course Mth 151 assignment #006006. Cards
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22:19:24 `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 --> C(4,2) * C(48,3) confidence assessment: 3
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22:20:25 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 --> Ok. self critique assessment: 3
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22:21:23 `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 --> C(4,2) * C(4,2) * 44 confidence assessment: 3
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22:22:30 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 --> Ok. self critique assessment: 3
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22:24:07 `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 --> C(4,2) * C(4,3) confidence assessment: 3
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22:24:41 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 --> Ok. self critique assessment: 3
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22:27:06 `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 --> 3 * C(4,2) * C(4,3) confidence assessment: 3
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22:27:42 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 --> Ok. self critique assessment: 3
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22:30:30 `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 --> 13 * 12 * C(4,2) * C(4,3) confidence assessment: 3
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22:31:18 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 --> Ok. self critique assessment: 3
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22:32:32 `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 --> 4 * C(13,5) confidence assessment: 3
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22:32:58 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 --> Ok. self critique assessment: 3
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22:35:21 `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 --> 4 * 4 * 4 * 4 * 4 confidence assessment: 3
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22:36:29 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 --> Ok. self critique assessment: 3
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22:37:21 `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 --> 10 * 4^5 confidence assessment: 3
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22:38:28 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 --> Ok. self critique assessment: 3
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assignment #007 007. Introduction to probability Liberal Arts Mathematics II 06-17-2007
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22:49:05 `q001. Note that there are 5 questions in this assignment. Suppose we toss two dice. How many possible outcomes are there for the numbers on the two dice? How many of these outcomes given a total greater than 4? What therefore is the probability that the total on a toss of two dice is greater than 4?
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RESPONSE --> 6 * 6 possible outcomes 36 - (6 less than four) = 30 greater than four 30 / 36 = .8333 = 83.3% confidence assessment: 3
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22:50:37 There are 6 possible outcomes on the first die and 6 on the second. The number of possible outcomes is therefore 6 * 6 = 36 (e.g., (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), etc.). The six outcomes (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1) are all the possible outcomes which are 4 or less. The remaining 36 - 6 = 30 outcomes are all greater than 4. It follows that the probability of obtaining a result greater than 4 is 30 / 36 = 5/6 or .833... or 83.33... %.
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RESPONSE --> Ok. self critique assessment: 3
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22:51:25 `q002. What are the odds that the total on a toss of two dice will be greater than 4?
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RESPONSE --> 30 : 6 confidence assessment: 3
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22:52:17 As seen in the previous question, there are 30 possible outcomes or the total is greater than 4 and 6 outcomes where the outcome is less than or equal to 4. The odds in favor of any event are expressed as odds = number in favor to number opposed. {}In this case the odds of a result greater than 6 are 30 to 6, which reduces to 5 to 1. These odds can also be expressed as 30 : 6 or 5 : 1.
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RESPONSE --> Ok. self critique assessment: 3
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22:55:51 `q003. Suppose we have three boxes, one containing balls numbered 1-15, another tiles labeled a-z, and another one ring for each of the seven colors of the rainbow. How many possibilities are there for the collection of items we obtain if we choose one item from each box? How many of these possibilities contain an odd number, a consonant and a 'blue-type' color (blue, indigo or violet)? If we choose one item from each box, what is the probability that our collection will contain an odd number, a consonant and a 'blue-type' color (blue, indigo or violet)?
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RESPONSE --> 15 * 26 * 7 possibilities from each box 8 * 21 * 3 possibilities of odd, consonant, and blue 504 / 2730 = .1846 = 18.5% confidence assessment: 3
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22:56:56 There are 15 * 26 * 7 possibilities for the collection obtained by choosing one item from each box. There are 8 odd numbers, 21 consonants if we include 'y' and three 'blue-type' colors. So there are 8 * 21 * 3 possible combinations consisting of an odd number, a consonant and a 'blue-type' color. The desired probability is therefore ( 8 * 21 * 3) / ( 15 * 26 * 7).
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RESPONSE --> Ok. self critique assessment: 3
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23:05:02 `q004. How many possible 5-card hands can be dealt from a 52-card deck? How many of these hands contain exactly one pair? What therefore is the probability that a hand dealt from a well-shuffled deck will contain exactly one pair? What are the odds in favor of such a deal resulting in a hand with exactly one pair?
