Complement of a Set Questions in English

(D) Using De Morgan's Law,$(A \cap B)^c = A^c \cup B^c$.
Therefore,$A \cap (A \cap B)^c = A \cap (A^c \cup B^c)$.
Applying the Distributive Law,we get $(A \cap A^c) \cup (A \cap B^c)$.
Since $A \cap A^c = \phi$,the expression becomes $\phi \cup (A \cap B^c) = A \cap B^c$.

(C) Using De Morgan's Law,we know that $(A \cup B)' = A' \cap B'$.
Therefore,$A \cap (A \cup B)' = A \cap (A' \cap B')$.
By the associative law of sets,this can be rewritten as $(A \cap A') \cap B'$.
Since $A \cap A' = \phi$,we have $\phi \cap B' = \phi$.
Thus,$A \cap (A \cup B)' = \phi$.

(B) By the definition of the complement of a set,$A'$ consists of all elements in the universal set $U$ that are not in $A$.
Therefore,the union of set $A$ and its complement $A'$ includes all elements of $A$ and all elements of $U$ that are not in $A$,which results in the universal set $U$.
Thus,$A \cup A' = U$.

(C) From the Venn diagram,the set $\{ (A - B) \cup (B - C) \cup (C - A) \}$ represents the elements that belong to exactly one of the sets $A, B,$ or $C$,or the elements that belong to exactly two of the sets.
Specifically,the region $\{ (A - B) \cup (B - C) \cup (C - A) \}$ covers all parts of the union $A \cup B \cup C$ except for the central intersection region $A \cap B \cap C$.
Therefore,the complement of this set within the universal set $U = A \cup B \cup C$ is the central region itself.
Thus,$\{ (A - B) \cup (B - C) \cup (C - A) \}' = A \cap B \cap C$.

(B) According to De Morgan's Law,the complement of the intersection of two sets is equal to the union of their complements.
Thus,$(A \cap B)' = A' \cup B'$.

(D) First,we find the number of elements in the union of sets $A$ and $B$ using the formula:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
$n(A \cup B) = 12 + 9 - 4 = 17$
Now,we find the number of elements in the complement of $(A \cup B)$ using the universal set $U$:
$n((A \cup B)^C) = n(U) - n(A \cup B)$
$n((A \cup B)^C) = 20 - 17 = 3$
Therefore,the correct option is $D$.

(B) The probability of an event $A$ not occurring is given by the complement rule:
$P(\text{not } A) = P(\bar{A}) = 1 - P(A)$.
Given $P(A) = 0.38$,
$P(\bar{A}) = 1 - 0.38 = 0.62$.

(B) We know that for any event $E$,$P(\bar{E}) = 1 - P(E)$.
Given $P(A) = 0.65$,so $P(\bar{A}) = 1 - 0.65 = 0.35$.
Given $P(B) = 0.15$,so $P(\bar{B}) = 1 - 0.15 = 0.85$.
Therefore,$P(\bar{A}) + P(\bar{B}) = 0.35 + 0.85 = 1.2$.

(D) We know that $P(A \cap \bar{B}) = P(A) - P(A \cap B)$.
Given $P(A) = 0.25$ and $P(A \cap B) = 0.14$.
Substituting the values,we get $P(A \cap \bar{B}) = 0.25 - 0.14 = 0.11$.
Since $0.11$ is not among the options $A, B,$ or $C$,the correct option is $D$.

(D) We know that $P(A \cap \overline{B})$ represents the probability of event $A$ occurring and event $B$ not occurring.
This can be expressed as $P(A \cap \overline{B}) = P(A) - P(A \cap B)$.
Given $P(A) = 0.25$ and $P(A \cap B) = 0.14$.
Substituting the values: $P(A \cap \overline{B}) = 0.25 - 0.14 = 0.11$.

(D) Given that $A$ is not a subset of $B$ $(A \not\subseteq B)$.
This implies that there exists at least one element $x$ such that $x \in A$ and $x \notin B$.
Since $x \notin B$,it follows that $x \in B^c$ (where $B^c$ is the complement of $B$ in $X$).
Thus,$x \in A \cap B^c$.
Since there exists at least one element in the intersection of $A$ and $B^c$,the sets $A$ and $B^c$ are non-disjoint.

(D) Using De Morgan's Law,$(A \cap B)^c = A^c \cup B^c$.
Substituting this into the expression:
$A \cap (A^c \cup B^c)$
Applying the distributive law:
$(A \cap A^c) \cup (A \cap B^c)$
Since $A \cap A^c = \phi$:
$\phi \cup (A \cap B^c) = A \cap B^c$.

