#$&* course MTH 152 2/20 7 005. Binary probabilities*********************************************
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Given Solution: There are 2 coins. Call one of them the first and the other the second coin. We can get Heads on the first and Heads on the second, which we will designate HH. Or we can get Heads on the first and Tails on the second, which we will designate HT. The other possibilities can be designated TH and TT. Thus there are 4 possible outcomes: HH, HT, TH and TT. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question:`q002. List the possible outcomes if a fair coin is flipped 3 times. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If a fair coin is flipped 3 times there are eight possible outcomes. Each coin flip may result in three heads or three tails (HHH, TTT), two heads and a tail (HHT), two tails and a head (TTH), a head, a tail, and another head (HTH), a tail, a head, and another tail (THT), tail then two heads (THH), and a head then two tails (HTT). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: There are 16 possible outcomes for a fair coin that is flipped 4 times. HHHH, TTTT, HHHT, TTTH, THTT, HTHH, HHTH, TTHT, THHH, HTTT, TTHH, HHTT, THTH, HTHT, HTTH, AND THHT confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We can follow the same strategy as in the preceding problem. We first list twice all the possibilities for 3 coins: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT HHH, HHT, HTH, HTT, THH, THT, TTH, TTT Then we append H to the front of one list and T to the front of the other: HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT Again we see why the number of possibilities doubles when the number of coins increases by one. With three coins we had 8 possible outcomes and with four coins we had 16 outcomes, twice as many as with two coins. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q004. If a fair coin is flipped 4 times, how many of the outcomes contain exactly two 'heads'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can easily see from the answer to the previous question, that exactly two heads occur in 6 of the possible outcomes. However, we can also use an equation to find this solution, noting that if two heads were to occur, they could occur in the first flip and the second, the first flip and the third, the first and the fourth, the second and the third, the second and the fourth, and the third and the fourth. These six possibilities can be expressed in sets as {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, and {3, 4}. Since we are choosing two numbers out of four possible flips, we have an equation of 4*3/2. Therefore, the equation also shows us that there are six possible outcomes in which exactly two heads occur. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 7 possible coin flips in which a certain number of outcomes with exactly three heads could occur. This means that we can create an equation of 7*6*5 for the chances of the three heads we are trying to find, divided by 3! since that is the number of choices we are making. Our equation of 7*6*5/3*2*1 is equal to 35, which means that there are 35 possible outcomes in which a coin flipped 7 times will contain exactly three heads. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Each of these problems can be solved using the same sort of equation. In order to find how many times a coin flipped 6 times would have no heads, we can create an equation of C(6,0), in order to find how many times a coin flipped 6 times would have one head, we can create an equation of C(6,1), and so on and so forth. Ultimately we have 8 equations of C(6,0), C(6,1), C(6,2), C(6,3), C(6,4), C(6,5), C(6,6), and C(6,7). You would solve each problem in the same way: for example, in order to find how many possible outcomes that would contain 2 heads, we would solve C(6,2)= 6*5/2*1, which is equal to 15 possible outcomes. There is one possible outcome in which there will be no heads, 6 possible outcomes in which there will be 1 head, 15 possible outcomes in which there will be 2 heads, 20 possible outcomes in which there will be 3 heads, 15 possible outcomes in which there will be 4 heads, 6 possible outcomes in which there will be 5 heads, 1 possible outcome in which there will be 6 heads, and no possible outcomes in which we can get seven heads, since we are only flipping the coin 6 times. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: There are four subsets for the set {a, b}. An empty set, { }, a set for a, {a}, a set for b, {b},and a set for a and b, {a, b}. There are 9 subsets for the set of {a, b, c}. An empty set for a and b, { }, an empty set for a, b, and c, { }, a set for a, {a}, a set for b, {b}, a set for c, {c}, a set for a and b, {a, b}, a set for a and c, {a, c}, a set for b and c, {b, c}, and a set for a, b, and c, {a, b, c}. There are 18 subsets for the set {a, b, c, d}. An empty set for a and b, { }, an empty set for a, b, and c, { }, an empty set for a, b, c, and d. { }, a set for a, {a}, a set for b, {b}, a set for c, {c}, a set for d, {d}, a set for a and b, {a, b}, a set for a, b, and c, {a, b, c}, a set for a, b, c, and d, {a, b, c, d}, a set for a and c, {a, c}, a set for b and c, {b, c}, a set for a and d, {a, d}, a set for b and d, {b, d}, a set for c and d, {c, d}, a set for a, b, and d, {a, b, d}, a set for a, c, and d, {a, c, d}, and a set for b, c, and d, {b, c, d}. