#$&* course MTH 152 Time: 3:34 PMDate: 7/22 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: 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. Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - If I had the money to buy a million lottery tickets I would more than likely set up at least 20% of it in bonds….But regardless, my first instinct is to choose the following path. - How many I would expect to win the$100 - 1,000,000 * .005 = 5000 - How many I would expect to win $1000 - 1,000,000 * .0002 = 200 - How many I would expect to win $10,000 - 1,000,000 * .00001 = 10 - Leading to a total sum of at least $800,000 assuming that each of the expected tickets were to be successful in winning, taken from the math of - 5000 * 100 = 500,000 - 200 * 1000 = 200,000 10 * 10000 = 100,000 = 800,000 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 $10,000 prize would be .00001 * 1,000,000 = 10. Self-critique: Nailed it. ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. In the lottery of the preceding problem, if you were given a million tickets how much total money would you expect to win? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - ~800,000. See my notes in the previous problem. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 $800,000 of these winnings. Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. In the lottery of the preceding problem, if you bought a million tickets for half a million dollars would you most likely come out ahead? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - You would definitely break even and make a profit of about 300,000. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - If one expected to win $800,000 from purchasing 1,000,000, so division tells us around .80 = 80 cents. - Similarly, 8,000,000/10,000,000 equates to the same quotient. - One would think that you would stand better average with a larger amount of tickets bought. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - $100 = .005 - $1000 = .0002 - $10,000 = .00001 - = .8 - confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. Self-critique: - So is this the monetary value of each ticket? Or what it would be worth in a manner of if it won? ------------------------------------------------ Self-critique rating:
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Given Solution: 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'. Self-critique: - ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - The first twenty numbers within the set {1, 2, 3, 4, 5, 6} are as follows. {1, 4, 5, 6, 2, 3, 2, 3, 5, 2, 3, 2, 1, 2, 3, 5, 4, 2, 4, 5} - The first pair gives the sum of 5 - The second pair gives the sum of 11 - The third gives sum of 5 - Fourth gives the sum of 5 - Fifth gives sum of 7 - Sixth give sum of 5 - Seventh gives sum of 3 - Eighth gives sum of 7 - Ninth gives sum of 6 - Tenth gives sum of 9. - So we will constrain the set of {1, 4, 5, 6, 2, 3, 2, 3, 5, 2, 3, 2, 1, 2, 3, 5, 4, 2, 4, 5} effectively to - {5, 11, 5, 5, 7, 5, 3, 7, 6, 9} confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 = 9. The totals we obtain are therefore 5, 11, 5, 5, 7, 5, 3, 8, 6, and 9. Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - We have 6 numbers being either 5, 6, or 7 with regards to the set of sums. - Thus there are 6/10 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. Self-critique: - How is this pertinent? Any number with representation to anything in the universe is imperative to me now, so I’d like a more elaborate summation. Does this relate to the very nature of chaos? To me, anything that attributes to something that goes against mathematical belief that regardless of the original probability 15/36, 6/10 is still a bigger number that previously expected. ------------------------------------------------ Self-critique rating: