course Mth 152 I wasn't able to check my e-mail for a response re: assignment 4. I don't know if the server was down tonight or not. But I did notice that assignment 4 is still missing from my port folio. Let me know if I should resend it. ?????????D?assignment #005
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22:36:54 `q001. Note that there are 10 questions in this assignment. List the possible outcomes if a fair coin is flipped 2 times.
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RESPONSE --> if the coin is flipped two times there is the possibility that the outcome will be: HT, HH,TT,TH four ways confidence assessment: 3
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22:37:04 There are 2 coins. Call one of them the first and the other the second coin. We can get Heads on the first and Heads on the second, which we will designate HH. Or we can get Heads on the first and Tails on the second, which we will designate HT. The other possibilities can be designated TH and TT. Thus there are 4 possible outcomes: HH, HT, TH and TT.
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RESPONSE --> o.k self critique assessment: 3
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22:40:58 `q002. List the possible outcomes if a fair coin is flipped 3 times.
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RESPONSE --> HHT, HHH, HTH, HTT, TTH, THT,TTT, THH Eight different ways confidence assessment: 2
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22:41:06 The possible results for the first 2 flips are HH, HT, TH and TT. We can obtain all possible results for 3 flips by appending either H or T to this list. We start out by writing the list twice: HH, HT, TH, TT HH, HT, TH, TT We then append H to each outcome in the first row, and T to each outcome in the second. We obtain HHH, HHT, HTH, HTT THH, THT, TTH, TTT Note that this process shows clearly why the number of possibilities doubles when the number of coins increases by one. With two coins we had 4 possible outcomes and with three coins we had 8 outcomes, twice as many as with two coins.
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RESPONSE --> o.k. self critique assessment: 3
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22:45:29 `q003. List the possible outcomes if a fair coin is flipped 4 times.
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RESPONSE --> HHHH, HHHT, HHTT, HTTT, TTTT, THHH, TTHH, HTHT, THTH, THHT,HTTH,HTHH,HTTT,THTT,TTHT, TTTH SIXTEEN DIFFERENT WAYS confidence assessment: 3
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22:45:39 We can follow the same strategy as in the preceding problem. We first list twice all the possibilities for 3 coins: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT HHH, HHT, HTH, HTT, THH, THT, TTH, TTT Then we append H to the front of one list and T to the front of the other: HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT Again we see why the number of possibilities doubles when the number of coins increases by one. With three coins we had 8 possible outcomes and with four coins we had 16 outcomes, twice as many as with two coins.
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RESPONSE --> o.k. self critique assessment: 3
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22:48:04 `q004. If a fair coin is flipped 4 times, how many of the outcomes contain exactly two 'heads'?
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RESPONSE --> HHTT, HTHT,HTTH,THHT, THTH,TTHH, six ways confidence assessment: 3
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22:48:16 The two 'heads' can occur in positions 1 and 2 (HHTT), 1 and 3 (HTHT), 1 and 4 (HTTH), 2 and 3 (THHT), 2 and 4 (THTH), or 3 and 4 (TTHH). These six possibilities can be expressed by the sets {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}. Thus the possibilities are represented by sets of two numbers chosen from the set {1, 2, 3, 4}. When choosing 2 numbers from a set of four, there are 4 * 3 / 2 possible combinations. Since in this case it doesn't matter in which order the two positions are picked, this will be the number of possible outcomes with exactly two 'heads'. The number of possibilities is thus C(4, 2) = 6.
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RESPONSE --> o.k. self critique assessment: 3
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22:50:49 `q005. If a fair coin is flipped 7 times, how many of the outcomes contain exactly three 'heads'?
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RESPONSE --> 35 ways confidence assessment: 1
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22:51:50 The possible positions for the three 'heads' can be numbered 1 through 7. We have to choose three positions out of these seven possibilities, and the order in which our choices occur is not important. This is equivalent to choosing three numbers from the set {1, 2, 3, 4, 5, 6, 7} without regard for order. This can be done in C(7,3) = 7 * 6 * 5 / 3! = 35 ways. There are thus 35 ways to obtain 3 'heads' on 7 flips.
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RESPONSE --> I worked this one out on paper. I knew that there had to be a better way of getting the answer. I understand how you arrived at the answer self critique assessment: 2
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23:05:43 `q006. If we flip a fair coin 6 times, in how many ways can we get no 'heads'? In how many ways can we get exactly one 'head'? In how many ways can we get exactly two 'heads'?
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RESPONSE --> The only way to get no heads would be one: T,T,T,T,T,T, One Head =( H,T,T,T,T,T,) (T,H,T,T,T,T) (T,T,H,T,T,T) (T,T,T,H,T,T) (T,T,T,T,H,T) (T,T,T,T,T,H) Six Ways Two Heads = (hhtttt) (tttthh) thhttt, tthhtt, tthtth, htttth, hthttt, tththt,ththtt, tththt, htthtt, ttthth, htttht,ttthht,thttth fifteen ways confidence assessment: 2
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23:07:40 In how many ways can we get exactly three 'heads'?
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RESPONSE --> twenty ways of getting exactly three heads. C(6,3) on the triangle confidence assessment: 2
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23:09:04 In how many ways can we get exactly four 'heads'? In how many ways can we get exactly five 'heads'?