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RESPONSE --> C(52,5) possible hands 48 * 44 * 40 possible 13 * C(4,2) * (48*44*40 / 3!) hands with one pair [13 * C(4,2) * (48*44*40 / 3!)] / C(52,5) probability of a pair confidence assessment: 3
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23:06:43 There are C(52, 5) possible hands. There are C(4,2) ways to get a pair of any given denomination and 13 denominations, and there are then 48 choices for the first of the remaining three cards, 44 for the second and 40 for the third. Any given combination of the three remaining cards can be chosen in any of 3! ways so there are 48 * 44 * 40 possible choices of these 3 cards. Thus there are 13 * C(4, 2) * ( 48 * 44 * 40 / 3!) hands containing of exactly one pair. The probability of exactly one pair is therefore [ 13 * C(4,2) * (48 * 44 * 40) / 3! ] / C(52,5). This expression is easily enough written out and reduced [ 13 * 6 * 48 * 44 * 40 / 3! ] / [ 52 * 51 * 50 * 49 * 48 / (5*4*3*2*1) ] = 13 * 6 * 48 * 44 * 40 * 5 * 4 * 3 * 2 * 1 / [ 52 * 51 * 50 * 49 * 48 * 3 * 2 * 1 ] = 6 * 44 * 4 * 5 * 4 / [ 4 * 51 * 5 * 49 ] = 6 * 44 * 4 / [ 51 * 49 ] = (24 * 44) / (51 * 49) = (8 * 44) / ( 17 * 49) = .42 approx. Further explanation: This builds on the ideas of permutations and combinations developed in previous assignments. To get a hand you have to 'choose' 5 of the 52 cards, and order doesn't matter. There are C(52, 5) ways of doing this To get a pair of 5's, for example, you have to choose 2 of the four 5's in the deck. There are C(4, 2) ways to do this. There are 13 denominations (2's, 3's, 4's, ..., Queens, Kings). The pair could be from anyone of these denominations so there are 13 * C(4,2) ways to get a pair. After choosing the pair, you can't choose another card of that denomination or you would no longer have a pair. That leaves only 48 cards from which to choose the third. You already have a pair so the next card can't match the denomination of the third, so you have only 44 cards from which to choose the fourth. Similar reasoning shows that there are only 40 cards from which to choose the fifth card. These last three cards could have been chosen in any of 3! orders. So the number of ways of choosing the last three cards is 48*44*40/3!. So by the fundamental counting principle, since we have to choose a pair and then choose three other cards not matching the denomination of the pair or of one another, the number of possible ways to accomplish this is 13 * C(4,2) * 48 * 44 * 40 / 3!.
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RESPONSE --> Ok. self critique assessment: 2
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23:12:47 `q005. If a fair coin is tossed five times, how many possible outcomes are there? How many of these outcomes will have exactly 3 'heads'? What therefore is the probability that on 5 tosses of a fair coin we will obtain exactly 3 'heads'?
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RESPONSE --> 2 * 2 * 2 * 2 * 2 = 32 total possible outcomes C(5,3) possibile outcomes with three heads C(5,3) / 32 = .3125 = 31.3% of 3 heads confidence assessment: 3
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23:14:21 On 5 flips there are C(5,3) = 10 possible outcomes with exactly 3 'heads'. There are 2^5 = 32 possible outcomes altogether. The probability of 3 'heads' on 5 flips is therefore 10 / 32 = 5/16 = .3125.
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RESPONSE --> Ok. self critique assessment: 3
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assignment #006 006. `query 6 Liberal Arts Mathematics II 06-17-2007
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23:26:58 Query 12.1.6 8 girls 5 boys What is the probability that the first chosen is a girl?
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RESPONSE --> 8 / 13 = .6153 = 61.5%
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23:27:42 ** Assuming the choice is completely random there are 13 possible choices, 8 of which are female so we have P(female) = 8 / 13 = .6154, approx. **
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RESPONSE --> Ok.
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23:29:55 Query 12.1.12 3 fair coins: Probability and odds of 3 Heads.
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RESPONSE --> 1 / 8 = .125 = 12.5%
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23:30:38 ** There are 8 equally likely possible outcomes when flipping 3 fair coins. You can list them: hhh, hht, hth, htt, thh, tht, tth, ttt. Or you can use the fact that there are 2 possibilities on each flip, therefore 2*2*2 = 2^3 = 8 possible outcomes. Only one of these outcomes, hhh, consists of 3 heads. The probability is therefore P(3 heads) = # of outcomes favorable/total number of possible outcomes = 1 / 8. The odds in favor of three heads are Odds ( 3 heads ) = # favorable to # unfavorable = 1 to 7. **
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RESPONSE --> Ok.
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23:32:00 Query 12.1.20 P(pink) from two pink parents (Rr and Rr) What is the probability of a pink offspring.
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RESPONSE --> 2 / 4 = .5 = 50%
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23:32:41 ** The genes R and r stand for the red and white genes. A pink offspring is either Rr or rR. RR will be red, rr white. R r R RR Rr r rR rr shows that {RR, Rr, rR, rr} is the set of equally likely outcomes. We season two of the four possible outcomes, rR and Rr, will be pink. So the probability of pink offspring is 2/4 = 1 / 2. **
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RESPONSE --> Ok.