(B) By the definition of set difference,the set $A - B$ consists of all elements that are in $A$ but not in $B$.
This is equivalent to the intersection of $A$ and the complement of $B$.
Therefore,$A - B = A \cap \bar{B}$.

(B) According to De Morgan's Law in set theory,the complement of the intersection of two sets is equal to the union of their complements.
Mathematically,this is expressed as $(A \cap B)' = A' \cup B'$.
Therefore,the correct option is $B$.

(C) The set difference $A - B$ is defined as the set of elements that are in $A$ but not in $B$.
$1$. $A - B = A \cap B'$ is a standard identity.
$2$. $A - (A \cap B) = A \cap (A \cap B)' = A \cap (A' \cup B') = (A \cap A') \cup (A \cap B') = \emptyset \cup (A \cap B') = A \cap B'$. Thus,$A - B = A - (A \cap B)$ is true.
$3$. $(A \cup B) - B = (A \cup B) \cap B' = (A \cap B') \cup (B \cap B') = (A \cap B') \cup \emptyset = A \cap B'$. Thus,$A - B = (A \cup B) - B$ is true.
$4$. $A - B'$ is the set of elements in $A$ that are not in $B'$,which means elements in $A$ that are in $B$. Thus,$A - B' = A \cap B$.
Since $A \cap B \neq A \cap B'$ in general,the statement $A - B = A - B'$ is false.

(C) The sample space of rolling a die is $S = \{1, 2, 3, 4, 5, 6\}$.
Event $A$ is 'getting a prime number',so $A = \{2, 3, 5\}$.
The event 'not $A$' is the complement of $A$,denoted as $A^c$ or $A'$.
$A^c = S - A = \{1, 2, 3, 4, 5, 6\} - \{2, 3, 5\} = \{1, 4, 6\}$.

(B) When two dice are thrown,the sample space $S$ contains $36$ outcomes:
$S = \{(x, y) : x, y \in \{1, 2, 3, 4, 5, 6\} \}$.
The event $A$ is defined as getting an even number on the first die:
$A = \{(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) \}$.
The complement event $A^{\prime}$ consists of all outcomes in $S$ that are not in $A$. This means the first die must show an odd number:
$A^{\prime} = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) \}$.
Comparing this with the definition of event $B$ (getting an odd number on the first die),we see that $A^{\prime} = B$.

(A) When two dice are thrown,the sample space $S$ contains $36$ outcomes:
$S = \{(x, y) : x, y \in \{1, 2, 3, 4, 5, 6\}\}$.
The event $B$ is defined as getting an odd number on the first die:
$B = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$.
The event $\text{not } B$ (denoted as $B'$) consists of all outcomes in the sample space $S$ that are not in $B$.
Since $B$ contains all outcomes where the first die is odd,$B'$ contains all outcomes where the first die is even.
Therefore,$B' = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$.
This is equivalent to event $A$.

(A) The complement of a set $A$ with respect to a universal set $U$,denoted by $A^{\prime}$,is defined as the set of all elements in $U$ that are not in $A$.
Given $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ and $A = \{1, 3, 5, 7, 9\}$.
We identify the elements of $U$ that are not present in $A$:
$A^{\prime} = U \setminus A = \{x : x \in U \text{ and } x \notin A\}$.
Removing the elements of $A$ from $U$,we get:
$A^{\prime} = \{2, 4, 6, 8, 10\}$.

(N/A) The universal set $U$ consists of all students in Class $XI$,which includes both boys and girls.
$A$ is the set of all girls in Class $XI$.
By definition,the complement of a set $A$,denoted by $A'$,is the set of all elements in $U$ that are not in $A$.
Therefore,$A' = U - A$ represents the set of all students in Class $XI$ who are not girls,which means $A'$ is the set of all boys in Class $XI$.

(A) Given $U = \{1, 2, 3, 4, 5, 6\}$,$A = \{2, 3\}$,and $B = \{3, 4, 5\}$.
$A' = U - A = \{1, 4, 5, 6\}$.
$B' = U - B = \{1, 2, 6\}$.
$A' \cap B' = \{1, 4, 5, 6\} \cap \{1, 2, 6\} = \{1, 6\}$.
$A \cup B = \{2, 3, 4, 5\}$.
$(A \cup B)' = U - (A \cup B) = \{1, 2, 3, 4, 5, 6\} - \{2, 3, 4, 5\} = \{1, 6\}$.
Since $(A \cup B)' = \{1, 6\}$ and $A' \cap B' = \{1, 6\}$,we have $(A \cup B)' = A' \cap B'$.
This verifies De Morgan's Law.

