portfolio #1 september 25, 2018 · portfolio #1 september 25, 2018 1 . 0 3 . 0 9 write out the...
TRANSCRIPT
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10107
Write the following set by listing its elements between braces
or
can be any real number that satisfies the equation the only real values that satisfy this
equation are -3 and -2 so the set the set consists only of the values -3 and -2 10122 Write the following set in set-builder notation Each term in the set increases by an odd number The difference between the first two terms is 3 the difference between the next two terms is 5 the difference between the next two terms is 7 etc Now that we have recognized the pattern our only challenge is expressing this in set builder notation The difference
between each successive square number is an odd number etc However none of the numbers in our set are square numbers To solve this problem we can define the nth term in our set to be equal to However the first term in our set is 3 meaning cannot equal anything less than 1 otherwise we would have a term in our set less than 3 can also not equal any number that is not a whole number so we must stipulate that is a member of the natural numbers Therefore we
can use the expression to express in set builder notation
10214 Sketch the following Cartesian product in the plane
There are an infinite number of values between 1 and 2 so it would be impossible for me to list all the products However in our set of points there are only three total values positioned at and Above I have notated three line segments as if they were three points expressing all the values from 1 to 2 as In actuality there are an infinite number of points expressed in the notation above
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10309 Write out the following set by listing its elements between braces
is a set with a cardinality of 2 could be one of three possibilities or However our question asks for the answer in the form of a set rather than individual sets of the possibilities This is expressed both in the wording of the problem and the set-builder notation used The set of all possible = 10416
find Because Because When you multiply a set with cardinality by a set with a cardinality the resulting set has a cardinality of
assuming both and are finite Therefore when you multiply the set by the set the
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
cardinality of the product set is equal to the cardinality of multiplied by the cardinality of and therefore equal to
thereforehellip
10508
Sketch the sets and Figure 1 a graph of the two sets
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 2 a graph of The bottom half of includes all terms that are also in however these are the only points that are elements in both sets so this is the only region shown shaded below
Figure 3 a graph of This graph shows all the points in both of the sets so it looks identical to the original graph of the two sets in the coordinate plane because it is a graph of all the points that are elements in one or both of the sets
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 4 a graph of Just as in only the bottom half of the circle is shaded because this represents all points in set that are also points in set in only the top half of set is shaded because this region represents all the points in set that are NOT in set There is a dotted line along the x axis because the points on the x axis are not included in because these points exist in set because the elements in this set are all values that are less than or equal to 1
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 5 a graph of This graph shows all points in set except for those that are also in the circle that is set There is a dotted line around the white because all points on the perimeter of the circle
are not included in because they are elements of set
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10605
Sketch the set on the coordinate plane On a separate drawing shade in the set Figure 7 a graph of all the points in set
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 8 a graph of all the points in The set has points in the plane of real numbers Therefore its complement must be as depicted below There is a dotted line because the points on the perimeter
of the circle and are included in set and therefore not elements of
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10709 Draw a Venn diagram for Figure 9 All elements included in both and are shaded except for elements also included in because is being subtracted from the intersection of and
Figure 10 a clearer representation of the same diagram
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10710 Draw a Venn diagram for First letrsquos take a look at what our diagram would look like for as depicted in figure 11
Now to find we must keep all the shaded region from the first graph and shade all of as well because we are finding the union of and This is depicted in figure 12 below
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Let S = a b c d e f and X sube S Let Pk(a) be the probability that a isin X given |X| = k Find P0(a) P1(a) P6(a) In general what is the probability that a randomly chosen subset of S will contain element a Explain why this is consistent with what you learned in sections 103 and 104 First letrsquos list out the number of possibilities for X with all cardinalities from 0 to 6 Elements for which rarr 1 if then no matter what is a subset of Elements for which rarr 6 If has a cardinality of 1 it has one element There are 6 total elements in so this one element has 6 different possibilities Therefore there are 6 possible when Elements for which rarr15 must include 2 of 6 possible elements There are 15 different ways to choose two elements from a set of 6 There are 6 possibilities for what the first element of is Once the first element has been chosen the second element has 5 possibilities (we can think of this as
) However we must then consider that it does not matter which element we designate to be the first element or the second element is equivalent to Therefore we must divide by the number of ways we can arrange 2 elements which is ways or just 2 ways The process we have just done is
it is the way to calculate how many ways you can choose 2 elements from a set of 6
Elements for which rarr20 This is equivalent to for the same reasons explained when finding all
possible X when
Elements for which rarr15
Elements for which rarr6 Elements for which rarr1 the only possible is For the only possible value of is so there the probability that a isin X is 0 For there are 6 possible each containing 1 of the 6 elements in S one of which is Therefore
For there are 15 possible We must now find how many of these contain element We know all where and a isin X will have the elements and one other element from S Since there
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
are 5 other elements in set that are not element there are 5 possible that contain = set of all possible where a isin X and Because there are 5 possible that contain element a when out of 15 total possible when
For there are 20 possible To find how many of these contain we can use the same logic as we used above For every such that and a isin X there will be the elements where
and and There are 20 total ways to pick 2 values for and as there are 5 possibilities for what can be and 4 possibilities for what can be once has been picked However saying there are 20 possibilities for when and a isin X would be double counting because it does not matter which value is and which value is (ie is equivalent to ) The total number of possible when and a isin X is 10 and there are 20 possible when
Therefore For there are 15 possible To find how many contain we are instead going to find all the
that do not contain a and list them = the set of all possible X where
Because there are 5 possible when and 15 possible when there are 10
where Therefore the probability that when is For there are 6 possible and only one where and The only value of X that
satisfies those two requirements is Therefore For there is only one possible and that contains all of the terms contained in set S Therefore
What is the probability that a randomly chosen subset of will contain Total subsets of Total subsets of such that =
Probability that a randomly chosen subset of will contain X = To find all the subsets of we can create a tree like the one depicted on page 12 of The Book of Proof We can define the first branch of our tree to be whether or not to include term a in our subset Our tree is split into two parts the top part includes only subsets that do not have a as an element the bottom part of our tree includes only subsets of that do have a as an element Both the ldquoyesrdquo branch and ldquonordquo branch are split into two more branches again when we add the next branching point whether or not b is an element in the subset These branches are split again and again as we do the same for each term in The important thing
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
