#$&* course Mth 152 10/7/2012 7:40PM 005. Binary probabilities
.............................................
Given Solution: There are 2 coins. Call one of them the first and the other the second coin. We can get Heads on the first and Heads on the second, which we will designate HH. Or we can get Heads on the first and Tails on the second, which we will designate HT. The other possibilities can be designated TH and TT. Thus there are 4 possible outcomes: HH, HT, TH and TT. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q002. List the possible outcomes if a fair coin is flipped 3 times. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You could double the outcomes you had in the previous problem but put H on half of them and T on the other half, so you would have HHH, TTH, HTH, THH, HHT, TTT, HTT, and THT as possible outcomes of flipping a coin 3 times. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q003. List the possible outcomes if a fair coin is flipped 4 times. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Doing the same as before, I listed the possible solutions for 3 flips twice and added a H to one and T to another, so all the possible combinations are HHHH, TTHH, HTHH, THHH, HHTH, TTTH, HTTH, THTH, HHHT, TTHT, HTHT, THHT, HHTT, TTTT, HTTT, and THTT. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: We can follow the same strategy as in the preceding problem. We first list twice all the possibilities for 3 coins: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT HHH, HHT, HTH, HTT, THH, THT, TTH, TTT Then we append H to the front of one list and T to the front of the other: HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT Again we see why the number of possibilities doubles when the number of coins increases by one. With three coins we had 8 possible outcomes and with four coins we had 16 outcomes, twice as many as with two coins. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q004. If a fair coin is flipped 4 times, how many of the outcomes contain exactly two 'heads'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: TTHH, HTTH, THTH, HTHT, THHT, and HHTT all have exactly two heads, so there are 6 possible outcomes. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q005. If a fair coin is flipped 7 times, how many of the outcomes contain exactly three 'heads'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are C(7, 3) = 7!/[(7 - 3)! * 3!] = 7 * 6 * 5* 4 * 3 * 2 * 1/[(4 * 3 * 2 * 1) * (3 * 2 * 1)] = 7 * 6 * 5/(3 * 2 * 1) = 35 outcomes. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The possible positions for the three 'heads' can be numbered 1 through 7. We have to choose three positions out of these seven possibilities, and the order in which our choices occur is not important. This is equivalent to choosing three numbers from the set {1, 2, 3, 4, 5, 6, 7} without regard for order. This can be done in C(7,3) = 7 * 6 * 5 / 3! = 35 ways. There are thus 35 ways to obtain 3 'heads' on 7 flips. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q006. If we flip a fair coin 6 times, in how many ways can we get no 'heads'? In how many ways can we get exactly one 'head'? In how many ways can we get exactly two 'heads'? In how many ways can we get exactly three 'heads'? In how many ways can we get exactly four 'heads'? In how many ways can we get exactly five 'heads'? In how many ways can we get exactly six 'heads'? In how many ways can we get exactly seven 'heads'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Zero heads would equal 1 because C(6, 0) = 1. One head would equal C(6, 1) = 6!/[(6 - 1)! * 1!] = 6. Two heads would equal C(6, 2) = 6!/[(6 - 2)! * 2!] = 6 * 5 * 4 * 3 * 2 * 1/[(4 * 3 * 2 *1) * (2 * 1)] = 15. Three heads would equal C(6, 3) = 6![(6 - 3)! * 3!] = 20. Four heads would equal C(6, 4) = 15. Five heads equal C(6, 5) = 6. Six heads equal C(6, 6) = 1, and there is no way to get 7 heads with 6 coins. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. STUDENT QUESTION Could you explain this a bit further? I am really confused how these answers are obtained. In the first two I obtained them easily using the counting principal, but afterwards it did not appear to work in obtaining the correct answer and I was very confused. INSTRUCTOR RESPONSE You aren't quite specific enough in this question to ensure that I'm answering it in the way you need. However I can expand on this in terms of the details you gave in your solution, and this should be helpful no only to you but to other students: To get 4 'heads' there are 6 * 5 * 4 * 3 possible ways to distribute their positions among the 6 flips to get them in order. This could be calculated as 6 ! / (6 - 4) ! = 6 ! / 2 ! = 6 * 5 * 4 * 3. There are 4! different orders in which the four positions of the 'heads' could have occurred, so there are 