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RESPONSE --> C(6,4,) = 15 C(6,5) = 6 confidence assessment: 3
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23:09:47 In how many ways can we get exactly six 'heads'?
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RESPONSE --> There would be only one way to get six heads h,h,h,h,h,h confidence assessment: 3
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23:11:49 In how many ways can we get exactly seven 'heads'?
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RESPONSE --> If we are only flipping the coin six times, there is no way to get exactly 7 heads. confidence assessment: 3
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23:12:23 The number of ways to get no 'heads' is C(6,0) = 1. The number of ways to get exactly one 'head' is C(6,1) = 6. The number of ways to get exactly two 'heads' is C(6,2) = 15. The number of ways to get exactly three 'heads' is C(6,3) = 20. The number of ways to get exactly four 'heads' is C(6,4) = 15. The number of ways to get exactly five 'heads' is C(6,5) = 6. The number of ways to get exactly six 'heads' is C(6,6) = 1. These numbers form the n = 6 row of Pascal's Triangle: 1 6 15 20 15 6 1 See your text for a description of Pascal's Triangle. Note also that these numbers add up to 64, which is 2^6, the number of possible outcomes when a coin is flipped 6 times.
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RESPONSE --> o.k. self critique assessment: 3
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23:21:51 `q007. List all the subsets of the set {a, b}. Then do the same for the set {a,b,c}. Then do the same for the set {a,b,c,d}.
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RESPONSE --> (a,b,) = (a), (b) (a,b,) three (a,b,c,) = (a),(b), (c), (a,b,) (a,c) (b,c) (a,b,c) seven (a,b,c,d) = fourteen confidence assessment: 1
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23:22:37 The set {a, b} has four subsets: the empty set { }, {a}, {b} and {a, b}. These four sets are also subsets of {a, b, c}, and if we add the element c to each of these four sets we get four different subsets of {a, b, c}. The subsets are therefore {}{ }, {a}, {b}, {a, b}, {c}, {a, c}, {b, c} and {a, b, c}. We see that the number of subsets doubles when the number of elements in the set increases by one. This seems similar to the way the number of possible outcomes when flipping coins doubles when we add a coin. The connection is as follows: To form a subset we can go through the elements of the set one at a time, and for each element we can either choose to include it or not. This could be done by flipping a coin once for each element of the set, and including the element if the coin shows 'heads'. Two different sequences of 'heads' and 'tails' would result in two different subsets, and every subset would correspond to exactly one sequence of 'heads' and 'tails'. Thus the number of possible subsets is identical to the number of outcomes from the coin flips.
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RESPONSE --> Why do we count the empty set as a subset? self critique assessment: 1
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23:33:16 `q008. How many subsets would there be of the set {a, b, c, d, e, f, g, h}?
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RESPONSE --> (a,b,c,d,e,f,g,h) there would be 4 subsets each containing 2 letters 2x2x2x2 = each one of the letters could be the start of 8 different subsets. so that would be 2x2x2x2x2x2x2x2= 256 confidence assessment: 2
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23:33:51 There are 2 possible subsets of the set {a}, the subsets being { } and {a, b}. The number doubles with each additional element. It follows that for a set of 2 elements there are 2 * 2 subsets (double the 2 subsets of a one-element set), double this or 2 * 2 * 2 subsets of a set with 3 elements, double this or 2 * 2 * 2 * 2 subsets of a set with 4 elements, etc.. There are thus 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 = 256 subsets of the given set, which has 8 elements. This number is also written as 2^8. }{More generally there are 2^n subsets of any set with n elements.
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RESPONSE --> o.k. I came up with the same answer, but I took the long way around to get to it. self critique assessment: 3
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23:47:09 `q009. How many 4-element subsets would there be of the set {a, b, c, d, e, f, g, h}?
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RESPONSE --> (a,b,c,d,e,f,g,h) 8x7x6x5= 1680 1680/ 3 x 8= 1680 / 24 = 70 confidence assessment: 1
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23:47:26 To form a 4-elements subset of the given 8-elements set, we have to choose 4 elements from the 8. Since the order of elements in a set does not matter, order will not matter in our choice. The number of ways to choose 4 elements from a set of 8, without regard for order, is C(8, 4) = 8 * 7 * 6 * 5 / ( 4 * 3 * 2 * 1) = 70.
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RESPONSE --> ????????? self critique assessment: 1
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23:51:15 `q010. How many subsets of the set {a,b,c,d} contain 4 elements? How many subsets of the set {a,b,c,d} contain 3 elements? How many subsets of the set {a,b,c,d} contain 2 elements? How many subsets of the set {a,b,c,d} contain 1 elements? How many subsets of the set {a,b,c,d} contain no elements?
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RESPONSE --> (a,b,c,d) 4 elements = 1 "" 3 elements = 4 "" 2 elements = 8 "" 1 element = 4 "" 0 element = 0 confidence assessment: 2
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23:51:23 The number of 4-element subsets is C(4,4) = 1. The number of 3-element subsets is C(4,3) = 4. The number of 2-element subsets is C(4,2) = 6. The number of 1-element subsets is C(4,1) = 4. The number of 0-element subsets is C(4,0) = 1. We note that these numbers form the n = 4 row 1 4 6 4 1 of Pascal's Triangle, and that they add up to 2^4 = 16, the number of possible subsets of a 4-element set.
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RESPONSE --> o.k. self critique assessment: 2
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