to remember is that the tree always remains symmetrical as shown below Therefore we are then left with one side of our tree that includes terms with a and another symmetrical side that includes all terms without a Therefor there are an equal number of subsets with a as there are without a Because 50 of subsets include
Bonus Hypothesis Letrsquos say we have set where and When the probability
that is equal to as long as there is only one element a in set Q Proof
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10309 Write out the following set by listing its elements between braces
is a set with a cardinality of 2 could be one of three possibilities or However our question asks for the answer in the form of a set rather than individual sets of the possibilities This is expressed both in the wording of the problem and the set-builder notation used The set of all possible = 10416
find Because Because When you multiply a set with cardinality by a set with a cardinality the resulting set has a cardinality of
assuming both and are finite Therefore when you multiply the set by the set the
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
cardinality of the product set is equal to the cardinality of multiplied by the cardinality of and therefore equal to
thereforehellip
10508
Sketch the sets and Figure 1 a graph of the two sets
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 2 a graph of The bottom half of includes all terms that are also in however these are the only points that are elements in both sets so this is the only region shown shaded below
Figure 3 a graph of This graph shows all the points in both of the sets so it looks identical to the original graph of the two sets in the coordinate plane because it is a graph of all the points that are elements in one or both of the sets
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 4 a graph of Just as in only the bottom half of the circle is shaded because this represents all points in set that are also points in set in only the top half of set is shaded because this region represents all the points in set that are NOT in set There is a dotted line along the x axis because the points on the x axis are not included in because these points exist in set because the elements in this set are all values that are less than or equal to 1
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 5 a graph of This graph shows all points in set except for those that are also in the circle that is set There is a dotted line around the white because all points on the perimeter of the circle
are not included in because they are elements of set
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10605
Sketch the set on the coordinate plane On a separate drawing shade in the set Figure 7 a graph of all the points in set
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 8 a graph of all the points in The set has points in the plane of real numbers Therefore its complement must be as depicted below There is a dotted line because the points on the perimeter
of the circle and are included in set and therefore not elements of
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10709 Draw a Venn diagram for Figure 9 All elements included in both and are shaded except for elements also included in because is being subtracted from the intersection of and
Figure 10 a clearer representation of the same diagram
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10710 Draw a Venn diagram for First letrsquos take a look at what our diagram would look like for as depicted in figure 11
Now to find we must keep all the shaded region from the first graph and shade all of as well because we are finding the union of and This is depicted in figure 12 below
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Let S = a b c d e f and X sube S Let Pk(a) be the probability that a isin X given |X| = k Find P0(a) P1(a) P6(a) In general what is the probability that a randomly chosen subset of S will contain element a Explain why this is consistent with what you learned in sections 103 and 104 First letrsquos list out the number of possibilities for X with all cardinalities from 0 to 6 Elements for which rarr 1 if then no matter what is a subset of Elements for which rarr 6 If has a cardinality of 1 it has one element There are 6 total elements in so this one element has 6 different possibilities Therefore there are 6 possible when Elements for which rarr15 must include 2 of 6 possible elements There are 15 different ways to choose two elements from a set of 6 There are 6 possibilities for what the first element of is Once the first element has been chosen the second element has 5 possibilities (we can think of this as
) However we must then consider that it does not matter which element we designate to be the first element or the second element is equivalent to Therefore we must divide by the number of ways we can arrange 2 elements which is ways or just 2 ways The process we have just done is
it is the way to calculate how many ways you can choose 2 elements from a set of 6
Elements for which rarr20 This is equivalent to for the same reasons explained when finding all
possible X when
Elements for which rarr15
Elements for which rarr6 Elements for which rarr1 the only possible is For the only possible value of is so there the probability that a isin X is 0 For there are 6 possible each containing 1 of the 6 elements in S one of which is Therefore
For there are 15 possible We must now find how many of these contain element We know all where and a isin X will have the elements and one other element from S Since there
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
are 5 other elements in set that are not element there are 5 possible that contain = set of all possible where a isin X and Because there are 5 possible that contain element a when out of 15 total possible when
For there are 20 possible To find how many of these contain we can use the same logic as we used above For every such that and a isin X there will be the elements where
and and There are 20 total ways to pick 2 values for and as there are 5 possibilities for what can be and 4 possibilities for what can be once has been picked However saying there are 20 possibilities for when and a isin X would be double counting because it does not matter which value is and which value is (ie is equivalent to ) The total number of possible when and a isin X is 10 and there are 20 possible when
Therefore For there are 15 possible To find how many contain we are instead going to find all the
that do not contain a and list them = the set of all possible X where
Because there are 5 possible when and 15 possible when there are 10
where Therefore the probability that when is For there are 6 possible and only one where and The only value of X that
satisfies those two requirements is Therefore For there is only one possible and that contains all of the terms contained in set S Therefore
What is the probability that a randomly chosen subset of will contain Total subsets of Total subsets of such that =
Probability that a randomly chosen subset of will contain X = To find all the subsets of we can create a tree like the one depicted on page 12 of The Book of Proof We can define the first branch of our tree to be whether or not to include term a in our subset Our tree is split into two parts the top part includes only subsets that do not have a as an element the bottom part of our tree includes only subsets of that do have a as an element Both the ldquoyesrdquo branch and ldquonordquo branch are split into two more branches again when we add the next branching point whether or not b is an element in the subset These branches are split again and again as we do the same for each term in The important thing
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
to remember is that the tree always remains symmetrical as shown below Therefore we are then left with one side of our tree that includes terms with a and another symmetrical side that includes all terms without a Therefor there are an equal number of subsets with a as there are without a Because 50 of subsets include
Bonus Hypothesis Letrsquos say we have set where and When the probability
that is equal to as long as there is only one element a in set Q Proof
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
cardinality of the product set is equal to the cardinality of multiplied by the cardinality of and therefore equal to
thereforehellip
10508
Sketch the sets and Figure 1 a graph of the two sets
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 2 a graph of The bottom half of includes all terms that are also in however these are the only points that are elements in both sets so this is the only region shown shaded below
Figure 3 a graph of This graph shows all the points in both of the sets so it looks identical to the original graph of the two sets in the coordinate plane because it is a graph of all the points that are elements in one or both of the sets
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 4 a graph of Just as in only the bottom half of the circle is shaded because this represents all points in set that are also points in set in only the top half of set is shaded because this region represents all the points in set that are NOT in set There is a dotted line along the x axis because the points on the x axis are not included in because these points exist in set because the elements in this set are all values that are less than or equal to 1
