course Mth 152 21:40:08`questionNumber 100000
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RESPONSE --> 1 - (.005 + .0002 + .00001) = .99479 confidence assessment: 1
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21:40:26 `questionNumber 100000 The probability of winning something is the sum .005 + .0002 + .00001 =.00521. The events of winning something and winning nothing are mutually exclusive, and they comprise all possible outcomes. It follows that the probability of winning something added to the probability winning nothing must give us 1, and that therefore Probability of winning nothing = 1 - .00521 = .99479.
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RESPONSE --> ok self critique assessment: 3
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21:43:10 `questionNumber 100000 `q002. In the same lottery , where the probability of winning $100 is .005, the probability of winning $1000 is .0002 and the probability of winning $10,000 is .00001, if you bought a million tickets how many would you expect to win the $100 prize? How many would you expect to win the $1000 prize? How many would you expect to win the $10,000 prize?
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RESPONSE --> 1,000,000*.005 = 5000 1,000,000*.0002 = 200 1,000,000*.00001 = 10 confidence assessment: 2
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21:43:16 `questionNumber 100000 The probability of winning the $100 prize is.005, so out of a million tries we would expect to win the $100 a total of .005 * 1,000,000 = 5,000 times. Similarly we would expect to win the $1000 prize a total of .0002 * 1,000,000 = 200 times. The expected number of times we would win the $10,000 prize would be .00001 * 1,000,000 = 10.
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RESPONSE --> ok self critique assessment: 3
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21:47:48 `questionNumber 100000 `q003. In the lottery of the preceding problem, if you were given a million tickets how much total money would you expect to win?
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RESPONSE --> 5000*100 = 500000 200*1000 = 200000 10*10000 = 100000 500000 + 200000 + 100000 = $800000 confidence assessment: 2
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21:48:00 `questionNumber 100000 As seen in preceding problem, you would expect to win $100 a total of 5,000 times for a total of $500,000, you would expect to win the $1000 prize 200 times for a total of $200,000, and you expect to win the $10,000 prize 10 times for total of $100,000. The expected winnings from a million tickets would therefore be the total $800,000 of these winnings.
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RESPONSE --> ok self critique assessment: 3
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21:50:13 `questionNumber 100000 `q004. In the lottery of the preceding problem, if you bought a million tickets for half a million dollars would you most likely come out ahead?
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RESPONSE --> Yes, because according to the assumed winnings you would gain around $300000, for you are only spending half a million on all of them. confidence assessment: 2
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21:50:23 `questionNumber 100000 You would expect on the average to win $800,000, and your probability of winning at least $500,000 would seem to be high. You would have a very good expectation of coming out ahead.
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RESPONSE --> ok self critique assessment: 3
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21:53:44 `questionNumber 100000 `q005. In the lottery of the preceding problem, how much would you expect to win, per ticket, if you bought a million tickets? Would the answer change if you bought 10 million tickets?
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RESPONSE --> 800000/1000000 = $.80 No confidence assessment: 2
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21:54:11 `questionNumber 100000 Your expected winnings would be $800,000 on a million tickets, which would average out to $800,000/1,000,000 = $.80, or 80 cents. If you bought 10 million tickets you expect to win 10 times as much, or $8,000,000 for an average of $8,000,000 / 10,000,000 = $.80, or 80 cents. The expected average wouldn't change. However you might feel more confident that your average winnings would be pretty close to 80 cents if you have 10 million chances that if you had 1 million chances.
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RESPONSE --> ok self critique assessment: 3
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21:57:08 `questionNumber 100000 `q006. If we multiply $100 by the probability of winning $100, $1000 by the probability of winning $1000, and $10,000 by the probability of winning $10,000, then add all these results, what is the sum? How does this result compare with the results obtained on previous problems, and why?
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RESPONSE --> (100*.005) + (1000*.0002) + (10000*.00001) = .8 This matches the assumed amount of winnings per ticket. confidence assessment: 2
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21:57:29 `questionNumber 100000 We get $100 * .005 + $1,000 * .0002 + $10,000 * .00001 = $.50 + $.20 + $.10 = $.80. This is the same as the average per ticket we calculated for a million tickets, or for 10 million tickets. This seems to indicate that a .005 chance of winning $100 is worth 50 cents, a .0002 chance of winning $1,000 is worth 20 cents, and a .00001 chance of winning $10,000 is worth 10 cents.
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RESPONSE --> ok self critique assessment: 3
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22:02:50 `questionNumber 100000 `q007. The following list of random digits has 10 rows and 10 columns: 3 8 4 7 2 3 0 8 3 9 1 8 3 7 3 2 9 1 0 3 4 3 3 0 2 1 4 9 8 2 4 3 4 9 9 2 0 1 3 9 8 3 4 1 3 0 5 3 9 7 2 4 7 4 5 3 7 2 1 8 3 6 9 0 2 5 9 5 2 3 4 5 8 5 8 8 2 9 8 5 9 3 4 6 7 4 5 8 4 9 4 1 5 7 9 2 9 3 1 2. Starting in the second column and working down the column, if we let even numbers stand for 'heads' and odd numbers for 'tails', then how many 'heads' and how many 'tails' would we end up with in the first eight flips? Answer the second question but starting in the fifth row and working across the row. Answer once more but starting in the first row, with the second number, and moving diagonally one space down and one to the right for each new number.