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The set {a, b} has four subsets: the empty set { }, {a}, {b} and {a, b}. These four sets are also subsets of {a, b, c}, and if we add the element c to each of these four sets we get four different subsets of {a, b, c}. The subsets are therefore {}{ }, {a}, {b}, {a, b}, {c}, {a, c}, {b, c} and {a, b, c}. We see that the number of subsets doubles when the number of elements in the set increases by one. This seems similar to the way the number of possible outcomes when flipping coins doubles when we add a coin. The connection is as follows: To form a subset we can go through the elements of the set one at a time, and for each element we can either choose to include it or not. This could be done by flipping a coin once for each element of the set, and including the element if the coin shows 'heads'. Two different sequences of 'heads' and 'tails' would result in two different subsets, and every subset would correspond to exactly one sequence of 'heads' and 'tails'. Thus the number of possible subsets is identical to the number of outcomes from the coin flips. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q008. How many subsets would there be of the set {a, b, c, d, e, f, g, h}? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The number of possible subsets doubles with each additional letter added to the original set. If a set of {a, b} has four possible subsets, a set of {a, b, c} has eight possible subsets, and a set of {a, b, c, d} has nineteen possible subsets, than a set of {a, b, c, d, e, f, g, h} has 256 possible subsets. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 are 8 letters in the original set, and we are trying to find how many 4-element subsets could be created of this 8 letter set. This gives us an equation of 8*7*6*5, and because the order does not matter, 8*7*6*5/4*3*2*1, which is equal to 70 possible subsets that would contain exactly 4-elements. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To form a 4-elements subset of the given 8-elements set, we have to choose 4 elements from the 8. Since the order of elements in a set does not matter, order will not matter in our choice. The number of ways to choose 4 elements from a set of 8, without regard for order, is C(8, 4) = 8 * 7 * 6 * 5 / ( 4 * 3 * 2 * 1) = 70. Self-critique: OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q010. How many subsets of the set {a,b,c,d} contain 4 elements? How many subsets of the set {a,b,c,d} contain 3 elements? How many subsets of the set {a,b,c,d} contain 2 elements? How many subsets of the set {a,b,c,d} contain 1 elements? How many subsets of the set {a,b,c,d} contain no elements? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We can use the same sort of equation in order to find the answer to all of these questions. There are 4 letters in the original set, so if we are trying to find out how many subsets will contain 4 elements we will have an equation of C(4,4). If we are trying to find how many subsets will contain 3 elements, we will have an equation of C(4,3), and so on and so forth. In order to solve these equations, we can create a problem like 4*3*2*1/4*3*2*1. This equation shows that there is only 1 possible subset which will contain 4 elements. When we solve the other equations in the same way, we find that there are 4 possible subsets that will contain 3 elements, 6 possible subsets that will contain 2 elements, 4 possible subsets that will contain 1 element, and 1 possible subset that will contain 0 elements. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: We can use the same sort of equation in order to find the answer to all of these questions. There are 4 letters in the original set, so if we are trying to find out how many subsets will contain 4 elements we will have an equation of C(4,4). If we are trying to find how many subsets will contain 3 elements, we will have an equation of C(4,3), and so on and so forth. In order to solve these equations, we can create a problem like 4*3*2*1/4*3*2*1. This equation shows that there is only 1 possible subset which will contain 4 elements. When we solve the other equations in the same way, we find that there are 4 possible subsets that will contain 3 elements, 6 possible subsets that will contain 2 elements, 4 possible subsets that will contain 1 element, and 1 possible subset that will contain 0 elements. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: #*&!