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 5 a graph of This graph shows all points in set except for those that are also in the circle that is set There is a dotted line around the white because all points on the perimeter of the circle
are not included in because they are elements of set
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10605
Sketch the set on the coordinate plane On a separate drawing shade in the set Figure 7 a graph of all the points in set
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 8 a graph of all the points in The set has points in the plane of real numbers Therefore its complement must be as depicted below There is a dotted line because the points on the perimeter
of the circle and are included in set and therefore not elements of
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10709 Draw a Venn diagram for Figure 9 All elements included in both and are shaded except for elements also included in because is being subtracted from the intersection of and
Figure 10 a clearer representation of the same diagram
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10710 Draw a Venn diagram for First letrsquos take a look at what our diagram would look like for as depicted in figure 11
Now to find we must keep all the shaded region from the first graph and shade all of as well because we are finding the union of and This is depicted in figure 12 below
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Let S = a b c d e f and X sube S Let Pk(a) be the probability that a isin X given |X| = k Find P0(a) P1(a) P6(a) In general what is the probability that a randomly chosen subset of S will contain element a Explain why this is consistent with what you learned in sections 103 and 104 First letrsquos list out the number of possibilities for X with all cardinalities from 0 to 6 Elements for which rarr 1 if then no matter what is a subset of Elements for which rarr 6 If has a cardinality of 1 it has one element There are 6 total elements in so this one element has 6 different possibilities Therefore there are 6 possible when Elements for which rarr15 must include 2 of 6 possible elements There are 15 different ways to choose two elements from a set of 6 There are 6 possibilities for what the first element of is Once the first element has been chosen the second element has 5 possibilities (we can think of this as
) However we must then consider that it does not matter which element we designate to be the first element or the second element is equivalent to Therefore we must divide by the number of ways we can arrange 2 elements which is ways or just 2 ways The process we have just done is
it is the way to calculate how many ways you can choose 2 elements from a set of 6
Elements for which rarr20 This is equivalent to for the same reasons explained when finding all
possible X when
Elements for which rarr15
Elements for which rarr6 Elements for which rarr1 the only possible is For the only possible value of is so there the probability that a isin X is 0 For there are 6 possible each containing 1 of the 6 elements in S one of which is Therefore
For there are 15 possible We must now find how many of these contain element We know all where and a isin X will have the elements and one other element from S Since there
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
are 5 other elements in set that are not element there are 5 possible that contain = set of all possible where a isin X and Because there are 5 possible that contain element a when out of 15 total possible when
For there are 20 possible To find how many of these contain we can use the same logic as we used above For every such that and a isin X there will be the elements where
and and There are 20 total ways to pick 2 values for and as there are 5 possibilities for what can be and 4 possibilities for what can be once has been picked However saying there are 20 possibilities for when and a isin X would be double counting because it does not matter which value is and which value is (ie is equivalent to ) The total number of possible when and a isin X is 10 and there are 20 possible when
Therefore For there are 15 possible To find how many contain we are instead going to find all the
that do not contain a and list them = the set of all possible X where
Because there are 5 possible when and 15 possible when there are 10
where Therefore the probability that when is For there are 6 possible and only one where and The only value of X that
satisfies those two requirements is Therefore For there is only one possible and that contains all of the terms contained in set S Therefore
What is the probability that a randomly chosen subset of will contain Total subsets of Total subsets of such that =
Probability that a randomly chosen subset of will contain X = To find all the subsets of we can create a tree like the one depicted on page 12 of The Book of Proof We can define the first branch of our tree to be whether or not to include term a in our subset Our tree is split into two parts the top part includes only subsets that do not have a as an element the bottom part of our tree includes only subsets of that do have a as an element Both the ldquoyesrdquo branch and ldquonordquo branch are split into two more branches again when we add the next branching point whether or not b is an element in the subset These branches are split again and again as we do the same for each term in The important thing
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
to remember is that the tree always remains symmetrical as shown below Therefore we are then left with one side of our tree that includes terms with a and another symmetrical side that includes all terms without a Therefor there are an equal number of subsets with a as there are without a Because 50 of subsets include
Bonus Hypothesis Letrsquos say we have set where and When the probability
that is equal to as long as there is only one element a in set Q Proof
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 2 a graph of The bottom half of includes all terms that are also in however these are the only points that are elements in both sets so this is the only region shown shaded below
Figure 3 a graph of This graph shows all the points in both of the sets so it looks identical to the original graph of the two sets in the coordinate plane because it is a graph of all the points that are elements in one or both of the sets
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 4 a graph of Just as in only the bottom half of the circle is shaded because this represents all points in set that are also points in set in only the top half of set is shaded because this region represents all the points in set that are NOT in set There is a dotted line along the x axis because the points on the x axis are not included in because these points exist in set because the elements in this set are all values that are less than or equal to 1
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 5 a graph of This graph shows all points in set except for those that are also in the circle that is set There is a dotted line around the white because all points on the perimeter of the circle
are not included in because they are elements of set
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10605
Sketch the set on the coordinate plane On a separate drawing shade in the set Figure 7 a graph of all the points in set
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 8 a graph of all the points in The set has points in the plane of real numbers Therefore its complement must be as depicted below There is a dotted line because the points on the perimeter
of the circle and are included in set and therefore not elements of
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10709 Draw a Venn diagram for Figure 9 All elements included in both and are shaded except for elements also included in because is being subtracted from the intersection of and
Figure 10 a clearer representation of the same diagram
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10710 Draw a Venn diagram for First letrsquos take a look at what our diagram would look like for as depicted in figure 11
Now to find we must keep all the shaded region from the first graph and shade all of as well because we are finding the union of and This is depicted in figure 12 below
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Let S = a b c d e f and X sube S Let Pk(a) be the probability that a isin X given |X| = k Find P0(a) P1(a) P6(a) In general what is the probability that a randomly chosen subset of S will contain element a Explain why this is consistent with what you learned in sections 103 and 104 First letrsquos list out the number of possibilities for X with all cardinalities from 0 to 6 Elements for which rarr 1 if then no matter what is a subset of Elements for which rarr 6 If has a cardinality of 1 it has one element There are 6 total elements in so this one element has 6 different possibilities Therefore there are 6 possible when Elements for which rarr15 must include 2 of 6 possible elements There are 15 different ways to choose two elements from a set of 6 There are 6 possibilities for what the first element of is Once the first element has been chosen the second element has 5 possibilities (we can think of this as
