course Mth 152 ÐK蘽ÊþÖŽ¨’²¢ã|ʼ팅assignment #010
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07:35:19 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 --> There would be 1 out of 6 chance of getting any of the numbers on a roll so 1/6 of $3.00 would be .50 1/6 of $2.00 be .33 and 1/6 of $1.00 would be .16. The sum of these three would be .99
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07:36:01 ** 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 --> o.k. I did follow through on the division to get the extra penny
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07:56:27 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 --> 18 red + 18 black + 1 zero = 37 possibilities for red it would be 18 out of 37 x $1.00 you would have to add that to the 19 out of 37(possible losses)x the possibility of losing a $1.00 18/37 x 1.00 = .486 19/37 x -$1.00 = -.513 -.513 + .486 = -.027
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07:56:35 ** 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 --> o.k.
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08:03:11 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 --> 1, 2, 3, 4, 5, 1 2 3 4 5 6 2 3 4 5 6 7 3 4 5 6 7 8 4 5 6 7 8 9 5 6 7 8 9 10
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08:03:30 ** 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 --> o.k.
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08:03:59 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> wasn't sure exactly what to do on that last problem
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