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23:34:35 Query 12.1.33 cystic fibrosis in 1 of 2K cauc, 1 in 250k noncauc What is the empirical probability, to 6 places, that a randomly chosen non-Caucasian newborn will have cystic fibrosis?
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RESPONSE --> 1 / 250000 = .000004
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23:34:45 ** There is 1 chance in 250,000 so the probability is 1 / 250,000 = 4 * 10^-6, or .000004. **
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RESPONSE --> Ok.
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23:35:49 12.1.40 Cc genes carrier, cc has disease; 2 carriers first child has disease **** What is the probability that the first child has the disease?
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RESPONSE --> 1 / 4 = .25 = 25%
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23:36:20 ** If cc has the disease, then the probability that the first child will have the disease is 1/4. **
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RESPONSE --> Ok.
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23:37:02 What is the sample space for this problem?
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RESPONSE --> [CC, Cc, cC, cc]
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23:37:28 ** The sample space is {CC, Cc. cC, cc}. **
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RESPONSE --> Ok.
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23:39:51 12.1.60 36 in class, 3 chosen **** What is the probability that the choice will be the given three people in any order?
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RESPONSE --> 6 / P(36,3) = 6 / 40000 = .00015
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23:41:01 ** There are P(36,3) possible ordered choices of 3 people out of the 36. P(36,3) = 36! / (36-3)! = 36! / 33! = (36*35*34*33*32*31*...*1) / (33*32*31*...*1) = 36*35*34=40,000 or so. The probability of any given choice is therefore 1 / P(36,3) = 1/40,000 = .000025, approx.. For any given set of three people there are six possible orders in which they can be chosen. So the probability of the three given people, in any order, is 6 * probability of a given order = 6 / P(36,3) = 6/40,000 = .00015. Alternatively we can say that we are choosing 3 of 36 people without regard for order, so there are C(36,3) possibilities and the probability of any one of them is 1 / C(36,3). **
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RESPONSE --> Ok.
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23:44:21 Query 12.1.75 digits 1, 2, ..., 5 rand arranged; prob even, prob digits 1 and 5 even What is the probability that the resulting number is even and how did you obtain your answer?
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RESPONSE --> P(2,2) = 2 choices for even P(5,2) = 20 choices 2 / 20
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23:45:05 ** The number will be even if it ends in 2 or 4. There are 5 possible ending numbers. So 2 of the 5 possible ending numbers are even and the probability of an even number is 2/5. We analyze in two ways the number of ways to choose a number with digits 1 and 5 even. First way: There are 5! = 120 possible arrangements of the 5 digits. There are only two possible even digits, from which we will choose digit 1 and digit 5. The order of our choice certainly matters, since a different choice will give us a different 5-digit number. So we are choosing 2 digits from a set of 2 digits, where order matters. We therefore have P(2, 2) = 2*1 / 0! = 2 ways to choose these digits. The remaining 3 digits will comprise digits 2, 3 and 4. We are therefore choosing 3 digits from a set of 3, in order. There are P(3, 3) = 3*2*1/0! = 6 ways to do so. To obtain our number we can choose digits 1 and 5, then digits 2, 3 and 4. There are P(2, 2) * P(3, 3) = 2 * 6 = 12 ways to do this. So the probability that digits 1 and 5 are even is 12 / 120 = 1/10. Second way: A simpler solution looks at just the possibilities for digits 1 and 5. There are P(2, 2) = 2 choices for which these digits are even, and P(5, 2) = 20 total choices for these two digits. The probability that both will be even is therefore 2/20 = 1/10, the same as before. **
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RESPONSE --> Ok.
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assignment #007 007. `query 7 Liberal Arts Mathematics II 06-17-2007
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23:54:35 12.2.6 single die, p(odd or <5). What is the probability of getting an odd result or a result < 5?
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RESPONSE --> 5 / 6
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23:57:48 ** there are 3 possible odd outcomes and four outcomes less that 5 which would add up to 7 outcomes, except that 2 of the outcomes < 5 are alrealdy odd and won't be counted. Thus the number of outcomes which are odd or less that 5 is 3 + 4 - 2 = 5 (this expresses the rule that n(A U B) = n(A) + n(B) - n(A ^ B), where U and ^ stand for union and intersection, respectively ). Thus the probability is 5/6. In terms of the specific sample space: The sample space for the experiment is {1, 2, 3, 4, 5, 6}. Success corresponds to events in the subset {1, 2, 3, 4, 5}. There are 6 elements in the sample space, 5 in the subset consisting of successful outcomes. Thus the probability is 5/6. **
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RESPONSE --> Ok.