(A) The complement of a set $A$,denoted by $A^{\prime}$,is defined as the set of all elements in the universal set $U$ that are not in $A$.
Given $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ and $A = \{1, 2, 3, 4\}$.
$A^{\prime} = U - A = \{x : x \in U \text{ and } x \notin A\}$.
Removing the elements of $A$ from $U$,we get $A^{\prime} = \{5, 6, 7, 8, 9\}$.

(A) The complement of a set $B$,denoted by $B^{\prime}$,is defined as the set of all elements in the universal set $U$ that are not in $B$.
Given $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ and $B = \{2, 4, 6, 8\}$.
$B^{\prime} = U - B = \{x : x \in U \text{ and } x \notin B\}$.
Removing the elements of $B$ from $U$,we get:
$B^{\prime} = \{1, 3, 5, 7, 9\}$.

(A) Given sets are $A = \{1, 2, 3, 4\}$ and $C = \{3, 4, 5, 6\}$.
First,find the union of sets $A$ and $C$:
$A \cup C = \{1, 2, 3, 4\} \cup \{3, 4, 5, 6\} = \{1, 2, 3, 4, 5, 6\}$.
The complement $(A \cup C)'$ is defined as $U \setminus (A \cup C)$,where $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
$(A \cup C)' = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \setminus \{1, 2, 3, 4, 5, 6\} = \{7, 8, 9\}$.

(A) Given sets are $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,$A = \{1, 2, 3, 4\}$,and $B = \{2, 4, 6, 8\}$.
First,find the union of sets $A$ and $B$:
$A \cup B = \{1, 2, 3, 4\} \cup \{2, 4, 6, 8\} = \{1, 2, 3, 4, 6, 8\}$.
Now,find the complement of $(A \cup B)$,which is $(A \cup B)' = U \setminus (A \cup B)$:
$(A \cup B)' = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \setminus \{1, 2, 3, 4, 6, 8\} = \{5, 7, 9\}$.

(C) The complement of a set $A$,denoted by $A'$,is the set of all elements in the universal set $U$ that are not in $A$.
Given $A = \{1, 2, 3, 4\}$ and $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,we have $A' = U \setminus A = \{5, 6, 7, 8, 9\}$.
The double complement of a set $A$ is given by the property $(A')' = A$.
Therefore,$(A')' = \{1, 2, 3, 4\}$.

(A) Given sets are $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,$B = \{2, 4, 6, 8\}$,and $C = \{3, 4, 5, 6\}$.
First,find the difference $B - C$,which contains elements present in $B$ but not in $C$.
$B - C = \{2, 8\}$.
The complement $(B - C)'$ is defined as $U \setminus (B - C)$,which contains all elements of $U$ except those in $\{2, 8\}$.
$(B - C)' = \{1, 3, 4, 5, 6, 7, 9\}$.

(A) The universal set is given by $U = \{a, b, c, d, e, f, g, h\}$.
The set $A$ is given by $A = \{a, b, c\}$.
The complement of set $A$,denoted by $A'$,is defined as the set of all elements in $U$ that are not in $A$.
$A' = U - A = \{x : x \in U \text{ and } x \notin A\}$.
Removing the elements of $A$ from $U$,we get $A' = \{d, e, f, g, h\}$.

(A) The universal set is $U = \{a, b, c, d, e, f, g, h\}$.
The given set is $B = \{d, e, f, g\}$.
The complement of set $B$,denoted as $B'$,is defined as the set of all elements in $U$ that are not in $B$.
$B' = U - B = \{x : x \in U \text{ and } x \notin B\}$.
Removing the elements of $B$ from $U$,we get $B' = \{a, b, c, h\}$.

(A) The universal set is $U = \{a, b, c, d, e, f, g, h\}$.
The given set is $C = \{a, c, e, g\}$.
The complement of a set $C$,denoted by $C'$,is defined as the set of all elements in $U$ that are not in $C$.
$C' = U - C = \{x : x \in U \text{ and } x \notin C\}$.
Removing the elements of $C$ from $U$,we get:
$C' = \{b, d, f, h\}$.