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RESPONSE --> H:4 T:4 H:3 T:5 H:4 T:4 confidence assessment: 2
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22:02:59 `questionNumber 100000 Using the second column, the first eight flips would be represented by the numbers in the second column, which are 8, 8, 3, 3, 3, 4, 6, and 5. According to the given rule this correspond to HHTTTHHT, total of four 'heads' and four 'tails'. Using the fifth row we have the numbers 8 3 4 1 3 0 5 3, which according to the even-odd rule would give us HTHTTHTT, or 3 'heads' and 5 'tails'. Using the diagonal scheme we get 8, 3, 0, 9, 0, 7, 5, 8 for HTHTHTTH, a total of four 'heads' and four 'tails'.
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RESPONSE --> ok self critique assessment: 3
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22:08:48 `questionNumber 100000 `q008. Using once more the table 3 8 4 7 2 3 0 8 3 9 1 8 3 7 3 2 9 1 0 3 4 3 3 0 2 1 4 9 8 2 4 3 4 9 9 2 0 1 3 9 8 3 4 1 3 0 5 3 9 7 2 4 7 4 5 3 7 2 1 8 3 6 9 0 2 5 9 5 2 3 4 5 8 5 8 8 2 9 8 5 9 3 4 6 7 4 5 8 4 9 4 1 5 7 9 2 9 3 1 2 let the each of numbers 1, 2, 3, 4, 5, 6 stand for rolling that number on a die-e.g., if we encounter 3 in our table we let it stand for rolling a 3. If any other number is encountered it is ignored and we move to the next. Starting in the fourth column and working down, then moving to the fifth column, etc., what are the numbers of the first 20 dice rolls we simulate? If we pair the first and the second rolls, what is the total? If we pair the third and fourth rolls, what is the total? If we continue in this way what are the 10 totals we obtain?
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RESPONSE --> 1,4,5,6,2,3,2,3,5,2,3,2,1,2,3,5,4,2,4,5 5,11,5,5,7,5,3,8,6,9 confidence assessment: 1
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22:09:35 `questionNumber 100000 The numbers we get in the fourth column are 7, 7, 0, 9, 1, 4, 0, 5, 6, 7, then in the fifth column we get 2, 3, 2, 9, 3, 5, 2, 8, 7, 9 and in the sixth column we get 3, 2, 1, 2, 0, 3, 5, 8, 4, 2. We hope to get 20 numbers between 1 and 6 from this list of 30 numbers, but we can be sure that this will be the case. If it is, we will add some numbers from the seventh column. Omitting any number on our current list not between 1 and 6 we get 1, 4, 5, 6 from the fourth column, then from the fifth column we get 2, 3, 2, 3, 5, 2 and from the sixth column we get 3, 2, 1, 2, 3, 5, 4, 2. This gives us only 18 numbers between 1 and 6, and we need 20. So we go to the seventh column, which starts with 0, 9, 4, 0, 5. The first number we encounter between 1 6 is 4. The next is 5. This completes our list. Our simulation therefore gives us the list 1, 4, 5, 6, 2, 3, 2, 3, 5, 2, 3, 2, 1, 2, 3, 5, 4, 2, 4, 5. This list represents a simulated experiment in which we row of a fair die 20 times. The first and second rolls were 1 and 4, which add up to 5.{} The second and third rolls were 5 and 6, which add up to 11. The remaining rolls give us 2 + 3 = 5, 2 + 3 = 5, 5 + 2 = 7, 3 + 2 = 5, 1 + 2 = 3, 3 + 5 = 8, 4 + 2 = 6, and 4 + 5 = 10. The totals we obtain our therefore 5, 11, 5, 5, 7, 5, 3, 8, 6, and 10.
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RESPONSE --> ok self critique assessment: 3
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22:10:03 `questionNumber 100000 `q009. According to the results of the preceding question, what proportion of the totals were 5, 6, or 7? How do these proportions compare to the expected proportions?
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RESPONSE --> ? confidence assessment: 0
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22:10:14 `questionNumber 100000 We obtain four 5's, one 6 and one 7. Thus 6 of our 10 results were 5, 6 or 7. We saw earlier that of the 36 possible outcomes of rolling two dice, four give us a total of 5, while five give us a total of 6 and six give the total of 7. If we add these numbers we see that 15 of the 36 possible outcomes in the sample space are 5, 6 or 7 for probability 15/36. Our simulation results in 6/10, a higher proportion than the probabilities would lead us to expect. However since the simulation resulted from random numbers it is certainly possible that this will happen, just as it is possible that if we rolled two dice 10 times 7 of the outcomes would be in this range.
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RESPONSE --> ok self critique assessment: 3