course Mth 152 7/2 around 11:55pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aA 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 ** `q 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: $1 * 18/37 + (-$1) * 18/37 + (-$1) * 1/37 = 0.486 – 0.486 + 0.027 = 0.027 Expected net winnings are app $0.03 Confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf your net gain is $1 for a win and -$1 for a loss the expected value is 18/37 * (+1) + 19/37 * (-1) = -$.027. ** `q 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I’m not sure how to go about solving this problem? The previous two problems dealt with plugging given numbers into a type of formula. I’m sure that this is solvable using the same method, I’m just not sure where to start? Confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aYou 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. ** Self-Critique: This makes sense, though I would have needed help getting started. But I didn’t think about finding the probabilities on my own, but that would make sense because this chapter is on probabilities. And I first I didn’t understand what each probability was being multiplied by, but I see now that they are each multiplied by the total that they are representing. (ex: 2/20 * 3 represents a 2/20 probability of the two cards adding to three). `q Query Add comments on any surprises or insights you experienced as a result of this assignment. The questions that are more obvious were not difficult at all for me, but the last question caused me a lot of problems because I didn’t take the information from other sections on probability and apply them together.