course Mth 152 }{OP¼|ޮ^џassignment #010
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18:37:30 `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 --> 1 - .005 - .0002 - .00001 = .995 confidence assessment: 3
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18:38:06 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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18:42:22 `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 * 1 million = 5000 $1000: .0002 * 1 million = 200 $10000: .00001 * 1 million = 10 confidence assessment: 3
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18:42:48 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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18:46:33 `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 --> $100 * 5,000 = $500,000 $1000 * 200 = $200,000 $10000 * 10 = $100,000 $800,000 confidence assessment: 3
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18:47:07 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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18:49:14 `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 --> You would be spending $500,000 and bringing in $800,000. confidence assessment: 3
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18:49:38 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 --> Ok. self critique assessment: 3
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18:51:47 `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 --> If you bought 10 million tickets the average win per ticket would be the same as with the million. confidence assessment: 3
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18:52:45 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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18:55:21 `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 = .50 $1000 * .0002 = .20 $10000 * .00001 = .10 $0.80 The same average per ticket as previous. confidence assessment: 3
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18:55:56 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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19:00:35 `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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19:01:47 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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19:09:47 `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 1 + 4 = 5 5 + 6 = 11 2 + 3 = 5 2 + 3 = 5 5 + 2 = 7 3 + 2 = 5 1 + 2 = 3 3 + 5 = 8 4 + 2 = 6 4 + 5 = 9 confidence assessment: 3
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19:11:26 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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19:17:14 `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 --> Four 5's One 6 One 7 6 / 10 confidence assessment: 3
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19:18:37 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
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ZVr䲐湚 assignment #011 011. Liberal Arts Mathematics II 06-26-2007
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19:21:01 014. mean vs median
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RESPONSE --> Mean is the average Median is the middle number
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19:24:36 `q001. Note that there are 8 questions in this assignment. {}{} What is the average, or mean value, of the numbers 5, 7, 9, 9, 10, 12, 13, and 15? On the average how 'far' is each number from this mean value?
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RESPONSE --> Mean: 10 Avg Dis: 20 / 8 = 2.5 confidence assessment: 3
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19:41:35 To get the mean value of the numbers, we first note that there are eight numbers. Then we had the numbers and divide by eight. We obtain 5 + 7 + 9 + 9 + 10 + 12 + 13 + 15 = 80. Dividing by 8 we obtain mean = 80 / 8 = 10. The difference between 5 and the mean 10 is 5; the difference between 7 and the mean 10 is 3; the difference between 9 and 10 is 1; the differences between 12, 13 and 15 and the mean 10 are 2, 3 and 5. So we have differences 5, 3, 1, 1, 0, 2, 3 and 5 between the mean and the numbers in the list. The average difference between the mean and the numbers in the list is therefore ave difference = ( 5 + 3 + 1 + 1 + 0 + 2 + 3 + 5 ) / 8 = 20 / 8 = 2.5.
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RESPONSE --> Ok. self critique assessment: 3
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19:42:01 `q002 What is the middle number among the numbers 13, 12, 5, 7, 9, 15, 9, 10, 8?
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RESPONSE --> 9 confidence assessment: 3
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19:43:07 It is easier to answer this question if we place the numbers in ascending order. Listed in ascending order the numbers are 5, 7, 8, 9, 9, 10, 12, 13, and 15. We see that there are 9 numbers in the list. If we remove the first 4 and the last 4 we are left with the middle number. So we remove the numbers 5, 7, 8, 9 and the numbers 10, 12, 13, and 15, which leaves the second '9' as the middle number.
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RESPONSE --> Ok. self critique assessment: 3
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19:44:11 `q003. On a list of 9 numbers, which number will be the one in the middle? Note that the middle number is called the 'median'.
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RESPONSE --> The 5th number. confidence assessment: 3
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19:45:32 If the 9 numbers are put in order, then we can find the middle number by throwing out the first four and the last four numbers on the list. We are left with the fifth number on the list. In general if we have an odd number n of number in an ordered list, we throw out the first (n-1) / 2 and the last (n-1) / 2 numbers, leaving us with the middle number, which is number (n-1)/2 + 1 on the list. So for example if we had 179 numbers on the list, we would throw out the first (179 - 1) / 2 = 178/2 = 89 numbers on the list and the last 89 numbers on the list, leaving us with the 90th number on the list. Note that 90 = (179 - 1) / 2 + 1, illustrating y the middle number in number (n-1)/2 + 1 on the list.
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RESPONSE --> Ok. self critique assessment: 3
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19:47:15 `q004. What is the median (the middle number) among the numbers 5, 7, 9, 9, 10, 12, 13, and 15?
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RESPONSE --> 9.5 confidence assessment: 3
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19:48:00 There are 8 numbers on this list. If we remove the smallest then the largest our list becomes 7, 9, 9, 10, 12, 13. If we remove the smallest and the largest from this list we obtain 9, 9, 10, 12. Removing the smallest and the largest from this list we are left with 9 and 10. We are left with two numbers in the middle; we don't have a single 'middle number'. So we do the next-most-sensible thing and average the two numbers to get 9.5. We say that 9.5 is the middle, or median, number.
