course Mth 152 `ǃΐɢIٓƪjɰassignment #010 010. `Query 10 Liberal Arts Mathematics II 03-02-2008
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13:47:47 Query 12.5.6 fair dice game pays $3 for 6, $2 for 5, $1 for 4. What is a fair price to pay for playing this game?
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RESPONSE --> 1/6 chance of rolling a 6. The payout is $3. 1/6 * $3 = .50 1/6 chance of rolling a 5. The payout is $2. 1/6 * $2 = .33 1/6 chance of rolling a 4. The payout is $1. 1/6 * $1 = .16 $1 * 1/6 + $2 * 1/6 + $3 * 1/6 = $1.00 Fair price is $1
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13:49:45 ** A 1 in 6 chance of getting $3 is worth 1/6 * $3 = $.50 . A 1 in 6 chance of getting $2 is worth 1/6 * $2 = $.33 1/3 . A 1 in 6 chance of getting $1 is worth 1/6 * $1 = $.16 2/3 . The total expectation is $1.00 * 1/6 + $2.00 * 1/6 + $3.00 * 1/6 = $1.00 So a fair price to pay is $1.00 **
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RESPONSE --> Correct answer.
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13:55:01 Query 12.5.10 expectation Roulette $1 bet 18 red, 18 black one zero What is the expected net value of a bet on red?
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RESPONSE --> $1 for win and -$1 for loss There are a total of 37 possibilities (18 + 18 + 1 = 37). 18 possible red with 19 non-red possibilities. 18/37 * $1 + 19/37 * -$1 = $.027 expected net value of a bet on red
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13:55:09 ** If your net gain is $1 for a win and -$1 for a loss the expected value is 18/37 * (+1) + 19/37 * (-1) = -$.027. **
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RESPONSE --> Correct answer.
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14:02:28 Query 12.5.20 exp sum of 2 of 5 cards 1-5. What is the expected sum of the numbers on the two cards drawn?
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RESPONSE --> With cards numbered 1-5, you cannot get a sum of 1 or 2. The lowest sum is 1 + 2 = 3 There are 2 ways to get 3 (1 + 2 = 3 and 2 + 1 = 3) There are 2 ways to get 4 (1 + 3 = 4 and 3 + 1 = 4) There are 4 ways to get 5 (1 + 4 = 5, 4 + 1 = 5, 2 + 3 = 5, 3 + 2 = 5) There are 4 ways to get 6 (1 + 5 = 6, 5 + 1 = 6, 2 + 4 = 6, 4 + 2 = 6) There are 4 ways to get 7 (2 + 5 = 7, 5 + 2 = 7, 4 + 3 = 7, 3 + 4 = 7) There are 2 ways to get 8 (3 + 5 = 8, 5 + 3 = 8) There are 2 ways to get 9 (4 + 5 = 9, 5 + 4 = 9) Add the total ways gives 20 possibilities: 2 + 2 + 4 + 4 + 4 + 2 + 2 = 20 2/20 * 3 + 2/20 * 4 + 4/20 * 5 + 4/20 * 6 + 4/20 * 7 + 2/20 * 8 + 2/20 * 9 = 120/20 = 6
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14:02:36 ** You can't get a sum of 1 on two cards. There is also no way to get a sum of two, since the lowest total possible is 1 + 2 = 3. There are 2 ways to get total 3. You can get 1 on the first and 2 on the second, or vice versa. There are 2 ways to get total 4. You can get 1 on the first and 3 on the second, or vice versa. There are 4 ways to get total 5. You can get 1 on the first and 4 on the second, or vice versa, or 2 on the first and 3 on the second, or vice versa. There are 4 ways to get total 6. You can get 1 on the first and 5 on the second, or vice versa, or 2 on the first and 4 on the second, or vice versa. There are 4 ways to get total 7. You can get 2 on the first and 5 on the second, or vice versa, or 4 on the first and 3 on the second, or vice versa. There are 2 ways to get total 8. You can get 3 on the first and 5 on the second, or vice versa. There are 2 ways to get total 9. You can get 4 on the first and 5 on the second, or vice versa. You can't get more than 9. There are 2+2+4+4+4+2+2 = 20 possibilities, so the probabilities are 2/20, 4/20, 5/20, etc.. The expected sum is therefore 2/20 * 3 + 2/20 * 4 + 4/20 * 5 + 4/20 * 6 + 4/20 * 7 + 2/20 * 8 + 2/20 * 9. This gives 120 / 20 = 6. **
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RESPONSE --> Correct answer.
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14:02:47 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> No surprises or comments.
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