) However we must then consider that it does not matter which element we designate to be the first element or the second element is equivalent to Therefore we must divide by the number of ways we can arrange 2 elements which is ways or just 2 ways The process we have just done is
it is the way to calculate how many ways you can choose 2 elements from a set of 6
Elements for which rarr20 This is equivalent to for the same reasons explained when finding all
possible X when
Elements for which rarr15
Elements for which rarr6 Elements for which rarr1 the only possible is For the only possible value of is so there the probability that a isin X is 0 For there are 6 possible each containing 1 of the 6 elements in S one of which is Therefore
For there are 15 possible We must now find how many of these contain element We know all where and a isin X will have the elements and one other element from S Since there
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
are 5 other elements in set that are not element there are 5 possible that contain = set of all possible where a isin X and Because there are 5 possible that contain element a when out of 15 total possible when
For there are 20 possible To find how many of these contain we can use the same logic as we used above For every such that and a isin X there will be the elements where
and and There are 20 total ways to pick 2 values for and as there are 5 possibilities for what can be and 4 possibilities for what can be once has been picked However saying there are 20 possibilities for when and a isin X would be double counting because it does not matter which value is and which value is (ie is equivalent to ) The total number of possible when and a isin X is 10 and there are 20 possible when
Therefore For there are 15 possible To find how many contain we are instead going to find all the
that do not contain a and list them = the set of all possible X where
Because there are 5 possible when and 15 possible when there are 10
where Therefore the probability that when is For there are 6 possible and only one where and The only value of X that
satisfies those two requirements is Therefore For there is only one possible and that contains all of the terms contained in set S Therefore
What is the probability that a randomly chosen subset of will contain Total subsets of Total subsets of such that =
Probability that a randomly chosen subset of will contain X = To find all the subsets of we can create a tree like the one depicted on page 12 of The Book of Proof We can define the first branch of our tree to be whether or not to include term a in our subset Our tree is split into two parts the top part includes only subsets that do not have a as an element the bottom part of our tree includes only subsets of that do have a as an element Both the ldquoyesrdquo branch and ldquonordquo branch are split into two more branches again when we add the next branching point whether or not b is an element in the subset These branches are split again and again as we do the same for each term in The important thing
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
to remember is that the tree always remains symmetrical as shown below Therefore we are then left with one side of our tree that includes terms with a and another symmetrical side that includes all terms without a Therefor there are an equal number of subsets with a as there are without a Because 50 of subsets include
Bonus Hypothesis Letrsquos say we have set where and When the probability
that is equal to as long as there is only one element a in set Q Proof
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 4 a graph of Just as in only the bottom half of the circle is shaded because this represents all points in set that are also points in set in only the top half of set is shaded because this region represents all the points in set that are NOT in set There is a dotted line along the x axis because the points on the x axis are not included in because these points exist in set because the elements in this set are all values that are less than or equal to 1
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 5 a graph of This graph shows all points in set except for those that are also in the circle that is set There is a dotted line around the white because all points on the perimeter of the circle
are not included in because they are elements of set
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10605
Sketch the set on the coordinate plane On a separate drawing shade in the set Figure 7 a graph of all the points in set
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 8 a graph of all the points in The set has points in the plane of real numbers Therefore its complement must be as depicted below There is a dotted line because the points on the perimeter
of the circle and are included in set and therefore not elements of
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10709 Draw a Venn diagram for Figure 9 All elements included in both and are shaded except for elements also included in because is being subtracted from the intersection of and
Figure 10 a clearer representation of the same diagram
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10710 Draw a Venn diagram for First letrsquos take a look at what our diagram would look like for as depicted in figure 11
Now to find we must keep all the shaded region from the first graph and shade all of as well because we are finding the union of and This is depicted in figure 12 below
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Let S = a b c d e f and X sube S Let Pk(a) be the probability that a isin X given |X| = k Find P0(a) P1(a) P6(a) In general what is the probability that a randomly chosen subset of S will contain element a Explain why this is consistent with what you learned in sections 103 and 104 First letrsquos list out the number of possibilities for X with all cardinalities from 0 to 6 Elements for which rarr 1 if then no matter what is a subset of Elements for which rarr 6 If has a cardinality of 1 it has one element There are 6 total elements in so this one element has 6 different possibilities Therefore there are 6 possible when Elements for which rarr15 must include 2 of 6 possible elements There are 15 different ways to choose two elements from a set of 6 There are 6 possibilities for what the first element of is Once the first element has been chosen the second element has 5 possibilities (we can think of this as
) However we must then consider that it does not matter which element we designate to be the first element or the second element is equivalent to Therefore we must divide by the number of ways we can arrange 2 elements which is ways or just 2 ways The process we have just done is
it is the way to calculate how many ways you can choose 2 elements from a set of 6
Elements for which rarr20 This is equivalent to for the same reasons explained when finding all
possible X when
Elements for which rarr15
Elements for which rarr6 Elements for which rarr1 the only possible is For the only possible value of is so there the probability that a isin X is 0 For there are 6 possible each containing 1 of the 6 elements in S one of which is Therefore
For there are 15 possible We must now find how many of these contain element We know all where and a isin X will have the elements and one other element from S Since there
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
are 5 other elements in set that are not element there are 5 possible that contain = set of all possible where a isin X and Because there are 5 possible that contain element a when out of 15 total possible when
For there are 20 possible To find how many of these contain we can use the same logic as we used above For every such that and a isin X there will be the elements where
and and There are 20 total ways to pick 2 values for and as there are 5 possibilities for what can be and 4 possibilities for what can be once has been picked However saying there are 20 possibilities for when and a isin X would be double counting because it does not matter which value is and which value is (ie is equivalent to ) The total number of possible when and a isin X is 10 and there are 20 possible when
Therefore For there are 15 possible To find how many contain we are instead going to find all the
that do not contain a and list them = the set of all possible X where
Because there are 5 possible when and 15 possible when there are 10
where Therefore the probability that when is For there are 6 possible and only one where and The only value of X that
satisfies those two requirements is Therefore For there is only one possible and that contains all of the terms contained in set S Therefore
What is the probability that a randomly chosen subset of will contain Total subsets of Total subsets of such that =