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RESPONSE --> Ok. self critique assessment: 3
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19:53:35 `q005. We saw that for the numbers 5, 7, 9, 9, 10, 12, 13, and 15, on the average each number is 2.5 units from the average. Are the numbers in the list 48, 48, 49, 50, 51, 53, 54, 55 closer or further that this, on the average, from their mean?
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RESPONSE --> Mean: 51 Avg Dis: 18 / 8 = 2.25 Closer confidence assessment: 3
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19:54:38 The mean of the numbers 48, 48, 49, 50, 51, 53, 54, and 55 is (48 + 48 + 49 + 50 + 51 + 53 + 54 + 55) / 8 = 408 / 8 = 51. 48 is 3 units away from the mean 51, 49 is 2 units away from the mean 51, 50 is 1 unit away from the mean 51, and the remaining numbers are 2, 3 and 4 units away from the mean of 51. So on the average the distance of the numbers from the mean is (3 + 3 + 2 + 1 + 0 + 2 + 3 + 4) / 8 = 18 / 8 = 2.25. This list of numbers is a bit closer, on the average, then the first list.
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RESPONSE --> Ok. self critique assessment: 3
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20:00:36 `q006. On a 1-10 rating of a movie, one group gave the ratings 1, 8, 8, 9, 9, 10 while another gave the ratings 7, 7, 8, 8, 9, 10. Find the mean (average) and the median (middle value) of each group's ratings. Which group would you say liked the movie better?
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RESPONSE --> Mean: 7.5 Median: 8.5 Mean: 8.17 Median: 8 The second group liked the movie better, mainly due to the score of 1 from the first group. confidence assessment: 3
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20:01:52 The mean of the first list is (1 + 8 + 8 + 9 + 9 + 10) / 6 = 45 / 6 = 7.5. The median is obtained a throwing out the first 2 numbers on the list and the last 2 numbers. This leaves the middle two, which are 8 and 9. The median is therefore 8.5. The mean of the numbers on the second list is (7 + 7 + 8 + 8 + 9 + 10) / 6 = 49 / 6 = 8 .16. The median of this list is found by removing the first 210 the last 2 numbers on the list, leaving the middle two numbers 8 and 8. The median is therefore 8. The first group had the higher median and the lower mean, while the second group had the lower median but the higher mean. Since everyone except one person in the first group scored the movie as 8 or higher, and since everyone in both groups except this one individual scored the movie 7 or higher, it might be reasonable to think that the one anomalous score of 1 is likely the result of something besides the quality of the movie. We might also note that this score is much further from the mean that any of the other scores, giving it significantly more effect on the mean than any other score. We might therefore choose to use the median, which limits the otherwise excessive weight given to this unusually low score when we calculate the mean. In this case we would say that the first group liked the movie better.
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RESPONSE --> Ok. self critique assessment: 3
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20:08:21 `q007. Suppose that in a certain office that ten employees make $700 per pay period, while five make $800 per pay period and the other two make $1000 per pay period. What is the mean pay per period in the office? What is the median?
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RESPONSE --> 10 * $700 = $7,000 5 * $800 = $4,000 2 * $1000 = $2,000 $13,000 / 17 = $764.71 Median: 700 confidence assessment: 3
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20:09:20 There are a total of 10 + 5 + 2 = 17 employees in the office. The total pay per pay period is 10 * $700 + 5 * $800 + 2 * $1000 = $13,000. The mean pay per period is therefore $13,000 / 17 = $823 approx.. The median pay is obtained by 'throwing out' the lowest 8 and the highest 8 in an ordered list, leaving the ninth salary. Since 10 people make $700 per period, this leaves $700 as the median.
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RESPONSE --> I didn't get a mean of $823. self critique assessment: 3
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20:11:25 `q008. In the preceding problem ten employees make $700 per pay period, while five make $800 per pay period and the other two make $1000 per pay period; we just found that the mean pay per period was $823. On the average, how much to the individual salaries differ from the mean?
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RESPONSE --> 10 * 123 5 * 23 2 * 177 $1630 / 17 = $95.88 confidence assessment: 3
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20:13:01 The mean was found in the preceding problem to be $823. The deviation of $700 from the mean is therefore $123, the deviation of $800 from the mean is $23 and the deviation of $1000 from the mean is $177. Since $700 is paid to 10 employees, $800 to five and $1000 to two, the total deviation is 10 *$123 + 5 * $23 + 2 * $177 = $1630. The mean deviation is therefore $1630 / 17 = $96, approx..
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RESPONSE --> Ok. self critique assessment: 3
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TIֿҠl assignment #010 010. `Query 10 Liberal Arts Mathematics II 06-26-2007
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20:23:26 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 --> $3 * 1 / 6 = $.50 $2 * 1 / 6 = $.33 $1 * 1 / 6 = $.17 $1.00
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20:25:51 ** 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 --> Ok.
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20:32:34 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 / 37 * 1 * 19 / 37 * -1 = $-.25
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20:32:53 ** 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 --> Ok.
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20:37:09 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 --> 120 / 20 = 6
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20:37:26 ** 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 --> Ok.
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20:37:33 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> None."