \$ 7 \) if the spinne stops on the number 3 , and \( \$ 10 \) if the spinner stops on the number 4 . He loses \( \$ 11 \) if the spinner stops on 5 or 6 . (a) Find the expected value of playing the game. (b) What can Juan expect in the long run, after playing the game many times? Juan can expect to gain money. He can expect to win \( \square \) dollars per spin. He can expect to lose \( \square \) dollars per spin. O Juan can expect to break even (neither gain nor lose money).
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To find the expected value of playing the game, we can calculate it using the outcomes and their respective probabilities. The outcomes and their associated wins/losses are as follows: - Winning \$1 with probability \( \frac{1}{6} \) (stopping on 1) - Winning \$4 with probability \( \frac{1}{6} \) (stopping on 2) - Winning \$7 with probability \( \frac{1}{6} \) (stopping on 3) - Winning \$10 with probability \( \frac{1}{6} \) (stopping on 4) - Losing \$11 with probability \( \frac{1}{6} \) (stopping on 5) - Losing \$11 with probability \( \frac{1}{6} \) (stopping on 6) Now, let's calculate the expected value \( E \): \[ E = \left(\frac{1}{6} \cdot 1\right) + \left(\frac{1}{6} \cdot 4\right) + \left(\frac{1}{6} \cdot 7\right) + \left(\frac{1}{6} \cdot 10\right) + \left(\frac{1}{6} \cdot (-11)\right) + \left(\frac{1}{6} \cdot (-11)\right) \] \[ E = \frac{1 + 4 + 7 + 10 - 11 - 11}{6} \] \[ E = \frac{0}{6} = 0 \] (a) The expected value of playing the game is \( \$0 \). (b) In the long run, after playing the game many times, Juan can expect to break even since the expected value is \( \$0 \) per spin. Thus, he will neither gain nor lose money overall.