Probability that a randomly chosen subset of will contain X = To find all the subsets of we can create a tree like the one depicted on page 12 of The Book of Proof We can define the first branch of our tree to be whether or not to include term a in our subset Our tree is split into two parts the top part includes only subsets that do not have a as an element the bottom part of our tree includes only subsets of that do have a as an element Both the ldquoyesrdquo branch and ldquonordquo branch are split into two more branches again when we add the next branching point whether or not b is an element in the subset These branches are split again and again as we do the same for each term in The important thing
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
to remember is that the tree always remains symmetrical as shown below Therefore we are then left with one side of our tree that includes terms with a and another symmetrical side that includes all terms without a Therefor there are an equal number of subsets with a as there are without a Because 50 of subsets include
Bonus Hypothesis Letrsquos say we have set where and When the probability
that is equal to as long as there is only one element a in set Q Proof
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 5 a graph of This graph shows all points in set except for those that are also in the circle that is set There is a dotted line around the white because all points on the perimeter of the circle
are not included in because they are elements of set
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10605
Sketch the set on the coordinate plane On a separate drawing shade in the set Figure 7 a graph of all the points in set
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 8 a graph of all the points in The set has points in the plane of real numbers Therefore its complement must be as depicted below There is a dotted line because the points on the perimeter
of the circle and are included in set and therefore not elements of
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10709 Draw a Venn diagram for Figure 9 All elements included in both and are shaded except for elements also included in because is being subtracted from the intersection of and
Figure 10 a clearer representation of the same diagram
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10710 Draw a Venn diagram for First letrsquos take a look at what our diagram would look like for as depicted in figure 11
Now to find we must keep all the shaded region from the first graph and shade all of as well because we are finding the union of and This is depicted in figure 12 below
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Let S = a b c d e f and X sube S Let Pk(a) be the probability that a isin X given |X| = k Find P0(a) P1(a) P6(a) In general what is the probability that a randomly chosen subset of S will contain element a Explain why this is consistent with what you learned in sections 103 and 104 First letrsquos list out the number of possibilities for X with all cardinalities from 0 to 6 Elements for which rarr 1 if then no matter what is a subset of Elements for which rarr 6 If has a cardinality of 1 it has one element There are 6 total elements in so this one element has 6 different possibilities Therefore there are 6 possible when Elements for which rarr15 must include 2 of 6 possible elements There are 15 different ways to choose two elements from a set of 6 There are 6 possibilities for what the first element of is Once the first element has been chosen the second element has 5 possibilities (we can think of this as
) However we must then consider that it does not matter which element we designate to be the first element or the second element is equivalent to Therefore we must divide by the number of ways we can arrange 2 elements which is ways or just 2 ways The process we have just done is
it is the way to calculate how many ways you can choose 2 elements from a set of 6
Elements for which rarr20 This is equivalent to for the same reasons explained when finding all
possible X when
Elements for which rarr15
Elements for which rarr6 Elements for which rarr1 the only possible is For the only possible value of is so there the probability that a isin X is 0 For there are 6 possible each containing 1 of the 6 elements in S one of which is Therefore
For there are 15 possible We must now find how many of these contain element We know all where and a isin X will have the elements and one other element from S Since there
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
are 5 other elements in set that are not element there are 5 possible that contain = set of all possible where a isin X and Because there are 5 possible that contain element a when out of 15 total possible when
For there are 20 possible To find how many of these contain we can use the same logic as we used above For every such that and a isin X there will be the elements where
and and There are 20 total ways to pick 2 values for and as there are 5 possibilities for what can be and 4 possibilities for what can be once has been picked However saying there are 20 possibilities for when and a isin X would be double counting because it does not matter which value is and which value is (ie is equivalent to ) The total number of possible when and a isin X is 10 and there are 20 possible when
Therefore For there are 15 possible To find how many contain we are instead going to find all the
that do not contain a and list them = the set of all possible X where
Because there are 5 possible when and 15 possible when there are 10
where Therefore the probability that when is For there are 6 possible and only one where and The only value of X that
satisfies those two requirements is Therefore For there is only one possible and that contains all of the terms contained in set S Therefore
What is the probability that a randomly chosen subset of will contain Total subsets of Total subsets of such that =
Probability that a randomly chosen subset of will contain X = To find all the subsets of we can create a tree like the one depicted on page 12 of The Book of Proof We can define the first branch of our tree to be whether or not to include term a in our subset Our tree is split into two parts the top part includes only subsets that do not have a as an element the bottom part of our tree includes only subsets of that do have a as an element Both the ldquoyesrdquo branch and ldquonordquo branch are split into two more branches again when we add the next branching point whether or not b is an element in the subset These branches are split again and again as we do the same for each term in The important thing
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
to remember is that the tree always remains symmetrical as shown below Therefore we are then left with one side of our tree that includes terms with a and another symmetrical side that includes all terms without a Therefor there are an equal number of subsets with a as there are without a Because 50 of subsets include
Bonus Hypothesis Letrsquos say we have set where and When the probability
that is equal to as long as there is only one element a in set Q Proof
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10605
Sketch the set on the coordinate plane On a separate drawing shade in the set Figure 7 a graph of all the points in set
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 8 a graph of all the points in The set has points in the plane of real numbers Therefore its complement must be as depicted below There is a dotted line because the points on the perimeter
of the circle and are included in set and therefore not elements of
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10709 Draw a Venn diagram for Figure 9 All elements included in both and are shaded except for elements also included in because is being subtracted from the intersection of and
Figure 10 a clearer representation of the same diagram
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10710 Draw a Venn diagram for First letrsquos take a look at what our diagram would look like for as depicted in figure 11
Now to find we must keep all the shaded region from the first graph and shade all of as well because we are finding the union of and This is depicted in figure 12 below
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Let S = a b c d e f and X sube S Let Pk(a) be the probability that a isin X given |X| = k Find P0(a) P1(a) P6(a) In general what is the probability that a randomly chosen subset of S will contain element a Explain why this is consistent with what you learned in sections 103 and 104 First letrsquos list out the number of possibilities for X with all cardinalities from 0 to 6 Elements for which rarr 1 if then no matter what is a subset of Elements for which rarr 6 If has a cardinality of 1 it has one element There are 6 total elements in so this one element has 6 different possibilities Therefore there are 6 possible when Elements for which rarr15 must include 2 of 6 possible elements There are 15 different ways to choose two elements from a set of 6 There are 6 possibilities for what the first element of is Once the first element has been chosen the second element has 5 possibilities (we can think of this as
