course Mth152 ???h€???????????assignment #010010. Expectation
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14:48:06 `q001. Note that there are 9 questions in this assignment. In a certain lottery the probability of winning $100 is .005, the probability of winning $1000 is .0002 and the probability of winning $10,000 is .00001. Otherwise you win nothing. What is the probability of winning nothing?
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RESPONSE --> In a chart we would have 0,1,2,3 chances of winning so we would have 1/4 chance of winning nothing. 0 0 1 100 .005 2 1000 .0002 3 10000 .00001 Although I am not sure how to get the probability of winning nothing. confidence assessment: 0
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14:50:43 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 --> I see . All I needed to do to get this answer was to get the sum of probability of winning and subtract it from 1 to get the probability of not winning. self critique assessment: 2
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14:57:35 `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 --> 100 *.005 = .5*1000000 = 500 000 1000 * .0002 = .2* 1000 000 = 200 000 10000 * .00001 = .1 * 1000 000 = 100 000 confidence assessment: 1
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14:59:51 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 --> I see. I should have multiplied the probability by 1000 000 itself instead. self critique assessment: 2
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15:03:05 `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 = 500 000 200 * 1000 = 200 000 10 * 10000 = 100 000 500 000 + 200 000 + 100 000 = 800 000 confidence assessment: 1
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15:03:30 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 --> Correct self critique assessment: 3
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15:05:32 `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 --> 800 000 - 500 000 = 300 000 ahead confidence assessment: 2
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15:06:38 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 --> correct self critique assessment: 3
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15:16:10 `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 --> 300 000 / 1000 000 = .3 per ticket 10 000 000 * .3 = 3 000 000 confidence assessment: 1
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15:18:14 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 --> I used the wrong amount but the right principle. I should have used 800 000 instead of just the profits. self critique assessment: 2
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15:22:26 `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 --> The sum is .8. 100*.005 + 1000*.0002 + 10000*.00001 .5 + .2 + .1 = .8 confidence assessment: 1
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15:23:23 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 --> Correct. I see that it is the same amount of .80. self critique assessment: 3
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15:29:31 `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 --> In the first 8 flips we would have 4 heads and 4 tails. In the 5th row we would have 3 heads and 7 tails. In the 1st row diagonally we would have 4 heads and 5 tails. confidence assessment: 2
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15:31:02 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 --> I see that I should have used 8 flips on all of the questions. I did use 8 on the first but then I used all on the rest. self critique assessment: 2
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15:38:26 `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 --> The first 20 rolls are : 1 4 5 6 2 3 2 3 5 2 3 2 1 2 3 5 4 2 4 5 The sum of the 10 totals are : 5 11 5 5 7 5 3 8 6 9 confidence assessment: 1
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15:41:08 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 --> Correct. Typo 4+5 = 9 not 10. self critique assessment: 3
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15:44:07 `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 --> 3/4 of the results were 5,6,7 confidence assessment: 1
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15:45:51 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 --> I see I simplified down 6/10 to 3/5. I understand that we would have 36 possibilities. self critique assessment: 2
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