6 * 5 * 4 * 3 / (4 * 3 * 2 * 1) unordered ways to obtain those positions. That could be calculated as 6 ! / (2 ! * 4 !) or, using the formula for combinations, 6 ! / ( (4-2)! * 4!). Whichever way you calculate it you get 15, which matches the r = 4 position of the n = 6 row of Pascal's triangle. Similar reasoning will confirm the results for 5 and 6 'heads'. Let me know if this doesn't answer your question, and if not tell me a little more about what you do and do not understand. STUDENT COMMENT The one with the answer one, and the one for the 4 times is the hardest for me for some reason. The others one were so easy INSTRUCTOR RESPONSE It's obvious that only one outcome has no 'heads' (that would be tttttt) and only one has six 'heads (that would be hhhhhh). So the answers to the first and last questions would both be 1, not 0. However that doesn't help with the calculation. For the calculation, remember that 0 ! = 1. So for example C(6, 0) = 6! / ( 0! * (6 - 0) !) = 6 ! / (1 * 6 !) = 1. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `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}. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The subsets for {a,b} are {a}, {b}, {a,b}, and { } which is an empty subset. To get the subsets for {a,b,c} I doubled the subsets for {a,b} and put {c} and half of them and the left the other half alone, So the subsets are {a}, {b}, {a,b}, {a,c}, {b,c}, {a,b,c}, {c} and { }. Doing the same as before I got the subsets for {a,b,c,d} which are { }, {a}, {b}, {a,b}, {a,c}, {b,c}, {a,b,c}, {c}, {a,d}, {b,d}, {a,b,d}, {d}, {a,c,d}, {b,c,d}, {a,b,c,d}, and {c,d}. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The set {a, b} has four subsets: the empty set { }, {a}, {b} and {a, b}. These four sets are also subsets of {a, b, c}, and if we add the element c to each of these four sets we get four different subsets of {a, b, c}. The subsets are therefore {}{ }, {a}, {b}, {a, b}, {c}, {a, c}, {b, c} and {a, b, c}. We see that the number of subsets doubles when the number of elements in the set increases by one. This seems similar to the way the number of possible outcomes when flipping coins doubles when we add a coin. The connection is as follows: To form a subset we can go through the elements of the set one at a time, and for each element we can either choose to include it or not. This could be done by flipping a coin once for each element of the set, and including the element if the coin shows 'heads'. Two different sequences of 'heads' and 'tails' would result in two different subsets, and every subset would correspond to exactly one sequence of 'heads' and 'tails'. Thus the number of possible subsets is identical to the number of outcomes from the coin flips. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q008. How many subsets would there be of the set {a, b, c, d, e, f, g, h}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The number of subsets doubles with each new element, and a subset with 1 element has 2 subset sets, so one with 8 would have 2^8 subsets, which equals 256. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q009. How many 4-element subsets would there be of the set {a, b, c, d, e, f, g, h}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There would be C(8, 4) = 8!/[(8 - 4)! * 4!)] = 70 subsets with 4 elements. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: To form a 4-elements subset of the given 8-elements set, we have to choose 4 elements from the 8. Since the order of elements in a set does not matter, order will not matter in our choice. The number of ways to choose 4 elements from a set of 8, without regard for order, is C(8, 4) = 8 * 7 * 6 * 5 / ( 4 * 3 * 2 * 1) = 70. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q010. How many subsets of the set {a,b,c,d} contain 4 elements? How many subsets of the set {a,b,c,d} contain 3 elements? How many subsets of the set {a,b,c,d} contain 2 elements? How many subsets of the set {a,b,c,d} contain 1 elements? How many subsets of the set {a,b,c,d} contain no elements? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There would be C(4, 3) = 4!/[(4 - 3)! * 3!)] = 4 * 3 * 2 * 1/[1 * (3 * 2 * 1)] = 4 subsets with 3 elements. There would be C(4, 2) = 4!/[(4 - 2)! * 2!)] = 6 subsets with 2 elements. There would be C(4, 1) = 4!/[(4 - 1)! * 1)] = 4 subsets with 1 element. And there would be C(4, 0) = 1 subset with 0 elements. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There would be C(4, 3) = 4!/[(4 - 3)! * 3!)] = 4 * 3 * 2 * 1/[1 * (3 * 2 * 1)] = 4 subsets with 3 elements. There would be C(4, 2) = 4!/[(4 - 2)! * 2!)] = 6 subsets with 2 elements. There would be C(4, 1) = 4!/[(4 - 1)! * 1)] = 4 subsets with 1 element. And there would be C(4, 0) = 1 subset with 0 elements. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!