) However we must then consider that it does not matter which element we designate to be the first element or the second element is equivalent to Therefore we must divide by the number of ways we can arrange 2 elements which is ways or just 2 ways The process we have just done is
it is the way to calculate how many ways you can choose 2 elements from a set of 6
Elements for which rarr20 This is equivalent to for the same reasons explained when finding all
possible X when
Elements for which rarr15
Elements for which rarr6 Elements for which rarr1 the only possible is For the only possible value of is so there the probability that a isin X is 0 For there are 6 possible each containing 1 of the 6 elements in S one of which is Therefore
For there are 15 possible We must now find how many of these contain element We know all where and a isin X will have the elements and one other element from S Since there
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
are 5 other elements in set that are not element there are 5 possible that contain = set of all possible where a isin X and Because there are 5 possible that contain element a when out of 15 total possible when
For there are 20 possible To find how many of these contain we can use the same logic as we used above For every such that and a isin X there will be the elements where
and and There are 20 total ways to pick 2 values for and as there are 5 possibilities for what can be and 4 possibilities for what can be once has been picked However saying there are 20 possibilities for when and a isin X would be double counting because it does not matter which value is and which value is (ie is equivalent to ) The total number of possible when and a isin X is 10 and there are 20 possible when
Therefore For there are 15 possible To find how many contain we are instead going to find all the
that do not contain a and list them = the set of all possible X where
Because there are 5 possible when and 15 possible when there are 10
where Therefore the probability that when is For there are 6 possible and only one where and The only value of X that
satisfies those two requirements is Therefore For there is only one possible and that contains all of the terms contained in set S Therefore
What is the probability that a randomly chosen subset of will contain Total subsets of Total subsets of such that =
Probability that a randomly chosen subset of will contain X = To find all the subsets of we can create a tree like the one depicted on page 12 of The Book of Proof We can define the first branch of our tree to be whether or not to include term a in our subset Our tree is split into two parts the top part includes only subsets that do not have a as an element the bottom part of our tree includes only subsets of that do have a as an element Both the ldquoyesrdquo branch and ldquonordquo branch are split into two more branches again when we add the next branching point whether or not b is an element in the subset These branches are split again and again as we do the same for each term in The important thing
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
to remember is that the tree always remains symmetrical as shown below Therefore we are then left with one side of our tree that includes terms with a and another symmetrical side that includes all terms without a Therefor there are an equal number of subsets with a as there are without a Because 50 of subsets include
Bonus Hypothesis Letrsquos say we have set where and When the probability
that is equal to as long as there is only one element a in set Q Proof
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Figure 8 a graph of all the points in The set has points in the plane of real numbers Therefore its complement must be as depicted below There is a dotted line because the points on the perimeter
of the circle and are included in set and therefore not elements of
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10709 Draw a Venn diagram for Figure 9 All elements included in both and are shaded except for elements also included in because is being subtracted from the intersection of and
Figure 10 a clearer representation of the same diagram
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10710 Draw a Venn diagram for First letrsquos take a look at what our diagram would look like for as depicted in figure 11
Now to find we must keep all the shaded region from the first graph and shade all of as well because we are finding the union of and This is depicted in figure 12 below
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Let S = a b c d e f and X sube S Let Pk(a) be the probability that a isin X given |X| = k Find P0(a) P1(a) P6(a) In general what is the probability that a randomly chosen subset of S will contain element a Explain why this is consistent with what you learned in sections 103 and 104 First letrsquos list out the number of possibilities for X with all cardinalities from 0 to 6 Elements for which rarr 1 if then no matter what is a subset of Elements for which rarr 6 If has a cardinality of 1 it has one element There are 6 total elements in so this one element has 6 different possibilities Therefore there are 6 possible when Elements for which rarr15 must include 2 of 6 possible elements There are 15 different ways to choose two elements from a set of 6 There are 6 possibilities for what the first element of is Once the first element has been chosen the second element has 5 possibilities (we can think of this as
) However we must then consider that it does not matter which element we designate to be the first element or the second element is equivalent to Therefore we must divide by the number of ways we can arrange 2 elements which is ways or just 2 ways The process we have just done is
it is the way to calculate how many ways you can choose 2 elements from a set of 6
Elements for which rarr20 This is equivalent to for the same reasons explained when finding all
possible X when
Elements for which rarr15
Elements for which rarr6 Elements for which rarr1 the only possible is For the only possible value of is so there the probability that a isin X is 0 For there are 6 possible each containing 1 of the 6 elements in S one of which is Therefore
For there are 15 possible We must now find how many of these contain element We know all where and a isin X will have the elements and one other element from S Since there
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
are 5 other elements in set that are not element there are 5 possible that contain = set of all possible where a isin X and Because there are 5 possible that contain element a when out of 15 total possible when
For there are 20 possible To find how many of these contain we can use the same logic as we used above For every such that and a isin X there will be the elements where
and and There are 20 total ways to pick 2 values for and as there are 5 possibilities for what can be and 4 possibilities for what can be once has been picked However saying there are 20 possibilities for when and a isin X would be double counting because it does not matter which value is and which value is (ie is equivalent to ) The total number of possible when and a isin X is 10 and there are 20 possible when
Therefore For there are 15 possible To find how many contain we are instead going to find all the
that do not contain a and list them = the set of all possible X where
Because there are 5 possible when and 15 possible when there are 10
where Therefore the probability that when is For there are 6 possible and only one where and The only value of X that
satisfies those two requirements is Therefore For there is only one possible and that contains all of the terms contained in set S Therefore
What is the probability that a randomly chosen subset of will contain Total subsets of Total subsets of such that =
Probability that a randomly chosen subset of will contain X = To find all the subsets of we can create a tree like the one depicted on page 12 of The Book of Proof We can define the first branch of our tree to be whether or not to include term a in our subset Our tree is split into two parts the top part includes only subsets that do not have a as an element the bottom part of our tree includes only subsets of that do have a as an element Both the ldquoyesrdquo branch and ldquonordquo branch are split into two more branches again when we add the next branching point whether or not b is an element in the subset These branches are split again and again as we do the same for each term in The important thing
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
to remember is that the tree always remains symmetrical as shown below Therefore we are then left with one side of our tree that includes terms with a and another symmetrical side that includes all terms without a Therefor there are an equal number of subsets with a as there are without a Because 50 of subsets include
Bonus Hypothesis Letrsquos say we have set where and When the probability
that is equal to as long as there is only one element a in set Q Proof
