Mike's game consists of selecting two balls, without replacement, from an urn containing 7 balls numbered 1 through 7. The amount that you win (in dollars) is the maximum of the numbers on the two balls. How much should be charged to play this game if the Mike wants to make $0.90, on average, each time the game is played?
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To calculate how much Mike should charge to play this game if he wants to make $0.90, on average, each time the game is played, we can first determine the expected value of the game.
There are 7 balls numbered 1 through 7 in the urn, and when you select two balls without replacement, the maximum number on the two balls will be the prize amount.
Let's calculate the probability of getting each possible maximum number:
Maximum number = 7: This can only happen when you draw the balls 7 and any other ball. There are 6 ways to choose the second ball. So, the probability is (1/7) * (6/6) = 1/7.
Maximum number = 6: This can happen in two ways: (6, 5) or (7, 6). So, the probability is (1/7) * (5/6) + (1/7) * (1/6) = 11/42.
Maximum number = 5: This can happen in three ways: (5, 4), (6, 4), or (7, 5). So, the probability is (1/7) * (4/6) + (1/7) * (4/6) + (1/7) * (2/6) = 17/42.
Maximum number = 4: This can happen in three ways: (4, 3), (5, 3), or (6, 3). So, the probability is (1/7) * (3/6) + (1/7) * (3/6) + (1/7) * (3/6) = 13/42.
Maximum number = 3: This can happen in two ways: (3, 2) or (4, 2). So, the probability is (1/7) * (2/6) + (1/7) * (2/6) = 4/21.
Maximum number = 2: This can happen in one way: (2, 1). So, the probability is (1/7) * (1/6) = 1/42.
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