(A) The universal set is given as $U = \{a, b, c, d, e, f, g, h\}$.
The set $D$ is given as $D = \{f, g, h, a\}$.
The complement of a set $D$,denoted by $D'$,is defined as the set of all elements in $U$ that are not in $D$.
$D' = U \setminus D = \{x : x \in U \text{ and } x \notin D\}$.
Removing the elements of $D$ from $U$,we get:
$D' = \{b, c, d, e\}$.

(N/A) Let $U = N$ be the set of natural numbers.
Given set $A = \{ x : x \text{ is an even natural number} \}$.
The complement of set $A$ is denoted by $A^\prime$ or $A^c$.
$A^\prime = U - A = \{ x : x \in N \text{ and } x \notin A \}$.
Since $N$ consists of all natural numbers,removing even natural numbers leaves only odd natural numbers.
Therefore,$A^\prime = \{ x : x \text{ is an odd natural number} \}$.

(N/A) Let $U = \mathbb{N}$ be the set of natural numbers.
The complement of a set $A$,denoted by $A^\prime$,is defined as $A^\prime = U - A$.
Given $A = \{ x : x \text{ is an odd natural number} \}$.
Therefore,$A^\prime = \{ x : x \in \mathbb{N} \text{ and } x \text{ is not an odd natural number} \}$.
Since a natural number is either odd or even,the complement of the set of odd natural numbers is the set of even natural numbers.
Thus,$A^\prime = \{ x : x \text{ is an even natural number} \}$.

(N/A) Let $U = N$ be the set of natural numbers.
The complement of a set $A$,denoted by $A'$,is defined as $A' = \{ x : x \in U \text{ and } x \notin A \}$.
Given $A = \{ x : x \text{ is a positive multiple of } 3 \}$.
Therefore,the complement $A' = \{ x : x \in N \text{ and } x \text{ is not a multiple of } 3 \}$.

(B) Let $U = N = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, \dots \}$ be the universal set of natural numbers.
Let $A = \{ x : x \text{ is a prime number} \} = \{ 2, 3, 5, 7, 11, \dots \}$.
The complement of set $A$,denoted by $A^\prime$,is defined as $A^\prime = U - A$.
$A^\prime = \{ x : x \in N \text{ and } x \text{ is not a prime number} \}$.
Since $1$ is neither prime nor composite,and all other natural numbers are either prime or composite,the set of non-prime natural numbers includes $1$ and all composite numbers.
Therefore,$A^\prime = \{ x : x \text{ is a composite number or } x = 1 \}$.

(N/A) Let $U = N$ be the set of natural numbers.
The given set is $A = \{ x : x \text{ is a natural number divisible by } 3 \text{ and } 5 \}$.
Since a number divisible by both $3$ and $5$ is divisible by their least common multiple,which is $15$,we can write $A = \{ x : x \text{ is a natural number divisible by } 15 \}$.
The complement of set $A$,denoted by $A'$,is the set of all elements in $U$ that are not in $A$.
Therefore,$A' = \{ x : x \in N \text{ and } x \text{ is not divisible by } 15 \}$.

(N/A) Let $U = \mathbb{N}$ be the set of natural numbers.
The complement of set $A$ is denoted by $A^\prime$ or $A^c$.
$A^\prime = U - A = \{ x : x \in \mathbb{N} \text{ and } x \notin A \}$.
Therefore,$A^\prime = \{ x : x \in \mathbb{N} \text{ and } x \text{ is not a perfect square} \}$.

(A) Let $U = N$ be the universal set of natural numbers.
The complement of a set $A$ is defined as $A' = \{ x : x \in U \text{ and } x \notin A \}$.
Given $A = \{ x : x \text{ is a perfect cube} \}$.
Therefore,the complement $A'$ is the set of all natural numbers that are not perfect cubes.
$A' = \{ x : x \in N \text{ and } x \text{ is not a perfect cube} \}$.

(N/A) Let the universal set $U = N$,where $N$ is the set of natural numbers.
Let $A = \{x: x+5=8\}$.
Solving the equation $x+5=8$,we get $x=3$.
So,$A = \{3\}$.
The complement of set $A$ is denoted by $A^\prime$ or $A^c$.
$A^\prime = U - A = \{x: x \in N \text{ and } x \neq 3\}$.

(N/A) Given the universal set $U = N$ (the set of natural numbers).
Let the set be $A = \{x: 2x + 5 = 9\}$.
Solving the equation: $2x + 5 = 9 \implies 2x = 4 \implies x = 2$.
So,$A = \{2\}$.
The complement of set $A$ is $A' = U - A = \{x: x \in N \text{ and } x \neq 2\}$.