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10709 Draw a Venn diagram for Figure 9 All elements included in both and are shaded except for elements also included in because is being subtracted from the intersection of and
Figure 10 a clearer representation of the same diagram
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10710 Draw a Venn diagram for First letrsquos take a look at what our diagram would look like for as depicted in figure 11
Now to find we must keep all the shaded region from the first graph and shade all of as well because we are finding the union of and This is depicted in figure 12 below
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Let S = a b c d e f and X sube S Let Pk(a) be the probability that a isin X given |X| = k Find P0(a) P1(a) P6(a) In general what is the probability that a randomly chosen subset of S will contain element a Explain why this is consistent with what you learned in sections 103 and 104 First letrsquos list out the number of possibilities for X with all cardinalities from 0 to 6 Elements for which rarr 1 if then no matter what is a subset of Elements for which rarr 6 If has a cardinality of 1 it has one element There are 6 total elements in so this one element has 6 different possibilities Therefore there are 6 possible when Elements for which rarr15 must include 2 of 6 possible elements There are 15 different ways to choose two elements from a set of 6 There are 6 possibilities for what the first element of is Once the first element has been chosen the second element has 5 possibilities (we can think of this as
) However we must then consider that it does not matter which element we designate to be the first element or the second element is equivalent to Therefore we must divide by the number of ways we can arrange 2 elements which is ways or just 2 ways The process we have just done is
it is the way to calculate how many ways you can choose 2 elements from a set of 6
Elements for which rarr20 This is equivalent to for the same reasons explained when finding all
possible X when
Elements for which rarr15
Elements for which rarr6 Elements for which rarr1 the only possible is For the only possible value of is so there the probability that a isin X is 0 For there are 6 possible each containing 1 of the 6 elements in S one of which is Therefore
For there are 15 possible We must now find how many of these contain element We know all where and a isin X will have the elements and one other element from S Since there
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
are 5 other elements in set that are not element there are 5 possible that contain = set of all possible where a isin X and Because there are 5 possible that contain element a when out of 15 total possible when
For there are 20 possible To find how many of these contain we can use the same logic as we used above For every such that and a isin X there will be the elements where
and and There are 20 total ways to pick 2 values for and as there are 5 possibilities for what can be and 4 possibilities for what can be once has been picked However saying there are 20 possibilities for when and a isin X would be double counting because it does not matter which value is and which value is (ie is equivalent to ) The total number of possible when and a isin X is 10 and there are 20 possible when
Therefore For there are 15 possible To find how many contain we are instead going to find all the
that do not contain a and list them = the set of all possible X where
Because there are 5 possible when and 15 possible when there are 10
where Therefore the probability that when is For there are 6 possible and only one where and The only value of X that
satisfies those two requirements is Therefore For there is only one possible and that contains all of the terms contained in set S Therefore
What is the probability that a randomly chosen subset of will contain Total subsets of Total subsets of such that =
Probability that a randomly chosen subset of will contain X = To find all the subsets of we can create a tree like the one depicted on page 12 of The Book of Proof We can define the first branch of our tree to be whether or not to include term a in our subset Our tree is split into two parts the top part includes only subsets that do not have a as an element the bottom part of our tree includes only subsets of that do have a as an element Both the ldquoyesrdquo branch and ldquonordquo branch are split into two more branches again when we add the next branching point whether or not b is an element in the subset These branches are split again and again as we do the same for each term in The important thing
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
to remember is that the tree always remains symmetrical as shown below Therefore we are then left with one side of our tree that includes terms with a and another symmetrical side that includes all terms without a Therefor there are an equal number of subsets with a as there are without a Because 50 of subsets include
Bonus Hypothesis Letrsquos say we have set where and When the probability
that is equal to as long as there is only one element a in set Q Proof
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
10710 Draw a Venn diagram for First letrsquos take a look at what our diagram would look like for as depicted in figure 11
Now to find we must keep all the shaded region from the first graph and shade all of as well because we are finding the union of and This is depicted in figure 12 below
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Let S = a b c d e f and X sube S Let Pk(a) be the probability that a isin X given |X| = k Find P0(a) P1(a) P6(a) In general what is the probability that a randomly chosen subset of S will contain element a Explain why this is consistent with what you learned in sections 103 and 104 First letrsquos list out the number of possibilities for X with all cardinalities from 0 to 6 Elements for which rarr 1 if then no matter what is a subset of Elements for which rarr 6 If has a cardinality of 1 it has one element There are 6 total elements in so this one element has 6 different possibilities Therefore there are 6 possible when Elements for which rarr15 must include 2 of 6 possible elements There are 15 different ways to choose two elements from a set of 6 There are 6 possibilities for what the first element of is Once the first element has been chosen the second element has 5 possibilities (we can think of this as
) However we must then consider that it does not matter which element we designate to be the first element or the second element is equivalent to Therefore we must divide by the number of ways we can arrange 2 elements which is ways or just 2 ways The process we have just done is
it is the way to calculate how many ways you can choose 2 elements from a set of 6
Elements for which rarr20 This is equivalent to for the same reasons explained when finding all
possible X when
Elements for which rarr15
Elements for which rarr6 Elements for which rarr1 the only possible is For the only possible value of is so there the probability that a isin X is 0 For there are 6 possible each containing 1 of the 6 elements in S one of which is Therefore
For there are 15 possible We must now find how many of these contain element We know all where and a isin X will have the elements and one other element from S Since there
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
are 5 other elements in set that are not element there are 5 possible that contain = set of all possible where a isin X and Because there are 5 possible that contain element a when out of 15 total possible when
For there are 20 possible To find how many of these contain we can use the same logic as we used above For every such that and a isin X there will be the elements where
and and There are 20 total ways to pick 2 values for and as there are 5 possibilities for what can be and 4 possibilities for what can be once has been picked However saying there are 20 possibilities for when and a isin X would be double counting because it does not matter which value is and which value is (ie is equivalent to ) The total number of possible when and a isin X is 10 and there are 20 possible when
Therefore For there are 15 possible To find how many contain we are instead going to find all the
that do not contain a and list them = the set of all possible X where
Because there are 5 possible when and 15 possible when there are 10
where Therefore the probability that when is For there are 6 possible and only one where and The only value of X that
satisfies those two requirements is Therefore For there is only one possible and that contains all of the terms contained in set S Therefore
What is the probability that a randomly chosen subset of will contain Total subsets of Total subsets of such that =
Probability that a randomly chosen subset of will contain X = To find all the subsets of we can create a tree like the one depicted on page 12 of The Book of Proof We can define the first branch of our tree to be whether or not to include term a in our subset Our tree is split into two parts the top part includes only subsets that do not have a as an element the bottom part of our tree includes only subsets of that do have a as an element Both the ldquoyesrdquo branch and ldquonordquo branch are split into two more branches again when we add the next branching point whether or not b is an element in the subset These branches are split again and again as we do the same for each term in The important thing
