#$&* course Mth 152 November 27 - 7:16pm 010. Expectation
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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: OKAY ------------------------------------------------ Self-critique rating: OKAY ********************************************* 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: To get the expectation of times we multiply the probability by the amount of money. $100 is .005 out of 1,000,000 = .005 *1,000,000 = 5,000 times $1000 is .0002 out of 1,000,000 = .0002 * 1,000,000 = 200 times $10,000 is .00001 out of 1,000,000 = .00001 * 1,000,000 = 10 times confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 $10,000 prize would be .00001 * 1,000,000 = 10. Self-critique: OKAY ------------------------------------------------ Self-critique rating: OKAY ********************************************* Question: `q003. In the lottery of the preceding problem, if you were given a million tickets how much total money would you expect to win? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It all has to do with the decimal point and how many times the tickets. Above .0002 * 1,000,000 tickets equaled 200 times. If we have a million tickets we need to know the information we were given in the previous problem. Do we have enough information to do this problem? I don’t understand how to do this problem after doing the previous problem. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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.
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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: OKAY - seems logical ------------------------------------------------ Self-critique rating: OKAY ********************************************* 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: Alright, so I understand this problem because the decimal comes into play again. With a million tickets we got 800,000. If we divide that by the million for each ticket that’s 80 cents a ticket. You could expect to win 80 cents per ticket then. If we bought 9 million more tickets that would be 9 times what we had or 10 million. Instead of $800,000 we would make $8,000,000. All we need to do is divide by the overall tickets then which is 10 million. You could still expect to win 80 cents per ticket because all we did was take the same amount and have it grow by it’s same amount 9 times. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OKAY ------------------------------------------------ Self-critique rating: OKAY ********************************************* 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: Let’s restate: $100 by the probability of winning $100 $1000 by the probability of winning $1000 $10,000 by the probability of winning $10,000 Add all results - what is the sum? Looking at the previous problems we have: $100 times .005 = .50 $1,000 times .0002 = .20 $ 10,000 times .00001 = .10 The sum of these is .80 Aha! I understand one of the problems I didn’t before. The sum is exactly the same as the per ticket price of a million and ten million tickets. 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 $10,000 is worth 10 cents. Self-critique: OKAY ------------------------------------------------ Self-critique rating: OKAY ********************************************* Question: `q007. The following list of random digits has 10 rows and 10 columns: 3 8 4 7 2 3 0 8 3 9 1 8 3 7 3 2 9 1 0 3 4 3 3 0 2 1 4 9 8 2 4 3 4 9 9 2 0 1 3 9 8 3 4 1 3 0 5 3 9 7 2 4 7 4 5 3 7 2 1 8 3 6 9 0 2 5 9 5 2 3 4 5 8 5 8 8 2 9 8 5 9 3 4 6 7 4 5 8 4 9 4 1 5 7 9 2 9 3 1 2. Starting in the second column and working down the column, if we let even numbers stand for 'heads' and odd numbers for 'tails', then how many 'heads' and how many 'tails' would we end up with in the first eight flips? Answer the second question but starting in the fifth row and working across the row. Answer once more but starting in the first row, with the second number, and moving diagonally one space down and one to the right for each new number. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: #1 (starting in the second column) The first 8 flips would be 8,8,3,3,3,4,6,5 That would translate to HHTTTHHT Which equals four heads and four tails #2 (fifth row) The first 8 flips would be 8,3,4,1,3,0,5,3 That would translate to HTHTTHTT Which equals three heads and five heads #3 (first row, second number, diagonal) The flips would be 8,3,0,9,0,7,5,8 That would translate to HTHTHTTH Which equals four heads and fours tails confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OKAY ------------------------------------------------ Self-critique rating: OKAY ********************************************* 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: Oh my goodness, what a problem. Each column has 10 numbers in it so we will do the first three rows starting with the 4th column as instructed. The 4th column is: 7, 7, 0, 9, 1, 4, 0, 5, 6, 7 The 5th column is: 2, 3, 2, 9, 3, 5, 2, 8, 7, 9 The 6th column is: 3, 2, 1, 2, 0, 3, 5, 8, 4, 2. Each number simulates a number from the die so we can only work with the numbers 1-6, which means we are going to have to delete or erase some numbers. The columns become this: The 4th column is: 1, 4, 5, 6 The 5th column is: 2, 3, 2, 3, 5, 2 The 6th column is: 3, 2, 1, 2, 3, 5, 4, 2 This doesn’t make the first 20 numbers anymore so we need to add some from the 7th column. The 7th column is 0, 9, 4, 0, 5, 7, 9, 2, 5, 9 Numbers from 1-6: 4, 5, 2, 5 Since we only need two more numbers we will add 4 and 5 onto our list to make 20! Our list is: 1, 4, 5, 6, 2, 3, 2, 3, 5, 2, 3, 2, 1, 2, 3, 5, 4, 2 plus 4 and 5. Now to the questions: If we pair the first and the second rolls, what is the total? 1 + 4 = 5 If we pair the third and fourth rolls, what is the total? 5+6 = 11 If we continue in this way what are the 10 totals we obtain? 2, 3 = 5 2, 3 = 5 5, 2 = 7 3, 2 = 5 1, 2 = 3 3, 5 = 8 4, 2 = 6 4, 5 = 9 Our sums are as follows: 5, 11, 5, 5, 7, 5, 3, 8, 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: Yay - okay! Although I think you meant third and fourth rolls when you said “The second and third rolls were 5 and 6, which add up to 11.” ------------------------------------------------ Self-critique rating: okay ********************************************* 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: The results were 5, 11, 5, 5, 7, 5, 3, 8, 6, and 9. Which gives us four 5’s, one 6, and one 7. These three numbers made up the majority of the results being 6 out of the 10 which is more than 50 percent. 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: OKay ------------------------------------------------ Self-critique rating: OKAY " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!