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00:05:14 Query 12.2.15 drawing neither heart nor 7 from full deck
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RESPONSE --> 36 / 52 = 9 / 13 4 to 9
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00:06:27 ** The sample space consists of the 52 cards in a full deck. There are 39 cards that aren't hearts, four 7's but only three aren't hearts so there are 36 cards that aren't hearts or seven. The probability is therefore 36/52 = 9/13. The odds in favor of the event are 16 to 36 (number favorable to number unfavorable), which in reduced form is 4 to 9. **
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RESPONSE --> Ok.
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00:14:29 12.2.24 prob of black flush or two pairs
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RESPONSE --> 2 * 1287 = 2574 possible black flushes 123, 552 possible two pairs 123552 + 2574 = 126126 of either the flush or pair 126126 / 2598960 = .0485 = 4.9%
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00:16:39 ** There are C(13,5) = 1287 ways to get a flush in a given suit--gotta choose the 5 cards from the 13 cards in that suit. There are two black suits so there are 2 * 1287 = 2574 possible black flushes. As the text tells you there are 123,552 ways to get two pairs. You can incidentally get this as 13 * C(4, 2) * 12 * C(4, 2) * C(44, 1) / 2 (2 of the 4 cards in any of the 13 denominations, then 2 of the 4 cards in any of the remaining 12 denominations, divide by 2 because the two denominations could occur in any order, then 1 of the 44 remaining cards not in either of the two denominations. There is no way that a hand can be both a black flush and two pairs, so there is no overlap to worry about (i.e., n(A and B) = 0 so n(A or B) = n(A) + n(B) - n(A and B) = n(A) + n(B) ). Thus there are 123,552 + 2574 = 126,126 ways to get one or the other. The probability is therefore 126,126 / 2,598,960 = .0485, approx. **
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RESPONSE --> Ok.
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00:21:33 12.2.33 x is sum of 2-digit numbers from {1, 2, ..., 5}; prob dist for random vbl x
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RESPONSE --> X P(X) 3 .1 4 .1 5 .2 6 .2 7 .2 8 .1 9 .1
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00:22:25 ** If 2 different numbers are chosen from the set (1, 2, 3, 4, 5} then the sum 3 can appear only as 1+2. 4 can appear only as 1+3, assuming numbers can't be repeated (so, for example, 2+2 is not allowed). 5 can occur as 1+4 or as 2+3. 6 can occur as 1+5 or as 2+3. 7 can occur as 2+5 or as 1+6. 8 can occur only as 3+5. 9 can occur only as 4+5. Of the 10 possible combinations, the sums 3, 4, 8 and 9 can occur only once each, so each has probability .1. The sums 5, 6 and 7 can occur 2 times each, so each has probability .2. The possible sums are as indicated in the table below. 1 2 3 4 5 1 3 4 5 6 2 5 6 7 3 7 8 4 9 This assumes selection without replacement. There are C(5, 2) = 10 possible outcomes, as can be verified by counting the outcomes in the table. 3, 4, 8 and 9 appear once each as outcomes, so each has probability 1/10. 5, 6 and 7 appear twice each as outcomes, so each has probability 2/10. x p(x) 3 .1 4 .1 5 .2 6 .2 7 .2 8 .1 9 .1 **
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RESPONSE --> Ok.
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00:30:28 Query 12.2.36 n(A)=a, n(S) = s; P(A')=? What is the P(A')?
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RESPONSE --> P(A') = n(A') / n(S) = (S - A) / S
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00:32:42 ** A' is everything that is not in A. There are a ways A can happen, and s possibilities in the sample space S, so there are s - a ways A' can happen. So of the s possibilities, s-a are in A'. Thus the probability of A' is P(A') = n(A') / n(S) = (s - a) / s. **
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RESPONSE --> Ok.
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00:33:16 Query 12.2.42 spinners with 1-4 and 8-10; prob product is even
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RESPONSE --> 10 / 12 = .8333 = 83.3%
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00:33:35 ** The first number can be 1, 2, 3 or 4. The second can be 8, 9 or 10. There are therefore 4 * 3 = 12 possible outcomes. The only way to get an odd outcome is for the two numbers to both be odd. There are only 2 ways that can happen (1 * 9 and 3 * 9). The other 10 products are all even. So the probability of an even number is 10 / 12 = 5/6 = .833... . Alternatively we can set up the sample space in the form of the table 8 9 10 1 8 9 10 2 16 18 20 3 24 27 30 4 32 36 40 We see directly from this sample space that 10 of the 12 possible outcomes are even. **
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RESPONSE --> Ok.
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