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
to remember is that the tree always remains symmetrical as shown below Therefore we are then left with one side of our tree that includes terms with a and another symmetrical side that includes all terms without a Therefor there are an equal number of subsets with a as there are without a Because 50 of subsets include
Bonus Hypothesis Letrsquos say we have set where and When the probability
that is equal to as long as there is only one element a in set Q Proof
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Let S = a b c d e f and X sube S Let Pk(a) be the probability that a isin X given |X| = k Find P0(a) P1(a) P6(a) In general what is the probability that a randomly chosen subset of S will contain element a Explain why this is consistent with what you learned in sections 103 and 104 First letrsquos list out the number of possibilities for X with all cardinalities from 0 to 6 Elements for which rarr 1 if then no matter what is a subset of Elements for which rarr 6 If has a cardinality of 1 it has one element There are 6 total elements in so this one element has 6 different possibilities Therefore there are 6 possible when Elements for which rarr15 must include 2 of 6 possible elements There are 15 different ways to choose two elements from a set of 6 There are 6 possibilities for what the first element of is Once the first element has been chosen the second element has 5 possibilities (we can think of this as
) However we must then consider that it does not matter which element we designate to be the first element or the second element is equivalent to Therefore we must divide by the number of ways we can arrange 2 elements which is ways or just 2 ways The process we have just done is
it is the way to calculate how many ways you can choose 2 elements from a set of 6
Elements for which rarr20 This is equivalent to for the same reasons explained when finding all
possible X when
Elements for which rarr15
Elements for which rarr6 Elements for which rarr1 the only possible is For the only possible value of is so there the probability that a isin X is 0 For there are 6 possible each containing 1 of the 6 elements in S one of which is Therefore
For there are 15 possible We must now find how many of these contain element We know all where and a isin X will have the elements and one other element from S Since there
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
are 5 other elements in set that are not element there are 5 possible that contain = set of all possible where a isin X and Because there are 5 possible that contain element a when out of 15 total possible when
For there are 20 possible To find how many of these contain we can use the same logic as we used above For every such that and a isin X there will be the elements where
and and There are 20 total ways to pick 2 values for and as there are 5 possibilities for what can be and 4 possibilities for what can be once has been picked However saying there are 20 possibilities for when and a isin X would be double counting because it does not matter which value is and which value is (ie is equivalent to ) The total number of possible when and a isin X is 10 and there are 20 possible when
Therefore For there are 15 possible To find how many contain we are instead going to find all the
that do not contain a and list them = the set of all possible X where
Because there are 5 possible when and 15 possible when there are 10
where Therefore the probability that when is For there are 6 possible and only one where and The only value of X that
satisfies those two requirements is Therefore For there is only one possible and that contains all of the terms contained in set S Therefore
What is the probability that a randomly chosen subset of will contain Total subsets of Total subsets of such that =
Probability that a randomly chosen subset of will contain X = To find all the subsets of we can create a tree like the one depicted on page 12 of The Book of Proof We can define the first branch of our tree to be whether or not to include term a in our subset Our tree is split into two parts the top part includes only subsets that do not have a as an element the bottom part of our tree includes only subsets of that do have a as an element Both the ldquoyesrdquo branch and ldquonordquo branch are split into two more branches again when we add the next branching point whether or not b is an element in the subset These branches are split again and again as we do the same for each term in The important thing
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
to remember is that the tree always remains symmetrical as shown below Therefore we are then left with one side of our tree that includes terms with a and another symmetrical side that includes all terms without a Therefor there are an equal number of subsets with a as there are without a Because 50 of subsets include
Bonus Hypothesis Letrsquos say we have set where and When the probability
that is equal to as long as there is only one element a in set Q Proof
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
are 5 other elements in set that are not element there are 5 possible that contain = set of all possible where a isin X and Because there are 5 possible that contain element a when out of 15 total possible when
For there are 20 possible To find how many of these contain we can use the same logic as we used above For every such that and a isin X there will be the elements where
and and There are 20 total ways to pick 2 values for and as there are 5 possibilities for what can be and 4 possibilities for what can be once has been picked However saying there are 20 possibilities for when and a isin X would be double counting because it does not matter which value is and which value is (ie is equivalent to ) The total number of possible when and a isin X is 10 and there are 20 possible when
Therefore For there are 15 possible To find how many contain we are instead going to find all the
that do not contain a and list them = the set of all possible X where
Because there are 5 possible when and 15 possible when there are 10
where Therefore the probability that when is For there are 6 possible and only one where and The only value of X that
satisfies those two requirements is Therefore For there is only one possible and that contains all of the terms contained in set S Therefore
What is the probability that a randomly chosen subset of will contain Total subsets of Total subsets of such that =
Probability that a randomly chosen subset of will contain X = To find all the subsets of we can create a tree like the one depicted on page 12 of The Book of Proof We can define the first branch of our tree to be whether or not to include term a in our subset Our tree is split into two parts the top part includes only subsets that do not have a as an element the bottom part of our tree includes only subsets of that do have a as an element Both the ldquoyesrdquo branch and ldquonordquo branch are split into two more branches again when we add the next branching point whether or not b is an element in the subset These branches are split again and again as we do the same for each term in The important thing
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
to remember is that the tree always remains symmetrical as shown below Therefore we are then left with one side of our tree that includes terms with a and another symmetrical side that includes all terms without a Therefor there are an equal number of subsets with a as there are without a Because 50 of subsets include
Bonus Hypothesis Letrsquos say we have set where and When the probability
that is equal to as long as there is only one element a in set Q Proof
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
to remember is that the tree always remains symmetrical as shown below Therefore we are then left with one side of our tree that includes terms with a and another symmetrical side that includes all terms without a Therefor there are an equal number of subsets with a as there are without a Because 50 of subsets include
Bonus Hypothesis Letrsquos say we have set where and When the probability
that is equal to as long as there is only one element a in set Q Proof
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Total number of possible when We know this because we are choosing
terms from a set of terms so this is equal to
Total number of where when I didnrsquot have time to prove this unfortunately I derived it through observation but this is all bonus so I thought I would keep it anyway The total number of when divided by the total number of where when will give us the probability that when This is equal tohellip
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden
AT Proof amp Structure Nico Candido Portfolio 1 September 25 2018
Works Cited Peer editing Dora Neiden