(N/A) Let the universal set be $U = N$,where $N$ is the set of all natural numbers.
The given set is $A = \{ x: x \in N \text{ and } x \ge 7 \}$.
The complement of set $A$,denoted by $A^\prime$,is defined as $A^\prime = \{ x: x \in U \text{ and } x \notin A \}$.
Since $A$ contains all natural numbers greater than or equal to $7$,its complement $A^\prime$ will contain all natural numbers less than $7$.
Therefore,$A^\prime = \{ x: x \in N \text{ and } x < 7 \}$.

The universal set is $U = N = \{1, 2, 3, 4, 5, \dots \}$.
Given set $A = \{ x : x \in N \text{ and } 2x + 1 > 10 \}$.
Solving the inequality: $2x > 9 \implies x > 4.5$.
Since $x \in N$,the set $A = \{5, 6, 7, 8, \dots \}$.
The complement of $A$ is $A' = U - A$.
$A' = \{ x : x \in N \text{ and } x \le 4.5 \}$.
Since $x \in N$,$A' = \{1, 2, 3, 4 \}$.

Given $U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,$A = \{2, 4, 6, 8\}$,and $B = \{2, 3, 5, 7\}$.
First,find $A \cap B = \{2\}$.
Then,$(A \cap B)^{\prime} = U \setminus \{2\} = \{1, 3, 4, 5, 6, 7, 8, 9\}$.
Next,find $A^{\prime} = U \setminus A = \{1, 3, 5, 7, 9\}$.
Find $B^{\prime} = U \setminus B = \{1, 4, 6, 8, 9\}$.
Then,$A^{\prime} \cup B^{\prime} = \{1, 3, 5, 7, 9\} \cup \{1, 4, 6, 8, 9\} = \{1, 3, 4, 5, 6, 7, 8, 9\}$.
Since $(A \cap B)^{\prime} = \{1, 3, 4, 5, 6, 7, 8, 9\}$ and $A^{\prime} \cup B^{\prime} = \{1, 3, 4, 5, 6, 7, 8, 9\}$,it is verified that $(A \cap B)^{\prime} = A^{\prime} \cup B^{\prime}$.

(N/A) To draw the Venn diagram for $(A \cup B)'$,follow these steps:
$1$. Draw a rectangle to represent the universal set $U$.
$2$. Draw two intersecting circles inside the rectangle to represent sets $A$ and $B$.
$3$. The region $A \cup B$ represents the union of sets $A$ and $B$,which includes all elements in $A$,in $B$,or in both.
$4$. The complement $(A \cup B)'$ represents all elements in the universal set $U$ that are $NOT$ in $A \cup B$.
$5$. Therefore,shade the entire region inside the rectangle $U$ that lies outside both circles $A$ and $B$.

(N/A) To draw the Venn diagram for $(A \cap B)^{\prime}$,follow these steps:
$1$. Draw a rectangle to represent the universal set $U$.
$2$. Draw two intersecting circles inside the rectangle to represent sets $A$ and $B$.
$3$. The intersection $A \cap B$ is the region common to both circles.
$4$. The complement $(A \cap B)^{\prime}$ represents all elements in the universal set $U$ that are $NOT$ in the intersection $A \cap B$.
$5$. Therefore,shade the entire region inside the rectangle except for the overlapping part (the intersection) of the two circles.

(A) The universal set $U$ consists of all triangles in a plane.
Set $A$ consists of all triangles that have at least one angle not equal to $60^{\circ}$.
The complement $A^{\prime}$ consists of all triangles in $U$ that are not in $A$.
This means $A^{\prime}$ contains all triangles where every angle is equal to $60^{\circ}$.
Since the sum of angles in a triangle is $180^{\circ}$,if every angle is $60^{\circ}$,the triangle must be equilateral.
Therefore,$A^{\prime}$ is the set of all equilateral triangles.

(A) By the definition of the complement of a set,$A^{\prime}$ contains all elements of the universal set $U$ that are not in $A$.
Therefore,the union of a set $A$ and its complement $A^{\prime}$ includes all elements of $A$ and all elements of $U$ not in $A$,which results in the universal set $U$.
Thus,$A \cup A^{\prime} = U$.

(A) We know that the complement of the empty set $\varnothing$ is the universal set $U$,i.e.,$\varnothing^{\prime} = U$.
Substituting this into the expression:
$\varnothing^{\prime} \cap A = U \cap A$.
Since $A$ is a subset of the universal set $U$,the intersection of $U$ and $A$ is $A$.
Therefore,$\varnothing^{\prime} \cap A = A$.