#$&* course MTH 152 Time: 10:52 PMDate: 7/16/2013 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `athere are 3 possible odd outcomes and four outcomes less that 5 which would add up to 7 outcomes, except that 2 of the outcomes < 5 are alrealdy odd and won't be counted. Thus the number of outcomes which are odd or less that 5 is 3 + 4 - 2 = 5 (this expresses the rule that n(A U B) = n(A) + n(B) - n(A ^ B), where U and ^ stand for union and intersection, respectively ). Thus the probability is 5/6. In terms of the specific sample space: The sample space for the experiment is {1, 2, 3, 4, 5, 6}. Success corresponds to events in the subset {1, 2, 3, 4, 5}. There are 6 elements in the sample space, 5 in the subset consisting of successful outcomes. Thus the probability is 5/6. ** Self-critique: - I started out a bit off, but I believe I’m on the same page as you presently. ------------------------------------------------ Self-critique Rating: Okay. See comment above. ********************************************* question: Query 12.2.15 drawing neither heart nor 7 from full deck YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - There are 52 cards total in a full deck of cards. - 13 of which are comprised of the “hearts” suit. - A ‘7’ card could be comprised of any of the four suits, so 4 cards. - Hearts: {A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K} = 13 7s: { 7 (hearts) , 7 (diamonds) , 7 (clubs) , 7 (spades) } = 4 - Thus only one card would be in union with the two sets given where A = Hearts and B = 7s. - - Hearts: {7} 7s: { 7 (hearts) } - - So by following the rule n(A U B) = n(A) + n(B) - n(A ^ B) n(A U B) = n(13) + n(4) - n(1) = 16 - Therefore there is a 16/52 = 4/13 =0.3076923077 = 31% chance of drawing neither a heart or a 7 from a full deck. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. ** Self-critique: - Note that I now reconsider my answer and make a small change by saying that there are 13 heart cards, but 52 - 13 = 39 cards that are not hearts and four 7’s, but three of those are either clubs/spades/diamonds so 39 - 3 = 36, thus there are 36/52 = 18/26 = 9/13 or .6923 = 69 % chance of getting neither hearts nor 7’s. ------------------------------------------------ Self-critique Rating: Okay. Re-directed myself, see comment above. ********************************************* question: 12.2.24 prob of black flush or two pairs YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - I first list the definition of a flush in poker, which is basically a hand consisting of all 5-cards being of the same suit in any order. So one could say this is the unordered collection of 5 cards from a 13 card suit. - C(13, 5) = 1287 combinations, but this is for just one suit in particular. To express the number of possibilities in each black suit it must be defined as 2 * C(13,5) = 2,574 - - ***Note that I say 2 * C(13,5) because when I calculated the number derived from C(13,5) I only had any one particular suit listed. There are 2 black suits, so therefore there are 2 * C(13,5) = 2574 possible outcomes*** - - I secondly list the definition of a two-pair hand in poker, which is basically a hand consisting of two pairs of cards with similar denominations but of different suits plus one card that isn’t of the denominations listed but can be any of the suits. - I try to reason out how many combinations of a two-pair hand could be made. - First I consider that for each pair, there are 13 denominations in one suit of which another pair in a different suit could be made. - I decide to write it as follows. - - H = {A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K} D = {A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K} C = {A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K} S = {A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K} - I ask myself “How many pairs can be made from a full deck of cards?” - Before I go any further, I glance downward to make sure I’m not flying off the rails. - I see that I’m going in the right direction, but I decide to interpret my words a little differently. - Deducing from the above listing of suits/denominations, there are C(4,2) ways to choose a pair of cards from four of the suits. Because these pairs can be any of the 13 denominations, this becomes 13 * C(4,2) to express the first choice of a pair (2) from any of the four suits (4) multiplied by the number of possible denominations (13). - - Because one of these 13 denominations has been chosen as a pair, I now list that for the second pair there are 12 possible denominations available so it becomes 12 * C(4,2). - Now the two pairs of cards from either of the 4 suits from any of the 13 denominations is expressed by saying { [13 * C(4,2) ] * [ 12 * C(4,2)] } / 2! - - The division here is for the purpose of showing that, for the two pairs, the denominations could be 2 cards of any possible order. (There are really only two denominations to be considered. Even though there are four cards in a two-pair hand that match, they will be of the same denomination but different suit, so 2 matching pairs. - - There was a collection of 52 cards, but 2 pairs of those cards (4) in two suits (4 *2) have already been chosen. So we say that there are 52 - (4 * 2) = 44 cards remaining in which to choose one. - So there are C(44,1) ways to choose the remaining card. - - Altogether we can express this as - - { [13 * C(4,2)] * [12 * C(4,2)] * C(44,1)} / 2! - - Which is reduced to 123,552 two-pair hands. Now we have numbers to work with. Since there are 2,574 possible black flushes and 123,552 two-pair hands, **** I find that it is impossible to have a black flush and a two pair hand at once, so I simply plan to add the number of black flushes with the number of two-pair hands, and divide the sum by the total number of hands. - 2 * C(13,5) + { [13 * C(4,2)] * [12 * C(4,2)] * C(44,1)} / 2! = 126,126 - 126,126 / C(52,5) = 126,126 / 2,598,960 = 0.0485294118 or 4.9% chance to get either. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThere 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. ** Self-critique: This one was a doozy for me, but I liked it. ------------------------------------------------ Self-critique Rating: Ok. Expect a question like this one for review; I may send a few extra questions from this assignment for you to analyze via Question Form/Email.v b ********************************************* question: 12.2.33 x is sum of 2-digit numbers from {1, 2, ..., 5}; prob dist for random vbl x YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - I wasn’t sure as to what the question had referred to, or why it would be useful. If the reason behind this isn’t mentioned in the rest of the section here, please provide an example of which this would be used. - I instead provided a few theories on what needed to be completed, as I was unsure how to answer. - I hypothesized the following: - - Random Variable ‘x’ Combination - 1 0 - 2 1, 1 - 3 1, 2 - 4 (1,3) , (2,2) - 5 (3,2) , (1,4) - - In the next table I had constructed, I provided a list of possible outcomes and their respective sum (1,1) = 2, (1,2) = 3 , (1, 3) = 4 . . . (5, 4) = 9 , (5, 5) = 10 - I recorded that there were 9 different numbers for the above tables {2, 3, 4, 5, 6, 7, 8, 9, 10} - I also recorded that there were 25 possible outcomes given by the above information. ********************************* - confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf 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 ** Self-critique: - See my comment in Your Solution. ------------------------------------------------ Self-critique Rating: See my comment above. ********************************************* question: Query 12.2.36 n(A)=a, n(S) = s; P(A')=? What is the P(A')? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Completely lost with this one. I do accept, however, that P(A’) is the denotation of everything that A is not, its “complement” confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aA' 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. ** Self-critique: Please provide further instruction.
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Given Solution: `aThe 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. ** Self-critique Self-critique Rating