course Mth 152 |茠ߋC~assignment #006 006. Cards Liberal Arts Mathematics II 03-02-2008
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09:29:11 `q001. Note that there are 8 questions in this assignment. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's?
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RESPONSE --> There are 52 cards in a deck and since there are 4- 5 cards and you only want 2 you take 2 of the 5 cards out leaving 50 cards to deal with. Since there are 10 possible hands out of 50 cards, there is 2/10 possibilities or 1/5 possiblities confidence assessment: 2
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09:36:25 In order to get a hand containing exactly two 5's we must select, without regard for order, two of the four 5's, then we must select the remaining 3 cards from the 48 cards that are not 5's. There are C(4,2) ways to select two 5's from the four 5's in the deck. There are C(48,3) ways to select the 3 remaining cards from the 48 cards which are not 5's. We must do both, so by the Fundamental Counting Principle there are C(4,2) * C(48, 3) ways to obtain exactly two 5's.
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RESPONSE --> I did not work it like that I took 2 of the #5 cards out and then got how many hands would be available in the 50 remaining cards and then got now many could contain 2 of the #5 cards but now that I type this explanation out, what I did does not make any sense to me becuase a 5 could have been in one of the hands and the other in one of the other hands. I see now that you work the 5 cards alone and then the remaining deck to get the probability. self critique assessment: 2
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09:43:57 `q002. Using a standard deck of cards, in how many ways is it possible to get a hand containing exactly two 5's and exactly two 9's?
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RESPONSE --> 52 cards in deck, there are 4 #5 cards and 4 #9 cards. C(4,2) for the 5 - 2 cards in the hand C(4,2) for the 9 - 2 cards in the hand C(44,1) for the remaining card to give a total of 5 cards for the hand. confidence assessment: 2
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09:44:25 There are C( 4, 2) ways to select exactly two 5's and C(4, 2) ways to select exactly two 9's. There are 44 remaining cards which are neither 9 nor 5. The total number of possible ways is therefore C(4, 2) * C(4, 2) * 44.
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RESPONSE --> That is how I worked it this time. It makes a lot more sense. self critique assessment: 2
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09:46:07 `q003. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three 9's?
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RESPONSE --> 52 cards in a deck 4 # 5s 4 # 9s C(4,2) - # 5 cards C(4,3) - # 9 cards C(4,2) * C(4,3) confidence assessment: 2
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09:47:11 There are C(4, 2) ways to get two 5's and C(4, 3) ways to get three 9's. It follows that the number of ways to get the specified 'full house' with two 5's and three 9's is C(4,2) * C(4,3).
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RESPONSE --> How I worked it too. I didn't incorporate any other cards other than the 9s and 5s since a hand consist of 5 cards. self critique assessment: 2
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09:48:59 `q004. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two 5's and three identical face cards?
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RESPONSE --> 52 cards in deck 4 # 5s 4 Kings 4 Queens 4 Jacks C(4,2) for #5s C(4,3) for King, Queen & Jack C(4,2) * C(4,3) confidence assessment: 2
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09:50:02 There are C(4,2) ways to get two 5's and C(4,3) ways to get three of any given face card. There are 3 possible face cards, so the number of ways to get a 'full house' consisting of two 5's and three identical face cards is 3 * C(4,2) * C(4,3).
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RESPONSE --> I started to do each of the face cards and mutiply each set but realized that there are only 5 cards in a hand and to use the other face cards, you would have more than just 5 cards in your hand. self critique assessment: 2
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09:51:22 `q005. Using a standard deck of cards, in how many ways is it possible to get a 'full house' consisting of two of one denomination and three of another?
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RESPONSE --> 52 cards in a deck 4 cards for each demonination C(4,2) for one demon. C(4,3) for another demon C(4,2) * C(4,3) confidence assessment: 2
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09:52:20 For any two denominations, e.g., a pair of 5's and three 9's, there are C(4,2) * C(4,3) different full houses. There are 13 possible choices for the pair, which leaves 12 possible choices for the three-of-a-kind, which to by the Fundamental Counting Principle makes 13 * 12 possible choices for the two denominations. Note that order does matter when choosing the denominations because one has to be the pair and the other the three-of-a-kind. Again by the Fundamental Counting Principle we conclude that there are 13 * 12 * C(4,2) * C(4,3) possible full houses.
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RESPONSE --> That is how I worked it. self critique assessment: 2
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09:53:21 `q006. Using a standard deck of cards, in how many ways is it possible to get a 'flush' consisting of five cards all of the same suit?
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RESPONSE --> 0 because there are only 4 matching cards confidence assessment: 2
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09:55:50 There are 13 cards in any given suit, so there are C(13,5) hands consisting of all 5 cards in that suit. There are 4 suits, so there are 4 * C(13,5) possible flushes.
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RESPONSE --> I did not know what a suit was, but I see how you worked it. Since there are 4 possible suits, you mutiple the hand by 4 self critique assessment: 2
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09:58:08 `q007. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of exactly one each of the denominations 5, 6, 7, 8 and 9?
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RESPONSE --> 52 cards in a deck 13 suits 4 #5 cards 4 #6 cards 4 #7 cards 4 #8 cards 4 #9 cards C(4,1) * C(4,1) * C(4,1) * C(4,1) * C(4,1) or could you do it 5*4*3*2*1???? confidence assessment: 2
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09:59:25 There are four 5's, four 6's, four 7's, four 8's and four 9's. So there are 4 * 4 * 4 * 4 * 4 = 4^5 possible straights from 5 to 9.
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RESPONSE --> I wasn't sure if you could do it with fundamental counting but I see that you can't. self critique assessment: 2
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10:01:39 `q008. Using a standard deck of cards, in how many ways is it possible to get a 'straight' consisting of five cards of consecutive denominations, assuming that the 'ace' can be either 'high' or 'low'?
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RESPONSE --> 4*4*4*4*4 = 4^5 you would be using one cards from each possible deomination confidence assessment: 2
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10:02:46 There are 10 possible denominations for the 'low' card of a straight (any card from ace through 10 can serve as the 'low' card of a straight). There are 4^5 possible straights for each possible low card. It follows that there are 10 * 4^5 possible straights.
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RESPONSE --> I did not allow for the face cards to get 10 possible demonations. self critique assessment: 2
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