course Phy 121 Important Message received at access site.
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10:18:38 Query 11.5.10 ways to get different number on each of two fair dice In how many ways can you get a different number on each of two fair dice? How would you obtain the answer to this question if you were marooned on a desert island with no book and no table?
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RESPONSE --> There are 35 different ways to get a sum of at least 3 on two different fair dice. The only possible sum lower than 3 is a 2, which is obtainable by only one method (aka snake eyes). There is a total possibility of 36 different rolls... but only 35 with a sum of at least 3. If I was on a desert island... and assuming I had some fair dice... I could either roll them or the smarter method would be to draw a table chart in the sand, with numbers 1 to 6 on the horizontal and numbers 1 to 6 on the vertical and compare... or I could simply multiply the total number and then solve the math to figure.
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10:19:27 ** On two fair dice you have 6 possible outcomes on the first and 6 on the second. By the Fundamental Counting Principle there are therefore 6 * 6 = 36 possible outcomes. We can list these outcomes in the form of ordered pairs: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6) (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6) (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) Of these 36 outcomes there are six that have the same number on both dice. It follows that the remaining 3 - 6 = 30 have different numbers. So there are 30 ways to get different numbers on the two dice. Note that your chance of getting different numbers is therefore 30 / 36 = 5/6 = .8333... .**
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RESPONSE --> Okay, I think I might have been answering a specific question or the wrong question. I was giving my answer for problem 10 on 11.5
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10:24:25 Query 11.5.12 bridge hands more than one suit How many bridge hands contain more than one suit?
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RESPONSE --> I cannot honestly answer this response. the question 11.5.12 is different in my book.... In my book the question is a different number on each die for 11.5.12 and I do not have any questions in the whole section dealing with the game of bridge... there are some dealing with poker but not bridge. And I never have played bridge so I honestly do not really know where to begin on solving this.
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10:25:12 ** There are 13 cards in a bridge hand. The number of possible bridge hands is therefore C(52, 13). There are 13 cards of each suit. The number of possible bridge hands with all cards in a given suit is therefore C(13, 13) = 1 (common sense is that there is only one way to get all 13 cards in a given suit, which is to get all the cards there are in that suit). Since there are 4 suits there are 4 * C(13, 13) = 4 * 1 = 4 possible one-suit hands. Of the number of hands having more than one suit is C(52, 13) - 4. **
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RESPONSE --> Okay, I've saved this for review... it was different than the materials presented in my textbook.
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10:32:40 11.5.20 # subsets of 12-elt set with from 3 to 9 elts? How many subsets contain from three to nine elements and how did you obtain your answer (answer in detail)?
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RESPONSE --> Okay, this is actually question 36 rather than question 20 in regards to my textbook.... I was confused in regards to finding a means inbetween 3 and 9 so I am going to accept this answer is wrong for the purpose of learning the instructor's means of solving.
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10:33:34 11.5.30 10200 ways to get a straight Verify that there are in fact 10200 ways to get a straight in a 5-card hand.
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RESPONSE --> Okay
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10:34:30 ** There are 9 choices for the denomination of the lowest card in a straight, which gives 36 cards that could be the low card. However but if aces can be high or low there are 40. There are then four choices for the next-higher card, four for the next after that, etc., giving 40*4*4*4*4 possibilities. **STUDENT COMMENT: I don 't understand this one . Idon't see where you get the 9 from. INSTRUCTOR RESPONSE: Cards run from 2 through 10, then the four face cards, then the ace. You need five consecutive cards to make a straight. The highest possible straight is therefore 10, Jack, Queen, King and Ace. The lowest is 2, 3, 4, 5, 6. The lowest card of the straight can be any number from 2 through 10. That is 9 possibilities. In some games the ace can be counted as the low card, below the 2, as well as the high card. In that case there would be one more possibility for a straight, which could then consists of denominations 1, 2, 3, 4, 5. *&*&
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RESPONSE --> I have copied this to study
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10:39:21 11.5.36 3-digit #'s from {0, 1, ..., 6}; how many mult of 25?
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RESPONSE --> If no digits may be used more than once then there are a total of... 10 025, 125, 150, 250, ,325, 350, 425, 450, 625, 650 (If not digits can be used more than once) If the digits may be used more than once there are a total of....20 025, 050, 100, 125, 150, 200, 225, 250, 300, 325, 350, 400, 425, 450, 500, 525, 550, 600, 625, 650
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10:39:58 ** A 3-digit number from the set has six choices for the first digit (can't start with 0) and 7 choices for each remaining digit. That makes 6 * 7 * 7 = 294 possibilities. A multiple of 25 is any number that ends with 00, 25, 50 or 75. SInce 7 isn't in the set you can't have 75, so there are three possibilities for the last two digits. There are six possible first digits, so from this set there are 6 * 3 = 18 possible 3-digit numbers which are multiples of 25. A listing would include 100, 125, 150, 200, 225, 250, 300, 325, 350, 400, 425, 450, 500, 525, 550, 600, 625, 650. Combinations aren't appropriate for two reasons. In the first place the uniformity criterion is not satisfied because different digits have different criteria (i.e., the first digit cannot be zero). In the second place we are not choosing object without replacement. The fundamental counting principle is the key here. STUDENT SOLUTION AND INSTRUCTOR RESPONSE: All I can come up with is C(7,2)=21. & choices of #s and the # must end in 0 or 5 making it 2 of the 7 choices INSTRUCTOR RESPONSE: Right reasoning on the individual coices but you're not choosing just any 3 of the 7 numbers (uniformity criterion isn't satisfied--second number has different criterion than first--so you wouldn't use permutations or combinations) and order does matter in any case so you wouldn't use combinations. You have 7 choices for the first and 2 for the second number so there are 7 * 2 = 14 multiples of 5. **
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RESPONSE --> Okay, I think I'm having trouble with this section. I plan to study it further before the test.
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10:44:13 Query 11.5.48 # 3-digit counting #'s without digits 2,5,7,8?
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RESPONSE --> Not sure what you mean regarding this question... I think you mean.... number of 3 digit counting numbers without digits 2,5,7,8? So do you mean.... 0,1,3,4,6,9? (aka 5 choices)? (this is different than question 48 of 11.5 in my textbook) Okay, the number of 3 digit counting numbers from a set of 5 choices for the fist digit (as you have stated we cannot start with 0) and 6 choices of each remaining (can use 0). That makes 5*6*6= 180 different possibilites.
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10:44:23 ** there are 5 possible first digits (1, 3, 4, 6, or 9) and 6 possibilities for each of the last two digits. This gives you a total of 5 * 6 * 6 = 180 possibilities. **
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RESPONSE --> Okay, got that one correct.
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10:45:57 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I feel that I need to study this a great deal more. I think I'm sorta on track but easily confused. Question... do some of the text books have entirely different questions for certain section numbers? Some of the questions asked to me are different than the corresponding number shows in my textbook (and thus my ""written down"" homework)? 2nd Question... would it be good to take the practice tests in the back for Chapter 11 to prep for the test?
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